The null hypothesis is that the population mean is equal to or less than 76, while the alternative hypothesis is that it is greater than 76. The standardized test statistic is 2.089, and the P-value is approximately 0.022. Therefore, we reject the null hypothesis and conclude that there is enough evidence to support the claim.
Explanation:The null hypothesis, denoted as H0, states that the population mean (mu) is equal to or less than 76. The alternative hypothesis, denoted as H1, states that the population mean (mu) is greater than 76. To test the claim, we can perform a one-sample t-test.
The standardized test statistic, also known as the t-value, can be calculated using the formula t = (x - mu) / (s / sqrt(n)), where x is the sample mean, mu is the population mean under the null hypothesis, s is the sample standard deviation, and n is the sample size. Plugging in the values x = 77.5, mu = 76, s = 3.3, and n = 29 into the formula gives us a t-value of 2.089 (rounded to two decimal places).
The P-value of the test statistic can be determined by comparing the t-value to the critical value of the t-distribution with (n - 1) degrees of freedom at the given level of significance (alpha). Since the alternative hypothesis is one-sided (mu > 76), we need to find the right-tail area of the t-distribution. Consulting a t-table or using statistical software, we find that the P-value is approximately 0.022 (rounded to three decimal places).
To make a decision, we compare the P-value to the significance level (alpha). If the P-value is less than alpha, we reject the null hypothesis. In this case, the P-value is 0.022, which is less than alpha = 0.01. Therefore, we reject the null hypothesis and conclude that there is enough evidence to support the claim that the population mean (mu) is greater than 76.
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The null and alternative hypotheses can be set, and the t-value can be calculated using the provided sample statistics. The P-value and the decision to reject or fail to reject the null hypothesis cannot be determined without further calculations.
Explanation:The null and alternative hypotheses can be defined as follows:
Null Hypothesis (H0): Mu is less than or equal to 76 (mu <= 76)Alternative Hypothesis (Ha): Mu is greater than 76 (mu > 76)The standardized test statistic (t-value) can be calculated using the formula:
t = (x - mu) / (s / sqrt(n))
Using the given sample statistics:
t = (77.5 - 76) / (3.3 / sqrt(29)) = 1.69 (rounded to two decimal places)
The P-value for the test statistic can be found using a t-distribution table or a statistical software. The P-value represents the probability of obtaining a t-value as extreme as the calculated one, assuming the null hypothesis is true. Since the P-value is not provided in the question, it cannot be determined without further calculations.
To decide whether to reject or fail to reject the null hypothesis, compare the P-value to the significance level (alpha). If the P-value is less than alpha, reject the null hypothesis. If the P-value is greater than or equal to alpha, fail to reject the null hypothesis.
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You believe that the mean BMI for college students is less than 25. The population standard deviation of BMI is unknown. You select a sample of size 30 and compute the sample mean to be 23.9 with a sample standard deviation of 3. What are the appropriate hypotheses? A. Null Hypothesis: μ< 25 Alt Hypothesis: μ= 25 B. Null Hypothesis: μ=23.9 Alt Hypothesis: μ < 23.9 C. Null Hypothesis: μ=25 Alt Hypothesis: μ>25 D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25
Answer:
D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25
Step-by-step explanation:
Previous concepts and notation
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".
[tex]\bar X= 23.9[/tex] represent the sample mean
[tex]s=3[/tex] represent the sample standard deviation
n=25 represent the sample selected
Solution
The claim that we want to test is that: "You believe that the mean BMI for college students is less than 25". So this statement needs to be on the Alternative hypothesis ([tex]\mu <25[/tex]) and the Null hypothesis would be the complement ([tex]\mu =25[/tex]) or ([tex]\mu \geq 25[/tex]).
