[tex]\vec F(x,y,z)=(x^2+y)\,\vec\imath+(y^2+x)\,\vec\jmath+ze^z\,\vec k[/tex]
We want to find [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. This means
[tex]\dfrac{\partial f}{\partial x}=x^2+y[/tex]
[tex]\dfrac{\partial f}{\partial y}=y^2+x[/tex]
[tex]\dfrac{\partial f}{\partial z}=ze^z[/tex]
Integrating both sides of the latter equation with respect to [tex]z[/tex] tells us
[tex]f(x,y,z)=e^z(z-1)+g(x,y)[/tex]
and differentiating with respect to [tex]x[/tex] gives
[tex]x^2+y=\dfrac{\partial g}{\partial x}[/tex]
Integrating both sides with respect to [tex]x[/tex] gives
[tex]g(x,y)=\dfrac{x^3}3+xy+h(y)[/tex]
Then
[tex]f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+h(y)[/tex]
and differentiating both sides with respect to [tex]y[/tex] gives
[tex]y^2+x=x+\dfrac{\mathrm dh}{\mathrm dy}\implies\dfrac{\mathrm dh}{\mathrm dy}=y^2\implies h(y)=\dfrac{y^3}3+C[/tex]
So the scalar potential function is
[tex]\boxed{f(x,y,z)=e^z(z-1)+\dfrac{x^3}3+xy+\dfrac{y^3}3+C}[/tex]
By the fundamental theorem of calculus, the work done by [tex]\vec F[/tex] along any path depends only on the endpoints of that path. In particular, the work done over the line segment (call it [tex]L[/tex]) in part (a) is
[tex]\displaystyle\int_L\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(4,0,0)=\boxed{1+3e^4}[/tex]
and [tex]\vec F[/tex] does the same amount of work over both of the other paths.
In part (b), I don't know what is meant by "df/dt for F"...
In part (c), you're asked to find the work over the 2 parts (call them [tex]L_1[/tex] and [tex]L_2[/tex]) of the given path. Using the fundamental theorem makes this trivial:
[tex]\displaystyle\int_{L_1}\vec F\cdot\mathrm d\vec r=f(0,0,0)-f(4,0,0)=-\frac{64}3[/tex]
[tex]\displaystyle\int_{L_2}\vec F\cdot\mathrm d\vec r=f(4,0,4)-f(0,0,0)=\frac{67}3+3e^4[/tex]
A sample of a given size is used to construct a 95% confidence interval for the population mean with a known population standard deviation. If a bigger sample had been used instead, then the 95% confidence interval would have been ________ and the probability of making an error would have been ________.
Answer:
narrower, reduced
Step-by-step explanation:
Given that a sample of a given size is used to construct a 95% confidence interval for the population mean with a known population standard deviation
we have confidence interval is margin of error on either side of mean.
Margin of error= Critical value * std error
Std error in turn is inversely proportinal to square root of n
For same confidence level, if sample size is increased we get a less margin of error and hence a narrower confidence interval.
when a confidence interval is narrowed, probability of making a type I error would be reduced.
Hence we get
If a bigger sample had been used instead, then the 95% confidence interval would have been __narrower______ and the probability of making an error would have been ___reduced_____.
A larger sample size would result in a narrower 95% confidence interval and a decreased probability of making an error. This is due to less variability and increased precision with increased sample size.
Explanation:If a larger sample size had been used to construct a 95% confidence interval for the population mean, the interval would have been narrower. This is because larger sample sizes typically result in less variability. In fact, as the sample size increases, the estimate of the population mean becomes more accurate, and thus the confidence interval around that estimate becomes tighter, containing less of the probability distribution.
The probability of making an error, or in other words, the margin of error, decreases with an increase in sample size. This reduction in the level of uncertainty is due to reduced variability seen with larger samples. By using a larger sample, the chances of the confidence interval not containing the population mean, thus making an error, decrease.
To give an example, if you initially have a 95% confidence interval of (67.02, 68.98) from a smaller sample and you then generate a bigger sample, you could get a confidence interval like (67.18, 68.82). This interval is narrower, representing the reduced variability and increased precision from the larger sample size.
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People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) people per hour, where t is measured in hours. The functions f and g are nonnegative and differentiable for all times t. Which of the following inequalities indicates that the rate of change of the number of people in the building is increasing at time t? o f (t) > 0 f (t)-9(t) > 0 o f (t)>0 of'(t)-g'(t) > 0
Answer:
The correct option is D) [tex]f'(t)-g'(t) > 0[/tex]
Step-by-step explanation:
Consider the provided information.
People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) people per hour,
The change of number of people in building is:
[tex]h(x)=f(t)-g(t)[/tex]
Where f(t) is people entering in building and g(t) is exiting from the building.
It is given that "The functions f and g are non negative and differentiable for all times t."
We need to find the the rate of change of the number of people in the building.
Differentiate the above function with respect to time:
[tex]h'(x)=\frac{d}{dt}[f(t)-g(t)][/tex]
[tex]h'(x)=f'(t)-g'(t)[/tex]
It is given that the rate of change of the number of people in the building is increasing at time t.
That means [tex]h'(x)>0[/tex]
Therefore, [tex]f'(t)-g'(t)>0[/tex]
Hence, the correct option is D) [tex]f'(t)-g'(t) > 0[/tex]
The rate of change of the number of people in the building is increasing at time t and with the help of this statement the correct option is D).
Given :
People are entering a building at a rate modeled by f (t) people per hour and exiting the building at a rate modeled by g (t) people per hour, where t is measured in hours. The functions f and g are nonnegative and differentiable for all times t.The change of the number of people in the building is given by:
[tex]h(x) = f(t) - g(t)[/tex]
To determine the inequality, differentiate the above equation with respect to time.
[tex]h'(x)=f'(t)-g'(t)[/tex]
Now, it is given that the rate of change of the number of people in the building is increasing at time t. That means:
[tex]h'(x)>0[/tex]
[tex]f'(t)-g'(t)>0[/tex]
Therefore, the correct option is D).
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A new species of sea crab has been discovered, and an experiment conducted to determine whether or not the animal can regulate its temperature. That the animal can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability. Ten of these sea crabs were exposed to ambient temperatures of 24 degrees Celsius. Their body temperatures were measured with the results below:24.33, 24.61, 24.67, 24.64, 24.42, 24.97, 25.23, 24.73, 24.90, 24.44For purposes of this example, assume that it is reasonable to regard these 10 crabs as a random sample from the population of all crabs of this species.a)- Calculate a point estimate of the population mean.b)-Construct and interpret a 99% confidence interval for m.c)-Does it appear from these data that the crabs are able to regulate their body temperature? Provide statistical justification for your response.
Answer:
a) [tex]\hat \mu =\bar X=24.694[/tex]
b) The 99% confidence interval would be given by (24.409;24.979)
c) Since the 99% confidence interval not contains the exposed to ambient temperatures of 24 degrees Celsius, and since the animals "that can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability", on this case we can conclude that the sea crabs are able to regulate their body temperature.