So the best option on this case is:
D. Null Hypothesis:μ=25 Alt Hypothesis:μ < 25
A poll showed that 48 out of 120 randomly chosen graduates of California medical schools last year intended to specialize in family practice. What is the width of a 90 percent confidence interval for the proportion that plan to specialize in family practice? Select one:a. ± .0736b. ± .0447c. ± .0876d. ± .0894
Answer:
± 0.0736
Step-by-step explanation:
Data provided in the question:
randomly chosen graduates of California medical schools last year intended to specialize in family practice, p = [tex]\frac{48}{120}[/tex] = 0.4
Confidence level = 90%
sample size, n = 120
Now,
For 90% confidence level , z-value = 1.645
Width of the confidence interval = ± Margin of error
= ± [tex]z\times\sqrt\frac{p\times(1-p)}{n}[/tex]
= ± [tex]1.645\times\sqrt\frac{0.4\times(1-0.4)}{120}[/tex]
= ± 0.07356 ≈ ± 0.0736
Hence,
The correct answer is option ± 0.0736
The correct option is a. [tex]\±0.0736.[/tex] The width of a [tex]90\%[/tex] confidence interval for the proportion that plan to specialize in family practice
To find the width of a confidence interval for a proportion, we can use the formula:
[tex]\[ \text{Width} = Z \times \sqrt{\frac{p(1-p)}{n}} \][/tex]
where:
[tex]\( Z \)[/tex] is the Z-score corresponding to the desired confidence level,
[tex]\( p \)[/tex] is the sample proportion,
[tex]\( n \)[/tex] is the sample size.
Given:
[tex]Sample\ proportion \( p = \frac{48}{120} = \frac{2}{5} = 0.4 \)[/tex],
[tex]Sample\ size \( n = 120 \)[/tex]
[tex]Confidence \ level = 90\%[/tex], which corresponds to a Z-score of approximately [tex]1.645.[/tex]
Substitute the values into the formula:
[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.4 \times 0.6}{120}} \][/tex]
[tex]\[ \text{Width} = 1.645 \times \sqrt{\frac{0.24}{120}} \][/tex]
[tex]\[ \text{Width} = 1.645 \times \sqrt{0.002} \][/tex]
[tex]\[ \text{Width} = 1.645 \times 0.0447 \][/tex]
[tex]\[ \text{Width} = 0.0736 \][/tex]
Since an instant replay system for tennis was introduced at a majorâ tournament, men challenged 1399 refereeâ calls, with the result that 411 of the calls were overturned. Women challenged 759 refereeâ calls, and 211of the calls were overturned.
Use a 0.01 significance level to test the claim that men and women have equal success in challenging calls.H0:p1=p2H1:p1 (does not equal) p2Identify the test statistic z=___
Answer: The value of test statistic is 0.696.
Step-by-step explanation:
Since we have given that
[tex]n_1=1399\\\\x_1=411\\\\p_1=\dfrac{x_1}{n_1}=\dfrac{411}{1399}=0.294[/tex]
Similarly,
[tex]n_2=759, x_2=211\\\\p_2=\dfrac{x_2}{n_2}=\dfrac{211}{759}=0.278[/tex]
At 0.01 level of significance.
Hypothesis are :
[tex]H_0:P_1=P_2=0.5\\\\H_a:P_1\neq P_2[/tex]
So, the test statistic value would be
[tex]z=\dfrac{(p_1-p_2)-(P_1-P_2)}{\sqrt{\dfrac{P_1Q_1}{n_1}+\dfrac{P_2Q_2}{n_2}}}\\\\z=\dfrac{0.294-0.278}{\sqrt{0.5\times 0.5(\dfrac{1}{1399}+\dfrac{1}{759})}}\\\\z=\dfrac{0.016}{0.023}=0.696[/tex]
Hence, the value of test statistic is 0.696.
A Common Measure of Student Achievement, The National Assessment of Educational Progress (NAEP) is the only assessment that measures what U.S. students know and can do in various subjects across the nation, states, and in some urban districts. Also known as The Nation's Report Card, NAEP has provided important information about how students are performing academically since 1969. NAEP is a congressionally mandated project administered by the National Center for Education Statistics (NCES) within the U.S. Department of Education and the Institute of Education Sciences (IES). NAEP is given to a representative sample of students across the country. Results are reported for groups of students with similar characteristics (e.g., gender, race and ethnicity, school location), not individual students. National results are available for all subjects assessed by NAEP. In the most recent year, The NAEP sample of 1077 young women had mean quantitative score of 275. Individual NAEP scores have a Normal distribution with standard deviation of 60. (This is indicating that the population standard deviation σσ is 60) Find a 99% Confidence Interval for the mean quantitative scores for young women. a) Check that the normality assumptions are met. ? b) What is the 99% confidence interval for the mean quantitative scores for young women? ____ ≤μ≤ _____ c) Interpret the confidence interval obtained in previous question
Answer:
a) By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.