Step-by-step explanation:
1) Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
Data: 24.33, 24.61, 24.67, 24.64, 24.42, 24.97, 25.23, 24.73, 24.90, 24.44
[tex]\bar X[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s represent the sample standard deviation
n=10 represent the sample size
Part a
A point of estimate for the population mean is the sample mean, given by this formula:
[tex]\bar X= \frac{\sum_{i=1}^n x_i}{n}[/tex]
If we apply this formula we got that [tex]\hat \mu =\bar X=24.694[/tex]
Part b
In order to calculate the confidence interval first we need to calculate the sample deviation given by this formula:
[tex]s=\frac{\sum_{i=1}^n (x_i -\bar x)}{n-1}[/tex]
If we use this formula we got that [tex]s=0.277[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=10-1=9[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that [tex]t_{\alpha/2}=3.25[/tex]
Now we have everything in order to replace into formula (1):
[tex]24.694-3.25\frac{0.277}{\sqrt{10}}=24.409[/tex]
[tex]24.694+3.25\frac{0.277}{\sqrt{10}}=24.979[/tex]
So on this case the 99% confidence interval would be given by (24.409;24.979)
Part c
Since the 99% confidence interval not contains the exposed to ambient temperatures of 24 degrees Celsius, and since the animals "that can maintain a body temperature different from the surroundings would be considered evidence of this regulating capability", on this case we can conclude that the sea crabs are able to regulate their body temperature.
Be sure to answer all parts. List the evaluation points corresponding to the midpoint of each subinterval to three decimal places, sketch the function and approximating rectangles and evaluate the Riemann sum to six decimal places if needed. f(x) = x2 + 4,[4, 5], n = 4. Give your answer in an ascending order. Evaluation points: , ,
Answer:
The Riemann Sum for [tex]\int\limits^5_4 {x^2+4} \, dx[/tex] with n = 4 using midpoints is about 24.328125.
Step-by-step explanation:
We want to find the Riemann Sum for [tex]\int\limits^5_4 {x^2+4} \, dx[/tex] with n = 4 using midpoints.
The Midpoint Sum uses the midpoints of a sub-interval:
[tex]\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)[/tex]
where [tex]\Delta{x}=\frac{b-a}{n}[/tex]
We know that a = 4, b = 5, n = 4.
Therefore, [tex]\Delta{x}=\frac{5-4}{4}=\frac{1}{4}[/tex]
Divide the interval [4, 5] into n = 4 sub-intervals of length [tex]\Delta{x}=\frac{1}{4}[/tex]
[tex]\left[4, \frac{17}{4}\right], \left[\frac{17}{4}, \frac{9}{2}\right], \left[\frac{9}{2}, \frac{19}{4}\right], \left[\frac{19}{4}, 5\right][/tex]
Now, we just evaluate the function at the midpoints:
[tex]f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(4\right)+\left(\frac{17}{4}\right)}{2}\right)=f\left(\frac{33}{8}\right)=\frac{1345}{64}=21.015625[/tex]
[tex]f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(\frac{17}{4}\right)+\left(\frac{9}{2}\right)}{2}\right)=f\left(\frac{35}{8}\right)=\frac{1481}{64}=23.140625[/tex]
[tex]f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(\frac{9}{2}\right)+\left(\frac{19}{4}\right)}{2}\right)=f\left(\frac{37}{8}\right)=\frac{1625}{64}=25.390625[/tex]
[tex]f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(\frac{19}{4}\right)+\left(5\right)}{2}\right)=f\left(\frac{39}{8}\right)=\frac{1777}{64}=27.765625[/tex]
Finally, use the Midpoint Sum formula
[tex]\frac{1}{4}(21.015625+23.140625+25.390625+27.765625)=24.328125[/tex]
This is the sketch of the function and the approximating rectangles.
The weights of bags of peas are normally distributed with a mean of 12.08 ounces and a standard deviation of 1.03 ounces. Bags in the upper 5% are too heavy and must be repackaged. What is the most that a bag can weigh and not need to be repackaged?
A. 13.88 ouncesB. 12.18 ouncesC. 10.28 ouncesD. 12.03 ounces.
Answer: A bag can weigh 13.77 ounces and not need to be repacked.
Step-by-step explanation:
Since we have given that
Mean = 12.08 ounces
Standard deviation = 1.03 ounces
Bags in the upper 5% are too heavy and must be repackaged.
Using the standard normal table.
z = 1.645
So, [tex]X=\mu+z\sigma\\\\X=12.08+1.645\times 1.03\\\\X=13.77\ ounces[/tex]
Hence, A bag can weigh 13.77 ounces and not need to be repacked.
Suppose that in your city 39% of the voters are registered as Democrats, 26% as Republicans, and 7% as members of other parties. Voters not aligned with any official party are termed "Independent." You are conducting a poll by calling registered voters at random. In your first three calls, what is the probability that you talk to
a) all Republicans?
b) no Democrats?
c) at least one Independent?
Answer:
a) 0.0176 or 1.76%
b) 0.2270 or 22.70%
c) 0.6268 or 62.68%
Step-by-step explanation:
Democrats: P(D) = 39%
Republicans: P(R) = 26%
Others: P(O) = 7%
Independent: P(I) = 100-39-26-7 =28%
a) Probability of talking to all Republicans in three calls:
[tex]P(N_R=3) =P(R)*P(R)*P(R)\\P(N_R=3) =0.26^3 = 0.0176[/tex]
a) Probability of talking to no democrats in three calls:
[tex]P(N_D=0) =(1-P(D)*(1-P(D)*(1-P(D)\\P(N_D=0) =(1-0.39)^3 = 0.2270[/tex]
c) Probability of talking to at least one independent in three calls:
[tex]P(N_I\geq1)=1-P(N_I=0)\\P(N_I\geq1) = 1-(1-0.28)*(1-0.28)*(1-0.28)\\P(N_I\geq1) = 0.6268[/tex]
a) The probability of talking to all Republicans in the first three calls is 1.7576%. b) The probability of talking to no Democrats in the first three calls is 22.6981%. c) The probability of talking to at least one Independent is 19.5643%.
Explanation:a) To find the probability of talking to all Republicans in the first three calls, we need to multiply the probability of talking to a Republican on each call. The probability of talking to a Republican on the first call is 26%, on the second call is also 26%, and on the third call is still 26%. Therefore, the probability of talking to all Republicans in the first three calls is 0.26 * 0.26 * 0.26 = 0.017576 or 1.7576%.
b) To find the probability of talking to no Democrats in the first three calls, we need to multiply the probability of not talking to a Democrat on each call. The probability of not talking to a Democrat on the first call is 1 - 39% = 61%, on the second call is also 61%, and on the third call is still 61%. Therefore, the probability of talking to no Democrats in the first three calls is 0.61 * 0.61 * 0.61 = 0.226981 or 22.6981%.
c) To find the probability of talking to at least one Independent, we need to subtract the probability of not talking to any Independents from 1. The probability of not talking to an Independent on the first call is 1 - 7% = 93%, on the second call is also 93%, and on the third call is still 93%. Therefore, the probability of not talking to any Independents in the first three calls is 0.93 * 0.93 * 0.93 = 0.804357 or 80.4357%. Since we want the probability of talking to at least one Independent, we subtract this result from 1: 1 - 0.804357 = 0.195643 or 19.5643%.