[tex]X \sim N(\mu, \sigma=60)[/tex]
And we are interested on the distribution for the sample mean [tex]\bar X[/tex], we know that distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for [tex]\bar X[/tex] is normal.
b) [tex]270.283\leq \mu \leq 279.717[/tex]
c) We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)
Step-by-step explanation:
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".
[tex]\bar X=275[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=60[/tex] represent the population standard deviation
n=1077 represent the sample size
Part a
By the central limit theorem we know that if the random variable X="quantitative scores for young women", and we know that this distribution is normal.
[tex]X \sim N(\mu, \sigma=60)[/tex]
And we are interested on the distribution for the sample mean [tex]\bar X[/tex], we know that distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And with that we have the assumptions required to apply the normal distribution to create the confidence interval since the distribution for [tex]\bar X[/tex] is normal.
Part b
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
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: "=-NORM.INV(0.005,0,1)".And we see that [tex]z_{\alpha/2}=2.58[/tex]
Now we have everything in order to replace into formula (1):
[tex]275-2.58\frac{60}{\sqrt{1077}}=270.283[/tex]
[tex]275+2.58\frac{60}{\sqrt{1077}}=279.717[/tex]
So on this case the 99% confidence interval would be given by (270.283;3279.717)
[tex]270.283\leq \mu \leq 279.717[/tex]
Part c
We are confident (99%) that the true mean for the quantitative scores for young women is between (270.283;3279.717)
Final answer:
A 99% confidence interval for the mean quantitative scores for young women, based on the NAEP data, is found to be between 270.27 and 279.73. We can state with 99% confidence that the true mean quantitative score for young women falls within this range.
Explanation:
We are asked to find a 99% confidence interval for the mean quantitative scores for young women, based on the National Assessment of Educational Progress (NAEP) sample data.
a) Check that the normality assumptions are met.
The normality assumption is met because individual NAEP scores are given to have a Normal distribution. Additionally, with a large sample size (more than 30), the Central Limit Theorem ensures the sampling distribution of the mean is approximately normal, regardless of the distribution of the population from which the sample is drawn.
b) What is the 99% confidence interval for the mean quantitative scores for young women?
Using the given data, the sample mean, \\bar{x}\, is 275 and the population standard deviation, \(\sigma\), is 60. Since the population standard deviation is given and the sample size is large (1077), we will use the z-distribution to find the 99% confidence interval.
The z-score corresponding to a 99% confidence level is approximately 2.576. Therefore, the margin of error (ME) can be calculated as:
ME = z * [tex](\(\sigma/\sqrt{n}\))[/tex]
ME = 2.576 * [tex](60/\sqrt{1077})[/tex] \approx 4.73
The 99% confidence interval is:
Lower limit =[tex]\(\bar{x}\)[/tex] - ME = 275 - 4.73 = 270.27
Upper limit =[tex]\(\bar{x}\)[/tex] + ME = 275 + 4.73 = 279.73
The 99% confidence interval for the mean quantitative scores for young women is 270.27 ≤ μ ≤ 279.73.
c) Interpret the confidence interval obtained in previous question
The interpretation of this confidence interval is that we are 99% confident that the true mean quantitative score for young women falls between 270.27 and 279.73.
For the given function, determine consecutive values of x between which each real zero is located. f(x) = –11x^4 – 5x^3 – 9x^2 + 12x + 10
Answer:
[-1, 0][0, 1]Step-by-step explanation:
Descartes' rule of signs tells you this function, with its signs {- - - + +}, having one sign change, will have one positive real root.
When odd-degree terms have their signs changed, the signs {- + - - +} have three changes, so there will be 1 or 3 negative real roots.
The constant term (10) tells us the y-intercept is positive. The sum of coefficients is -11 -5 -9 +12 +10 = -3, so f(1) < 0 and there is a root between 0 and 1.
When odd-degree coefficients change sign, the sum becomes -11 +5 -9 -12+10 = -17, so there is a root between -1 and 0.
__
Synthetic division by (x+1) gives a quotient of -11x^3 +6x^2 -15x +27 -17/(x+1), which has alternating signs, indicating -1 is a lower bound on real roots.
Real roots are located in the intervals [-1, 0] and [0, 1].
_____
The remaining roots are complex. All roots are irrational.
The attached graph confirms that roots are in the intervals listed here. Newton's method iteration is used to refine these to calculator precision. Dividing them from f(x) gives a quadratic with irrational coefficients and complex roots.
You want to estimate the mean time college students spend watching online videos each day. The estimate must be within 2 minutes of the population mean. Determine the required sample size to construct a 99% confidence interval for the population mean. Assume the population standard deviation is 2.4 minutes.
Answer: 10
Step-by-step explanation:
We know that the formula to find the sample size is given by :-
[tex]n= (\dfrac{z_{\alpha/2\cdot \sigma}}{E})^2[/tex]
, where [tex]\sigma[/tex] = population standard deviation.
[tex]z_{\alpha/2}[/tex] = Two -tailed z-value for [tex]{\alpha[/tex] (significance level)
E= margin of error.
Given : Confidence level : C =99%=0.99
i.e. [tex]1-\alpha=0.99[/tex]
⇒Significance level :[tex]\alpha=1-0.99=0.01[/tex]
By using z-value table ,Two -tailed z-value for [tex]\alpha=0.01 [/tex]:
[tex]z_{\alpha/2}=2.576[/tex]
E= 2 minutes
[tex]\sigma=\text{2.4 minutes}[/tex]
The required sample size will be :-
[tex]n= (\dfrac{2.576\cdot 2.4}{2})^2\\\\= (3.0912)^2\\\\=9.55551744\approx10[/tex]
Hence, the required sample size = 10
To estimate the mean time college students spend watching online videos each day within 2 minutes of the population mean at a 99% confidence level, at least 10 students should be sampled assuming a population standard deviation of 2.4 minutes.
To determine the required sample size, we use the formula for the sample size of a mean:
n = (z*σ/E)^2
Where n is the sample size, z is the z-score corresponding to the desired confidence level, σ is the population standard deviation, and E is the maximum error allowed (margin of error).
For a 99% confidence interval, the z-score (from the standard normal distribution) is approximately 2.576. The population standard deviation σ is given as 2.4 minutes, and the margin of error E is 2 minutes. Using these values:
n = (2.576*2.4/2)^2
n ≈ (6.1824/2)^2
n ≈ (3.0912)^2
n ≈ 9.554
Since the sample size must be a whole number, we would round up to ensure the sample size is large enough to achieve the desired margin of error, which means at least 10 students should be surveyed to construct the 99% confidence interval for the mean time spent watching online videos.
It is important to note that the larger the sample size, the more accurate the estimate of the population mean will be.
Find the absolute maximum and absolute minimum values of f on the given interval.
a) f(t)= t sqrt(36-t^2) [-1,6]
absolute max=
absolute min=
b) f(t)= 2 cos t + sin 2t [0,pi/2]
absolute max=
absolute min=
Answer:
absolute max= (4.243,18)
absolute min =(-1,-5.916)
absolute max=(pi/6, 2.598)
absolute min = (pi/2,0)
Step-by-step explanation:
a) [tex]f(t) = t\sqrt{36-t^2} \\[/tex]
To find max and minima in the given interval let us take log and differentiate
[tex]log f(t) = log t + 0.5 log (36-t^2)\\Y(t) = log t + 0.5 log (36-t^2)[/tex]
It is sufficient to find max or min of Y
[tex]y'(t) = \frac{1}{t} -\frac{t}{36-t^2} \\\\y'=0 gives\\36-t^2 -t^2 =0\\t^2 =18\\t = 4.243,-4.243[/tex]
In the given interval only 4.243 lies
And we find this is maximum hence maximum at (4.243,18)
Minimum value is only when x = -1 i.e. -5.916
b) [tex]f(t) = 2cost +sin 2t\\f'(t) = -2sint +2cos2t\\f"(t) = -2cost-4sin2t\\[/tex]
Equate I derivative to 0
-2sint +1-2sin^2 t=0
sint = 1/2 only satisfies I quadrant.
So when t = pi/6 we have maximum
Minimum is absolute mini in the interval i.e. (pi/2,0)
If a 25 in. by 32 in. window has a wooden frame 2 inches wide surrounding it, what is the total area of the window and frame?
1044 inches
1036 sq. inches
580 inches
1044 sq. inches
Answer:
1044 sq. inches
Step-by-step explanation:
The frame adds 2 inches on each side of the window, so the width becomes 29 in. and the length becomes 36 in.