There are 15 balls in an urn. 7 of the balls are red, and 8 of the balls are blue. Drawing a ball with replacement means that you take a ball out, note its color, and then return it. Drawing withoutreplacement means that you take a ball out and do not put it back.
(a) What is the probability that if you draw one ball, it will be blue?(b) If you draw two balls without replacement, what is the probability that you draw a red and then a blue?(c) If you draw two balls with replacement, what is the probability that you draw a red and then a blue?(d) If you draw three balls without replacement, what is the probability that you draw at least two red balls?(e) If you draw three balls with replacement, what is the probability that you draw at least two red balls?(f) We play a game where I draw a ball first without replacement, and then you draw one. I draw a blue ball, and then you draw a red ball. Are these events independent?
Answer:
Dependent
Step-by-step explanation:
Given that there are 15 balls in an urn. 7 of the balls are red, and 8 of the balls are blue
when we do with replacement each time probability for red or blue remains the same.
a) the probability that if you draw one ball, it will be blue=[tex]\frac{8}{15}[/tex]
(b) If you draw two balls without replacement, the probability that you draw a red and then a blue
=[tex]\frac{7}{15} *\frac{8}{14} \\=\frac{4}{15}[/tex]
(c) If you draw two balls with replacement, the probability that you draw a red and then a blue
=[tex]\frac{7}{15} *\frac{8}{15} \\=\frac{56}{225}[/tex]
(d) If you draw three balls without replacement, the probability that you draw at least two red balls
=P(2 red, 1 blue)+P(3 blue)
=[tex]\frac{7C2*8C1}{15C3} +\frac{7C3}{15C3} \\=\frac{203}{455} =\frac{29}{65}[/tex]
(e) If you draw three balls with replacement, the probability that you draw at least two red balls
= P(X≥2) where x is binomial with n=3 and p = 7/15
= 0.4506
(f) We play a game where I draw a ball first without replacement, and then you draw one. I draw a blue ball, and then you draw a red ball.
No when first ball is drawn without replacement the next probability for blue and red would be changed. So dependent.
The probability of various scenarios involving drawing balls from an urn is calculated, including with and without replacement. The events of drawing a blue ball and a red ball after each other are examined in terms of their independence.
Explanation:In order to answer the given question, we need to calculate the probability for each scenario:
(a) The probability of drawing a blue ball with replacement is
8/15
.
(b) The probability of drawing a red and then a blue ball without replacement is
(7/15) * (8/14)
.
(c) The probability of drawing a red and then a blue ball with replacement is
(7/15) * (8/15)
.
(d) The probability of drawing at least two red balls without replacement can be calculated by finding the probability of drawing two red balls and the probability of drawing three red balls, and then adding them together:
(7/15) * (6/14) + (7/15) * (6/14) * (5/13)
.
(e) The probability of drawing at least two red balls with replacement can be calculated by finding the probability of drawing two red balls and the probability of drawing three red balls, and then adding them together:
(7/15) * (7/15) + (7/15) * (7/15) * (7/15)
(f) No, the events of you drawing a blue ball and the student drawing a red ball after that are
not independent
because drawing the blue ball affects the probability of drawing a red ball for the student.
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Which statements are true about the solid, which has a length of
6 inches, a width of 3 inches, and a height of 2 inches?
Check all that apply.
The perimeter of one of its faces is 8 inches.
The perimeter of one of its faces is 10 inches.
The area of one of its faces is 6 inches.
The area of one of its faces is 12 in.2.
The volume of the solid is 36 in.3.
Answer:
Step-by-step explanation:
The solid is a cuboid with length of 6 inches, a width of 3 inches, and a height of 2 inches.
The volume is l× w×h = 6×3×2 = 36inches^3
Therefore, the following statements are true about the Solid
1) The volume of the solid is 36 in.3.
2) The perimeter of one of its faces is 10 inches.(2width + 2 height = 2×2 + 2×3 = 10 inches)
3) The area of one of its faces is 6 inches^2. (width × height = 3×2 = 6 inches)
4) The area of one of its faces is 12 in.2.(length × height = 6×2 = 12 inches^2)
Answer : The correct options are:
The perimeter of one of its faces is 10 inches.
The area of one of its faces is 6 inch².
The area of one of its faces is 12 inch².
The volume of the solid is 36 inches³.
Step-by-step explanation :
Formula used for perimeter of one face (rectangle) is:
Perimeter = 2(l+b)
Perimeter = 2(b+h)
Perimeter = 2(l+h)
Formula used for area of one face (rectangle) is:
Area = l × b
Area = b × h
Area = l × h
Formula used for volume of solid (cuboid) is:
volume of solid = l × b × h
Given:
l = length = 6 inches
b = width = 3 inches
h = height = 2 inches
Now we have to calculate the perimeter of one face.
Perimeter = 2(l+b)
Perimeter = 2(6+3) = 18 inch
Perimeter = 2(b+h)
Perimeter = 2(3+2) = 10 inch
Perimeter = 2(l+h)
Perimeter = 2(6+2) = 16 inch
Now we have to calculate the area of one face.
Area = l × b
Area = 6 × 3 = 18 inch²
Area = b × h
Area = 3 × 2= 6 inch²
Area = l × h
Area = 6 × 2 = 12 inch²
Now we have to calculate the volume of solid.
volume of solid = l × b × h
volume of solid = 6 × 3 × 2
volume of solid = 36 inches³
The volume of solid is, 36 inches³
A random sample of 85 group leaders, supervisors, and similar personnel at General Motors revealed that, on average, they spent 6.5 years in a particular job before being promoted. The standard deviation of the sample was 1.7 years. Construct a 95% confidence interval.
Final answer:
To construct the 95% confidence interval for the average time spent in a particular job at General Motors, we calculated the margin of error and then added and subtracted it from the sample mean, resulting in a confidence interval of approximately (6.1331, 6.8669).
Explanation:
To construct a 95% confidence interval for the average time spent in a particular job at General Motors, we utilize the sample mean, standard deviation, and the sample size to compute the margin of error and the confidence interval. Given that the sample mean is 6.5 years and the standard deviation is 1.7 years with a sample size of 85, we can calculate the confidence interval as follows:
Firstly, determine the critical value (z*) for a 95% confidence level. Since the population standard deviation is unknown and the sample size is less than 30, we use the t-distribution. For a 95% confidence level and 84 degrees of freedom (n-1), the critical value (using a t-table or software) is approximately 1.989.Compute the standard error of the mean by dividing the standard deviation by the square root of the sample size. Standard error (SE) = 1.7 / sqrt(85) ≈ 0.1845.Calculate the margin of error (MOE) by multiplying the critical value with the standard error. MOE = 1.989 * 0.1845 ≈ 0.3669.Finally, construct the confidence interval by adding and subtracting the margin of error from the sample mean. Lower limit = 6.5 - 0.3669 ≈ 6.1331 and upper limit = 6.5 + 0.3669 ≈ 6.8669. Therefore, the 95% confidence interval is (6.1331, 6.8669).This confidence interval suggests that we are 95% confident that the true average time spent in the job before promotion at General Motors lies between approximately 6.1 and 6.9 years.
Given the following null and alternative hypotheses H0: μ1 ≥ μ2 HA: μ1 < μ2 Together with the following sample information (shown below). Assuming that the populations are normally distributed with equal variances, test at the 0.10 level of significance whether you would reject the null hypothesis based on the sample information. Use the test statistic approach. Sample 1 Sample 2 n1 = 14 n2 = 18 x-bar1 = 565 x-bar2 = 578 s1 = 28.9 s2 = 26.3
Answer:
Null hypothesis: [tex]\mu_1 \geq \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 <\mu_2[/tex]
[tex]t=-1.329[/tex]
[tex]p_v =P(t_{30}<-1.329) =0.0969[/tex]
With the p value obtained and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.
Step-by-step explanation:
When we have two independnet samples from two normal distributions with equal variances we are assuming that
[tex]\sigma^2_1 =\sigma^2_2 =\sigma^2[/tex]
And the statistic is given by this formula:
[tex]t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}[/tex]
Where t follows a t distribution with [tex]n_1+n_2 -2[/tex] degrees of freedom and the pooled variance [tex]S^2_p[/tex] is given by this formula:
[tex]\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}[/tex]
This last one is an unbiased estimator of the common variance [tex]\simga^2[/tex]
The system of hypothesis on this case are:
Null hypothesis: [tex]\mu_1 \geq \mu_2[/tex]
Alternative hypothesis: [tex]\mu_1 <\mu_2[/tex]
Or equivalently:
Null hypothesis: [tex]\mu_1 - \mu_2 \geq 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2<0[/tex]
Our notation on this case :
[tex]n_1 =14[/tex] represent the sample size for group 1
[tex]n_2 =18[/tex] represent the sample size for group 2
[tex]\bar X_1 =565[/tex] represent the sample mean for the group 1
[tex]\bar X_2 =578[/tex] represent the sample mean for the group 2
[tex]s_1=28.9[/tex] represent the sample standard deviation for group 1
[tex]s_2=26.3[/tex] represent the sample standard deviation for group 2
First we can begin finding the pooled variance:
[tex]\S^2_p =\frac{(14-1)(28.9)^2 +(18 -1)(26.3)^2}{14 +18 -2}=753.882[/tex]
And the deviation would be just the square root of the variance:
[tex]S_p=27.457[/tex]
And now we can calculate the statistic:
[tex]t=\frac{(565 -578)-(0)}{27.457\sqrt{\frac{1}{14}}+\frac{1}{18}}=-1.329[/tex]
Now we can calculate the degrees of freedom given by:
[tex]df=14+18-2=30[/tex]
And now we can calculate the p value using the altenative hypothesis:
[tex]p_v =P(t_{30}<-1.329) =0.0969[/tex]
So with the p value obtained and using the significance level given [tex]\alpha=0.1[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 10% of significance the mena of the group 1 is significantly lower than the mean for the group 2.
A new type of fertilizer is being tested on a plot of land in an orange grove, to see whether it increases the amount of fruit produced. The mean number of pounds of fruit on this plot of land with the old fertilizer was 409 pounds. Agriculture scientists believe that the new fertilizer may decrease the yield. State the appropriate null and alternate hypotheses
a. The null hypothesis is ____________b. The alternate hypothesis is ___________
Answer:
a) Null Hypothesis:[tex]\mu \geq 409[/tex]
b) Alternative hypothesis:[tex]\mu <409[/tex]
Step-by-step explanation:
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
On this case we are insterested on the claim that "Agriculture scientists believe that the new fertilizer may decrease the yield". So the system of hypothesis for this case would be:
a) Null Hypothesis:[tex]\mu \geq 409[/tex]
b) Alternative hypothesis:[tex]\mu <409[/tex]
Suppose a basketball player has made 217 out of 302 free throws. If the player makes the next 3 free throws, I will pay you $23. Otherwise you pay me $15. Step 1 of 2 : Find the expected value of the proposition. Round your answer to two decimal places. Losses must be expressed as negative values.
Answer:
-$0.90
Step-by-step explanation:
There are only two possible outcomes, winning $23 (W) or losing $15 (L). Therefore:
[tex]P(W) + P(L) = 1[/tex]
The probability of the player making his next 3 free throws (P(W)) is:
[tex]P(W) = \frac{217}{302}*\frac{217}{302}*\frac{217}{302}\\P(W) = 0.37098[/tex]
The probability of the player NOT making his next 3 free throws (P(L)) is:
[tex]P(L) = 1 - P(W) = 1 - 0.37098\\P(L) = 0.62902[/tex]
Expected value (EV) is given by the payoff of each outcome multiplied by its probability:
[tex]EV = (23*0.37098) -(15*0.62902)\\EV = -\$0.90[/tex]
The expected value of the proposition is -$0.90
A group of friends has gotten very competitive with their board game nights. They have found that overall, they each have won an average of 18 games, with a population standard deviation of 6 games. If a sample of only 2 friends is selected at random from the group, select the expected mean and the standard deviation of the sampling distribution from the options below. Remember to round to the nearest whole number.
Answer: [tex]\mu_x=18\text{ hours}[/tex]
[tex]\sigma_x=4\text{ hours}[/tex]
Step-by-step explanation:
We know that mean and standard deviation of sampling distribution is given by :-
[tex]\mu_x=\mu[/tex]
[tex]\sigma_x=\dfrac{\sigma}{\sqrt{n}}[/tex]
, where [tex]\mu[/tex] = population mean
[tex]\sigma[/tex] =Population standard deviation.
n= sample size .
In the given situation, we have
[tex]\mu=18\text{ hours}[/tex]
[tex]\sigma=6\text{ hours}[/tex]
n= 2
Then, the expected mean and the standard deviation of the sampling distribution will be :_
[tex]\mu_x=\mu=18\text{ hours}[/tex]
[tex]\sigma_x=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{2}}=4.24264068712\approx4[/tex] [Rounded to the nearest whole number]
Hence, the the expected mean and the standard deviation of the sampling distribution :
[tex]\mu_x=18\text{ hours}[/tex]
[tex]\sigma_x=4\text{ hours}[/tex]
A trade magazine routinely checks the drive through service times of fast food restaurants. a 90% confidence interval that results from examining 653 customers in one fast food chains drive through has a lower bound of 178.2 secinds and an upper bound of 181.6 seconds.