Area = width × length
So the total area of the window and the frame is:
A = 29 × 36 = 1044 in²
Choose the graph that represents the following system of inequalities:
y ≤ −3x + 1
y ≤ x + 3
In each graph, the area for f(x) is shaded and labeled A, the area for g(x) is shaded and labeled B, and the area where they have shading in common is labeled AB.
Correct graph is C
Step-by-step explanation:
Given two inequalities are:
1. [tex]y\leq -3x+1[/tex]
2.[tex]y\leq x+3[/tex]
Step1 : Remove the inequalities
1. [tex]y =-3x+1[/tex]
2.[tex]y = x+3[/tex]
Step2 : Finding intersection points of equations
By solving linear equation
[tex]y =-3x+1 =x+3 [/tex]
[tex]-3x+1 =x+3 [/tex]
[tex] x = -0.5[/tex]
Replacing value of x in any equations
we get,
[tex]y = x+3[/tex]
[tex]y =-0.5+3[/tex]
[tex]y = 2.5[/tex]
Therefore, Point of intersection is (-0.5,2.5)
Step3: Test of origin (0,0)
Here, If inequalities holds true for origin then, shades the graph towards the origin.
For equation 1.
[tex]y\leq -3x+1[/tex]
[tex]0\leq -3(0)+1[/tex]
[tex]0\leq +1[/tex]
True, Shade graph towards origin.
For equation 2.
[tex]y\leq x+3[/tex]
[tex]0\leq 0+3[/tex]
[tex]0\leq 3[/tex]
True, Shade graph towards origin.
Thus, Correct graph is C
Answer:
c
Step-by-step explanation:
took test
A plant in the manufacturing sector is concerned about its sulfur dioxide emissions. The weekly sulfur dioxide emissions follow a normal distribution with a mean of 1000 ppm (parts per million) and a standard deviation of 25. Recently, a "cleaner" technology has been adopted. In such a scenario, the CEO would like to investigate whether there has been a significant change in the emissions and has hired you for advice. In other words, the CEO wants to know if the mean level of emissions is different from 1000. Suppose that you are given a sample of data on weekly sulfur dioxide emissions for that plant. The sample size is 50 and x = 1006 ppm. What is the value of the test statistic?a. 2.26b. 1.70c. 4.78d. 2.59
Answer:
The correct option is b) 1.70
Step-by-step explanation:
Consider the provided information.
The weekly sulfur dioxide emissions follow a normal distribution with a mean of 1000 ppm (parts per million) and a standard deviation of 25.
Thus, μ=1000 and σ = 25
The CEO wants to know if the mean level of emissions is different from 1000.
Therefore the null and alternative hypothesis is:
[tex]H_0:\mu =1000[/tex] and [tex]H_a:\mu \neq1000[/tex]
The sample size is n = 50 and [tex]\bar x=1006[/tex] ppm.
Now calculate the test statistic by using the formula: [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Substitute the respective values in the above formula.
[tex]z=\frac{1006-1000}{\frac{25}{\sqrt{50}}}[/tex]
[tex]z=\frac{6}{\frac{25}{\sqrt{50}}}=1.697\approx 1.70[/tex]
Hence, the correct option is b) 1.70
Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at 0.5 ppm. Below are cadmium levels for a random sample of the edible mushroom Boletus pinicoloa.0.24 0.92 0.59 0.19 0.62 0.330.16 0.25 0.77 0.59 1.33 0.32a) Perform a hypothesis test at the 5% significance level to determine if the meancadmium level in the population of Boletus pinicoloa mushrooms is greater than thegovernment’s recommended limit of 0.5 ppm. Suppose that the standard deviation ofthis population’s cadmium levels is o( = 0.37 ppm. Note that the sum of the data is 6.31 ppm. For this problem, be sure to: State your hypotheses, compute your test statistic, give the critical value.(b) Find the p-value for the test.
Answer:
hi
Step-by-step explanation:
A deck of cards contains 52 cards. They are divided into four suits: spades, diamonds, clubs and hearts. Each suit has 13 cards: ace through 10, and three picture cards: Jack, Queen, and King. Two suits are red in color: hearts and diamonds. Two suits are black in color: clubs and spades.
Use this information to compute the probabilities asked for below and leave them in fraction form. All events are in the context that three cards are dealt from a well-shuffled deck without replacement.
a. The first and second cards are both hearts.
b. The third card is an eight.
c. None of the three cards is an ace.
Answer:
a) 13/52 and the second 12/51
b) Two solutions:
b.1 if we did not picked up an eight in the first two cards 4/50
b.2 there is an eight i the two previous 3/59
c ) (48/52)*(47/51)*(46/50)
Step-by-step explanation:
Condition: Cards are taken out without replacement
a) Probability of first card is heart
There are 52 cards and 4 suits with the same probability , so you can compute this probability in two ways
we have 13 heart cards and 52 cards then probability of one heart card is 13/52 = 0.25
or you have 4 suits, to pick up one specific suit the probability is 1/4 = 0,25
Now we have a deck of 51 card with 12 hearts, the probability of take one heart is : 12/51
b) There are 4 eight (one for each suit ) P = 4/50 if neither the first nor the second card was an eight of heart, if in a) previous we had an eight, then this probability change to 3/50
c) The probability of the first card different from an ace is 48/52 , the probability of the second one different of an ace is 47/51 and for the thirsd card is 46/50. The probability of none of the three cards is an ace is
(48/52)*(47/51)*(46/50)
A production process fills containers by weight. Weights of containers are approximately normally distributed. Historically, the standard deviation of weights is 5.5 ounces. (This standard deviation is therefore known.) A quality control expert selects n containers at random. How large a sample would be required in order for the 99% confidence interval for to have a length of 2 ounces?
a. n 15
b. n 16
c. n 201
d. n 226
Answer:
option (c) n = 201
Step-by-step explanation:
Data provided in the question:
Standard deviation, s = 5.5 ounce
Confidence level = 99%
Length of confidence interval = 2 ounces
Therefore,
margin of error, E = (Length of confidence interval ) ÷ 2
= 2 ÷ 2
= 1 ounce
Now,
E = [tex]\frac{zs}{\sqrt n}[/tex]
here,
z = 2.58 for 99% confidence interval
n = sample size
thus,
1 = [tex]\frac{2.58\times5.5}{\sqrt n}[/tex]
or
n = (2.58 × 5.5)²
or
n = 201.3561 ≈ 201
Hence,
option (c) n = 201
Need help
6n - 3(2n - 5)
SHOW ALL WORK!
6n - 3(2n-5)
mutiply the bracket by -3
(-3)(2n)= -6n
(-3)(-5)= 15
6n-6n+15
0+15
answer:
15
The proportion of students at a college who have GPA higher than 3.5 is 19%. a. You take repeated random samples of size 25 from that college and find the proportion of student who have GPA higher than 3.5 for each sample. What is the mean and the standard error of the sampling distribution of the sample proportions?
Answer:
[tex]\mu_{\hat{p}}=0.19[/tex]
[tex]\sigma_{\hat{p}}=0.0785[/tex]
Step-by-step explanation:
We know that the mean and the standard error of the sampling distribution of the sample proportions will be :-
[tex]\mu_{\hat{p}}=p[/tex]
[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{p(1-p)}{n}}[/tex]
, where p=population proportion and n= sample size.
Given : The proportion of students at a college who have GPA higher than 3.5 is 19%.
i.e. p= 19%=0.19
The for sample size n= 25
The mean and the standard error of the sampling distribution of the sample proportions will be :-
[tex]\mu_{\hat{p}}=0.19[/tex]
[tex]\sigma_{\hat{p}}=\sqrt{\dfrac{0.19(1-0.19)}{25}}\\\\=\sqrt{0.006156}=0.0784601809837\approx0.0785[/tex]
Hence , the mean and the standard error of the sampling distribution of the sample proportions :
[tex]\mu_{\hat{p}}=0.19[/tex]
[tex]\sigma_{\hat{p}}=0.0785[/tex]
Suppose a basketball player is practicing shooting, and has a prob-ability .95 of making each of his shots. Also assume that his shots are in-dependent of one another. Using the Poisson distribution, approximate theprobability that there are at most 2 misses in the first 100 attempts
Answer:
0.082
Step-by-step explanation:
Number of attempts = n = 100
Since there are only two outcomes and in-dependent of each other, the probability of missing a shot = 1 - Probability of making each shot
p = 1 - 0.95 = 0.05
Possion Ratio (λ) = np where n is the number of events and p is the probability of the shot missing
λ = 100 x 0.05 = 5
Define X such that X = Number of misses and X ≅ Poisson (λ = 5)
P [X ≤ 2] = P [X = 0] + P [X = 1] + P [X = 2]
P [X ≤ 2] = e⁻⁵ + e⁻⁵ x 5 + e⁻⁵ x 5²/2!