What does this mean?
a.) one can be 90% confident that the mean deive through serivce time of this fast food chain is 179.9 seconds.
b.) the mean drive through service time of this fast food chain is 179.9 seconds 90% of the time.
c.) there is a 90 % probability that the mean drive through service time of this fast food chain is between 178.2 seconds and 181.6 seconds.
d.) one can be 90% confident that the mean drive through service time of this fast food chain is between 178.2 seconds and 181.6 seconds.
Answer: d) one can be 90% confident that the mean drive through service time of this fast food chain is between 178.2 seconds and 181.6 seconds.
Step-by-step explanation:
The interpretation of 90% confidence interval says that a person can be 90% confident that the true population mean lies in it.
Given : A 90% confidence interval that results from examining 653 customers in one fast food chains drive through has a lower bound of 178.2 seconds and an upper bound of 181.6 seconds.
i.e. 90% confidence interval for population mean drive through service time of fast food restaurants is between 178.2 seconds and 181.6 seconds.
It means a person can be 90% confident that the mean drive through service time of this fast food chain is between 178.2 seconds and 181.6 seconds.
Hence, the correct answer is option d) .one can be 90% confident that the mean drive through service time of this fast food chain is between 178.2 seconds and 181.6 seconds.
A prisoner is trapped in a cell containing 3 doors. The first door leads to a tunnel that returns him to his cell after 2 days’ travel. The second leads to a tunnel that returns him to his cell after 4 days’ travel. The third door leads to freedom after 1 day of travel. If it is assumed that the prisoner will always select doors 1, 2, and 3 with respective probabilities .5, .3, and .2, what is the expected number of days until the prisoner reaches freedom?
1. Repeat problem, assuming the prisoner remembers previously chosen doors, and does not re-choose them. Assume the probabilities for the other doors are proportionally larger.
2. Repeat problem but now suppose there is another cell, and that door 1 takes him to the other cell after 2 days of travel. For the other cell, there are two doors, one of which leads to freedom after 3 days of travel, and the other leads back to the prisoner’s original cell after 3 days of travel; each door is equally likely.
Answer:
dont escape lol
Step-by-step explanation:
the answer is 2
Consider the differential equation ay′′ + by′ + cy = 0, where b2 − 4ac < 0 and the characteristic equation has complex roots λ ± iμ. Substitute the functions u(t) = eλt cosμt and v(t) = eλt sinμt for y in the differential equation and thereby confirm that they are solutions.
Answer:
Please see attachment
Step-by-step explanation:
A CPU manufacturer is interested in studying the relationship between clock speed and the operating temperature that results at that clock speed for a particular CPU model. Let x be the clock speed in MHz and let Y be the temperature in ^{\circ}C . The following data was collected:i xi yi1 350 31.42 360 35.63 370 41.84 380 51.05 390 56.86 400 62.87 410 67.4a) Find the equation of the regression line.b) Estimate the temperature for clock speed x = 430 MHz.c) Find the 95% confidence interval for \beta .d) Compute the coefficient of determination R^{2} ?. Is this a high quality fit?
Final answer:
To find the equation of the regression line, calculate the slope and y-intercept using the given formulas with the given data points.
Explanation:
To find the equation of the regression line, we need to calculate the slope and the y-intercept. The slope, b, can be found using the formula:
b = (nΣ(xy) - ΣxΣy) / (nΣ(x^2) - (Σx)^2)
where n is the number of data points, Σ represents summation, x and y are the given data points. The y-intercept, a, can be calculated using the formula:
a = (Σy - bΣx) / n
Substituting the values from the given data into these formulas will give you the equation of the regression line.
Marie is saving money for home repairs. To date, she has saved $1,329. She needs at least $1,569 for the repairs. She plans to set aside $40 per week to add to her current savings. If this situation is modeled by the inequality below, how many more weeks, x, does she need to continue saving in order to have enough money for the repairs? $1,329 + $40x > $1,569
Answer:
x > 6
Step-by-step explanation:
Subtract $1329 from both sides, then divide by $40.
$1,329 + $40x > $1,569
$40x > $240
x > 6
Marie needs to save for more than 6 more weeks.
_____
Comment on the answer
The given inequality is written using the > symbol. Marie would have exactly enough to cover the projected cost after 6 weeks, but it is appropriate from a finance point of view for her to save more than the required amount. It will take more than 6 weeks for her to do that.
The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance in the pounds of the impurities. The sample variance was found to be 6.62. Find a 95% confidence interval for the population variance in the pounds of impurities. State any assumption you need to make to be able to answer this problem.
Answer:
The 95% confidence interval for the variance in the pounds of impurities would be [tex] 3.829 \leq \sigma^2 \leq 14.121[/tex].
Step-by-step explanation:
1) Data given and notation
[tex]s^2 =6.62[/tex] represent the sample variance
s=2.573 represent the sample standard deviation
[tex]\bar x[/tex] represent the sample mean
n=20 the sample size
Confidence=95% or 0.95
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
The Chi square distribution is the distribution of the sum of squared standard normal deviates .
2) Calculating the confidence interval
The confidence interval for the population variance is given by the following formula:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=20-1=19[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.025,19)" "=CHISQ.INV(0.975,19)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=32.852[/tex]
[tex]\chi^2_{1- \alpha/2}=8.907[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(19)(6.62)}{32.852} \leq \sigma^2 \frac{(19)(6.62)}{8.907}[/tex]
[tex] 3.829 \leq \sigma^2 \leq 14.121[/tex]
So the 95% confidence interval for the variance in the pounds of impurities would be [tex] 3.829 \leq \sigma^2 \leq 14.121[/tex].
The confidence interval for the population variance in the pounds of impurities is calculated using the sample variance, the degrees of freedom, and the critical values from a chi-squared distribution. The values are then plugged into the formulas to give the 95% confidence interval for the population variance.
Explanation:The subject of this question is in the field of statistics and it asks about the calculation of the 95% confidence interval for the population variance. In this case, we are trying to estimate the variance in the pounds of impurities in fertilizer bags based on a sample of twenty 100-pound bags with a variance of 6.62.
To solve this, we make the assumption that the population from which the bags are sampled follows a normal distribution. The confidence interval for the variance of a normal distribution can be found using the chi-squared distribution. The formula for this assumes that our samples are independent and identically distributed.
For a 95% confidence interval and d.f. (degrees of freedom) = n - 1 = 20 - 1 = 19, the upper and lower critical values (in this case for a chi-squared distribution, rounded to three decimal places) are 8.231 and 32.852 respectively. We can then use these values in the following formulas:
Lower Limit = sample variance * (d.f. / upper critical value) = 6.62*(19/32.852)
Upper Limit = sample variance * (d.f. / lower critical value) = 6.62*(19/8.231)
Finally, evaluate these to get our confidence interval.
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Help with this problem
I got the answer (4,12) (interval notation) 4
Answer:
(4, 12)
Step-by-step explanation:
Simplify (remember to flip the sign when dividing or multiplying by a negative number).
2 − 4 |8 − x| > -14
-4 |8 − x| > -16
|8 − x| < 4
There are two solutions. The first:
8 − x < 4
-x < -4
x > 4
The second solution:
8 − x > -4
-x > -12
x < 12
Therefore, the interval is (4, 12).
Amachineworksforanexponentiallydistributedtimewithrateμandthenfails. A repair crew checks the machine at times distributed according to a Poisson process with rate λ; if the machine is found to have failed then it is immediately replaced. Find the expected time between replacements of machines.
Answer:[tex]\frac{1}{\mu }+\frac{1}{\lambda }[/tex]
Step-by-step explanation:
If a machine is replaced at some time t, then the expected time until next failure is [tex]\frac{1}{\mu }[/tex]
and the time between the checks is exponentially distributed with rate \lambda, the expected time until next failure is [tex]\frac{1}{\lambda }[/tex]
Because of memory less property of the exponential The answer is
[tex]\frac{1}{\mu }+\frac{1}{\lambda }[/tex]
The ASQ (American Society for Quality) regularly conducts a salary survey of its membership, primarily quality management professionals. Based on the most recently published mean and standard deviation, a quality control specialist calculated thez-score associated with his own salary and found it was minus2.50. What is his salary?A. 2.5 standard deviations below the average salary.B. 2 and a half times less than the average salary.C. 2.5 standard deviations above the average salary.D. 2 and a half times more than the average salary.
Answer:
A. 2.5 standard deviations below the average salary.
Step-by-step explanation:
Hello!
Usually, when you calculate a probability of an X₀ value from a normally distributed variable you need to standardize it to reach the proper value. In this case, you already have the Z-value and need to reverse the standardization to obtain the X₀ value.
Z= X₀ - μ
δ
δ*Z = X₀ - μ
X₀= (δ*Z) + μ
replace Z= -2.50
X₀= (δ*-2.50) + μ
X₀= μ - (δ*2.50)
You can replace the Z value any time. I've just choose to do it at the end because it's more confortable for me.
I hope it helps!
(1 point) The effectiveness of a new bug repellent is tested on 18 subjects for a 10 hour period. Based on the number and location of the bug bites, the percentage of exposed surface area protected from bites was calculated for each of the subjects. Assume the population is normally distributed. The results were as follows: x¯¯¯=92 %, s=8% The new repellent is considered effective if it provides a percent repellency of more than 89%. Using α=0.025, construct a hypothesis test with null hypothesis μ≤0.89 and alternative hypothesis μ>0.89 to determine whether the mean repellency of the new bug repellent is greater than 89% by computing the following: (a)The degree of freedom is df= .
Answer:
z=1.59
If we compare the p value and the significance level given [tex]\alpha=0.025[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we fail to reject the null hypothesis, so we can conclude that the mean repellency of the new bug repellent is greater than 89% at 0.025 of signficance.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=0.92[/tex] represent the mean effectiveness of a new bug repellent for the sample
[tex]\sigma=0.08[/tex] represent the population standard deviation for the sample
[tex]n=18[/tex] sample size
[tex]\mu_o =0.89[/tex] represent the value that we want to test
[tex]\alpha=0.025[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to determine if the mean repellency of the new bug repellent is greater than 89% or 0.89, the system of hypothesis would be:
Null hypothesis:[tex]\mu \leq 0.89[/tex]
Alternative hypothesis:[tex]\mu > 0.89[/tex]
We don't know the population deviation, and the sample size <30, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{0.92-0.89}{\frac{0.08}{\sqrt{18}}}=1.59[/tex]
Calculate the P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=18-1=17[/tex]
Since is a one-side upper test the p value would be:
[tex]p_v =P(t_{17}>1.59)=0.065[/tex]
In Excel we can use the following formula to find the p value "=1-T.DIST(1.59;17;TRUE)"
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.025[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we fail to reject the null hypothesis, so we can conclude that the mean repellency of the new bug repellent is greater than 89% at 0.025 of signficance.
The probability distribution for damage claims paid by the Newton Automobile Insurance Company on collision insurance follows. Payment ($) Probability 0 0.85 500 0.04 1,000 0.04 3,000 0.03 5,000 0.02 8,000 0.01 10,000 0.01
(a) Use the expected collision payment to determine the collision insurance premium that would enable the company to break even. If required, round your answers to two decimal places. If your answer is zero, enter "0". x f(x) xf(x) 0 500 1000 3000 5000 8000 10000 Total
(b) The insurance company charges an annual rate of $520 for the collision coverage. What is the expected value of the collision policy for a policyholder? (Hint: It is the expected payments from the company minus the cost of coverage.) If required, enter negative value as negative number. $ Why does the policyholder purchase a collision policy with this expected value? The input in the box below will not be graded, but may be reviewed and considered by your instructor.
Answer:
a) $430
b) -$90
Step-by-step explanation:
Given a random variable X with possible values
[tex]\large x_1,x_2,...x_n[/tex]
with respective probabilities of occurrence
[tex]\large P(x_1),P(x_2),...P(x_n)[/tex]
then the expected value E(X) of X is
[tex]\large x_1*P(x_1)+x_2*P(x_2)+...+x_n*P(x_n)[/tex]
a)
The expected collision payment would then be
0*0.85 + 500*0.04 + 1000*0.04 + 3000*0.03 + 5000*0.02 + 8000*0.01 + 10000*0.01 = $430
So the insurance premium that would enable the company to break even is $430
b)
The expected value of the collision policy for a policyholder is the expected payments from the company minus the cost of coverage:
$430 - $520 = -$90
Why does the policyholder purchase a collision policy with this expected value?
Because the policyholder does not know what the probability of having an accident is in her particular case.
Besides, it is better to have the policy and not need it than to need it and not have it.
The probability computed shows that the collision insurance premium that would enable the company to break even is $430.
How to calculate the probability?The collision insurance premium that would enable the company to break even will be calculated thus:
= (0 × 0.85) + (500 × 0.04) + (1000 × 0.04) + (3000 × 0.03) + (5000 × 0.02) + (8000 × 0.01 + (10000 × 0.01)
= $430
The expected value of the collision policy for a policyholder will be:
= $430 - 520
= -$90
In conclusion, the policyholder purchase a collision policy with this expected because he doesn't know the probability of having an accident.
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The two intervals (113.4, 114.6) and (113.1, 114.9) are confidence intervals for μ = mean resonance frequency (in hertz) for all tennis rackets of a certain type. The two intervals were calculated using the same sample data. (a) What is the value of the sample mean (in hertz) resonance frequency?
Answer:
114
Step-by-step explanation:
Given that two intervals are confidence intervals for μ = mean resonance frequency (in hertz) for all tennis rackets of a certain type.
They are
i) [tex](113.4, 114.6)[/tex] and
ii) [tex](113.1, 114.9)[/tex]
We know that confidence interval has centre as the mean value
Hence we find the average of lower and upper bounds to find out the sample mean.
Sample mean in i) [tex]\frac{113.4+114.6}{2} =114[/tex]
Sample mean in ii) [tex]\frac{113.1+114.9}{2} =114[/tex]
Thus we find that value of sample mean =114
The value of the sample mean resonance frequency is 114 hertz.
Explanation:The value of the sample mean (in hertz) resonance frequency can be found by taking the average of the lower and upper bounds of the confidence intervals. For the first interval (113.4, 114.6), the sample mean is (113.4 + 114.6) / 2 = 114 hertz. For the second interval (113.1, 114.9), the sample mean is (113.1 + 114.9) / 2 = 114 hertz. So the value of the sample mean resonance frequency is 114 hertz.
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The combined math and verbal scores for females taking the SAT-I test are normally distributed with a mean of 998 and a standard deviation of 202 (based on date from the College Board). If a college includes a minimum score of 1025 among its requirements, what percentage of females do not satisfy that requirement?
Approximately 55.39% of females would not meet a college requirement of scoring at least 1025 on the SAT. This is found by calculating the z-score for the minimum required score and looking up the corresponding percentile.
Explanation:To answer this question, we need to calculate the z-score — a statistic that tells us how many standard deviations away a score is from the mean. This formula is Z = (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation.
In this case, X is 1025 (the minimum required score), μ is 998 (the average SAT score), and σ is 202 (the standard deviation). So the z-score is Z = (1025 - 998) / 202 = 0.1336.
This z-score is positive, which means the minimum required score for the college is above the average score. Next, we check a z-score table or use a statistical calculator to find the percentage of scores that fall below this z-score, which gives us the percentage of females not meeting the requirement. For Z = 0.1336, the percentage is approximately 55.39% (using a statistical calculator or z-table). Therefore, roughly 55.39% of females would not meet a college requirement of scoring at least 1025 on the SAT.
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To find the percentage of females who do not satisfy the minimum score requirement of 1025 on the SAT-I test, we calculate the z-score for the minimum score and find the area under the normal distribution curve. Approximately 55.22% of females do not satisfy the minimum score requirement.
Explanation:To find the percentage of females who do not satisfy the minimum score requirement of 1025 on the SAT-I test, we need to calculate the z-score for the minimum score and then find the area under the normal distribution curve to the left of that z-score. The formula for calculating the z-score is z = (x - μ) / σ, where x is the minimum score, μ is the mean, and σ is the standard deviation. So, z = (1025 - 998) / 202 = 0.1337.
Next, we can use a z-table or a calculator to find the area under the normal distribution curve to the left of a z-score of 0.1337. The area represents the percentage of females who do not satisfy the minimum score requirement. Using a z-table, we can find that the area is approximately 0.5522 or 55.22%.
Therefore, approximately 55.22% of females do not satisfy the minimum score requirement of 1025 on the SAT-I test.
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What is the p-value? -- Researcher Jessie is studying how the fear of going to the dentist affects an adult's actual number of visits to the dentist. She asks a random sample of adults whether or not they fear going to the dentist and also how many times they have gone in the past 10 years. She would like to assess if the average number of visits made by adults who fear going to the dentist (Group 1) is higher than the average number of visits for those who don't have that fear (Group 2), that is, test H0: μ1 = μ2 versus Ha: μ1 > μ2, using a 10% significance level. Her random sample of adults resulted in 14 stating they feared going to the dentist and 31 stated they did not fear going to the dentist. The first sample mean was 1.71 pooled standard errors above the second sample mean. Jessie has asked you to provide a complete sketch of the p-value that she can include in her report. You can use the shiny app in R which will give the exact p-value or make a complete sketch by hand and provide the bounds for the p-value using the T table.
2) Jessie also remembers some condition about a normal model required for her two populations of responses. She asks you to check this condition for her. You recall that a QQ plot helps to assess if a population of responses can be considered normally distributed. How many QQ plots do you need to make in this case?
None, since the total sample size of 45 adults is large you can just use the CLT.
One, for the 45 responses (number of visits) reported by the 45 adults that were surveyed.
Two, one for the 14 responses (number of visits) by the adults who fear going to the dentist and one for the 31 responses (number of visits) by the adults who do not fear going to the dentist.
Answer:
Two, one for the 14 responses (number of visits) by the adults who fear going to the dentist and one for the 31 responses (number of visits) by the adults who do not fear going to the dentist.
Step-by-step explanation:
Hello!
1)
You want to test if the average visits to the dentist of people who fear to visit it are greater than the average visits of people that don't fear it.
In this case, the statistic to use is a pooled Student t-test. The reason I've to choose this test is that one of your sample sizes is small (n₁= 14) and the t-test is more accurate for small samples. Even if the second sample is greater than 30, if both variables are normally distributed, the pooled t-test is the one to use.
H₀: μ₁ = μ₂
H₁: μ₁ > μ₂
α: 0.10
t= (X₁[bar]-X₂[bar]) - (μ₁ - μ₂) ~ t[tex]_{n₁+n₂-2}[/tex]
Sₐ√(1/n₁+1/n₂)
Where
X₁[bar] and X₂[bar] are the sample means of both groups
Sₐ is the pooled standard deviation
This is a one-tailed test, you will reject the null hypothesis to big numbers of t. Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis), and in this case, is also one-tailed.
P(t[tex]_{n₁+n₂-2}[/tex] ≥ t[tex]_{H0}[/tex]) = 1 - P(t[tex]_{n₁+n₂-2}[/tex] < t[tex]_{H0}[/tex])
Where t[tex]_{H0}[/tex] is the value of the calculated statistic.
Since you didn't copy the data of both samples, I cannot calculate it.
2)
Well there was one sample taken and separated in two following the criteria "fears the dentist" and "doesn't fear the dentist" making two different samples, so this is a test for two independent samples. To check if both variables are normally distributed you need to make two QQplots.
I hope it helps!
The p-value represents the probability of the observed data under the null hypothesis. Jessie's study regarding dentist visits requires a p-value sketch based on a test statistic of 1.71 standard errors. Two QQ plots are necessary to check the normality condition of the sample data for both groups.
Explanation:The p-value is a statistical measure that indicates the probability of the observed data or something more extreme occurring under the assumption that the null hypothesis is true. In Researcher Jessie's study, the null hypothesis (H0) is that the mean number of dentist visits for adults who fear going to the dentist (Group 1) is equal to the mean number of visits for those who do not fear going to the dentist (Group 2).
The alternative hypothesis (Ha) is that the mean for Group 1 is greater than the mean for Group 2. Jessie has found that the sample mean for Group 1 is 1.71 pooled standard errors above the mean for Group 2. To sketch the p-value, we need to find the area to the right of the test statistic (1.71 standard errors above the mean) on a normal distribution curve. This area represents the p-value, which we could find using statistical software or by consulting a T table with appropriate degrees of freedom.
Regarding the condition about the normal model, we need to check whether the data is normally distributed to correctly apply the t-test. We would create two QQ plots: one for the 14 responses from adults who fear going to the dentist, and one for the 31 responses from adults who do not.
The QQ plots will help determine if the data for each group deviates from a normal distribution, which is important given the sample sizes are less than 30 and normality cannot be assumed based on the Central Limit Theorem (CLT).
At Maple Grove Middle School, the record for the 100 m dash is 12.45 seconds. The
table show the length of time it took for the four fastest students to complete the 100-m
dash this year relative to the school record.
Based on the results in the table, which statement is true?
Student
Pablo
Time(s)
-0.12
Cindy
0.34
Eddie
-0.15
Sam
0.21
A. Pablo and Eddie both broke the school record this year.
B. The order of the runners from fastest to slowest was Cindy, Sam, Pablo, and Eddie.
C. Sam ran the 100-m dash faster than Cindy.
D. Eddie was the fastest runner in the race.
Answer:
Step-by-step explanation:
the record for the 100 m dash is 12.45 seconds. Since the length of times in the table is relative to the record time, 12.45 seconds,
Pablo's time - 12.45 = -0.12
Pablo's time = -0.12 + 12.45 = 12.33 seconds.
Cindy's time - 12.45 = 0.34
Cindy's time = 0.34 + 12.45 = 12.79 seconds
Eddie's time - 12.45 = -0.15
Eddie's time = -0.15 + 12.45 = 12.3 seconds.
Sam's time - 12.45 = 0.21
Sam's time = 0.21 + 12.45 = 12.66 seconds.
The following statements are true
A. Pablo and Eddie both broke the school record this year.
C. Sam ran the 100-m dash faster than Cindy.
D. Eddie was the fastest runner in the race.
Answer:
A. Pablo and Eddie both broke the school record this year.
Step-by-step explanation:
Current record is 12.45seconds
Pablo record is -0.12 seconds which is 12.45 – 0.12 = 12.33 seconds
Eddie’s record is -0.15 seconds which is 12.45 – 0.15 = 12.30 seconds.
Calamity Jane goes to the bank to make a withdrawal, and is equally likely to find 0 or 1 customers ahead of her. The service time of the customer ahead, if present, is exponentially distributed with parameter λ. What is the CDF of Jane’s waiting time?
Answer:
[tex]F_{T}(t)=\frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})[/tex]
Step-by-step explanation:
The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".
The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:
[tex]P(X=x)=\lambda e^{-\lambda x}[/tex]
We are interested on the cumulative distribution function (CDF) of Jane’s waiting time. Let T the random variable that represent the Jane’s waiting time, and t possible value for the random variable T.
For [tex]t\geq 0[/tex] the CDF for the waiting time is given by:
[tex]F_{T}(t) =P(T\leq t)[/tex]
For this case we can use the total rule of probability and the conidtional probability. We have two options, find 0 or 1 customer ahead of Jane, and for each option we have one possibility in order to find the CDF, like this:
[tex]F_{T}(t)=P(T\leq t|0 customers)P(0 customers)+ P(T\leq t|1 customers)P(1 customers)[/tex]
On this case the probability that we need to wait if we have o customers ahead is [tex]P(T\leq t|0 customers)=1[/tex] since if we don't have a customer ahead, so on this case Jane will not wait. And assuming that the probability of find 0 or 1 customers ahead of Jane is equal we have [tex]P(0 customers)=P(1 customers)=\frac{1}{2}[/tex]. And replacing we have:
[tex]F_{T}(t)=\frac{1}{2} +\frac{1}{2}\int_{0}^{t} \lambda e^{-\lambda \tau} d\tau = \frac{1}{2}+\frac{1}{2}[-e^{-\lambda t}+e^{-\lambda 0}]= \frac{1}{2} +\frac{1}{2}(1-e^{-\lambda t})[/tex]
Jane's waiting time scenario can be split into two cases, when there are no customers ahead of her, and when there is one customer ahead of her. For these two cases, the Cumulative Distribution Function (CDF) of Jane's waiting time can be calculated as [tex]0.5 + 0.5 \times (1 - e^{-\lambda w})[/tex]. This answer is derived using the law of total probability and the definition of the CDF for the exponential distribution.
Explanation:The question asks for the cumulative distribution function (CDF) of Jane’s waiting time at the bank. This is a probability question related to the exponential distribution and cumulative distribution function. The situation can be represented using two sub-situations: one, when there is no customer ahead (happens with 0.5 probability), Jane has no waiting time, and two, when there is a customer (happens with 0.5 probability) and Jane has to wait for an exponentially distributed time with parameter λ.
The cumulative distribution function (CDF) of an exponential random variable is given by [tex]F(x) = 1 - e^{-\lambda x}[/tex] when x ≥ 0. Using the law of total probability, the cumulative distribution function (CDF) of Jane's waiting time W can be obtained as [tex]P(W \leq w) = P(0 \text{ customer}) \cdot P(W \leq w | 0 \text{ customer}) + P(1 \text{ customer}) \cdot P(W \leq w | 1 \text{ customer}) = 0.5 \cdot 1 + 0.5 \cdot (1 - e^{-\lambda w})[/tex]when w ≥ 0. Or in other words, Jane's waiting time has a CDF of [tex]0.5 + 0.5 \times (1 - e^{-\lambda w})[/tex]
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Are you smarter than a second grader? A random sample of 60 second graders in a certain school district are given a standardized mathematics skills test. The sample mean score is = 52. Assume the standard deviation of test scores is σ = 15. The nationwide average score on this test is 50. The school superintendent wants to know whether the second graders in her school district have greater math skills than the nationwide average.
This is a one-sample z-test in statistics. You need to set null and alternative hypotheses, calculate the z-score, and compare it to the critical value to test if the mean score of second grade students in the school district significantly exceeds the nationwide average.
Explanation:The subject matter here falls within Statistics, a branch of Mathematics. You want to test whether the mean score of second grade students in the school district is significantly greater than the nationwide average. To solve this, we perform a one-sample z-test, because we know the population standard deviation (σ = 15).
To start, we set the null hypothesis that the mean score for the district is equal to the nationwide average of μ0 = 50. The alternative hypothesis, according to the superintendent's belief, is that the mean score for the district is greater than 50.
You then calculate the z-score, which is the ratio of the difference between the sample mean and the population mean to the standard deviation of the population divided by the square root of the sample size (z = (X-bar - μ0) / (σ / √n). Here, X-bar is 52, n is 60 and σ is 15. After calculating the z-score, you would compare it to the critical z-value for the selected level of significance (typically 0.05 for a two-tailed test). If the calculated z score is larger than the critical z value, you would reject the null hypothesis and accept the alternative hypothesis, thus concluding that the second graders in the district have greater math skills than the nationwide average.
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