P [X ≤ 2] = e⁻⁵ [1 + 5 + 5²/2!]
P [X ≤ 2] = e⁻⁵ x 12.25 = 0.082
The required probability that there are at most 2 misses in the first 100 attempts is 0.082
In the word balloons the ratio of vowels consonants is
Answer:
3 : 5
Step-by-step explanation:
vowels: a o o (3 of them)
consonants: b l l n s (5 of them)
The ratio of vowels to consonants is 3 : 5.
Louise is collecting can tabs for charity. She already has 35 collected and intends to collect 4 every week. The equation for the number of can tabs y she has collected is y = 4x + 35, where x is the number of weeks. What do the slope and y-intercept represent?
Answer:
Step-by-step explanation:
The equation for the number of can tabs y that she has collected is
y = 4x + 35
where x is the number of weeks. To plot the graph, values for different number of weeks are inputted to get the corresponding y values.
Comparing the equation with the slope intercept form,
y = mx + c,
m represents slope.
So slope of the equation is 4. It represents the rate at which the total number of can tabs that she has collected is increasing with respect to the increase in the number of weeks. So the total number increases by 4 each week
The y intercept is the point at x = 0
It is equal to 35. It represents the number of can tabs that she had initially collected before she started collecting 4 can tabs weekly. At this point, her weekly collection is 0
Evaluate the integral Integral ∫ from (1,2,3 ) to (5, 7,-2 ) y dx + x dy + 4 dz by finding parametric equations for the line segment from (1,2,3) to (5,7,- 2) and evaluating the line integral of of F = yi + x j+ 3k along the segment. Since F is conservative, the integral is independent of the path.
[tex]\vec F(x,y,z)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k[/tex]
is conservative if there is a scalar function [tex]f(x,y,z)[/tex] such that [tex]\nabla f=\vec F[/tex]. This would require
[tex]\dfrac{\partial f}{\partial x}=y[/tex]
[tex]\dfrac{\partial f}{\partial y}=x[/tex]
[tex]\dfrac{\partial f}{\partial z}=3[/tex]
(or perhaps the last partial derivative should be 4 to match up with the integral?)
From these equations we find
[tex]f(x,y,z)=xy+g(y,z)[/tex]
[tex]\dfrac{\partial f}{\partial y}=x=x+\dfrac{\partial g}{\partial y}\implies\dfrac{\partial g}{\partial y}=0\implies g(y,z)=h(z)[/tex]
[tex]f(x,y,z)=xy+h(z)[/tex]
[tex]\dfrac{\partial f}{\partial z}=3=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=3z+C[/tex]
[tex]f(x,y,z)=xy+3z+C[/tex]
so [tex]\vec F[/tex] is indeed conservative, and the gradient theorem (a.k.a. fundamental theorem of calculus for line integrals) applies. The value of the line integral depends only the endpoints:
[tex]\displaystyle\int_{(1,2,3)}^{(5,7,-2)}y\,\mathrm dx+x\,\mathrm dy+3\,\mathrm dz=\int_{(1,2,3)}^{(5,7,-2)}\nabla f(x,y,z)\cdot\mathrm d\vec r[/tex]
[tex]=f(5,7,-2)-f(1,2,3)=\boxed{18}[/tex]
Use a(t) = -32 ft/sec^2 as the acceleration due to gravity. (Neglect air resistance.) With what initial velocity must an object be thrown upward (from ground level) to reach the top of a national monument (580 feet)? (Round your answer to three decimal places.) Use a(t) = -32 ft/sec^2 as the acceleration due to gravity. (Neglect air resistance.) A balloon, rising vertically with a velocity of 16 feet per second, releases a sandbag at the instant when the balloon is 32 feet above the ground.
(a) How many seconds after its release will the bag strike the ground? (Round your answer to two decimal places.) sec
(b) At what velocity will it strike the ground? (Round your answer to three decimal places.)
Answer:
Initial velocity 192.666 ft/sec
a) 1.41 sec
Step-by-step explanation:
Equations for
V(f) = V₀ ± at
V(f)² =( V₀)² ± 2 a*d
d = V₀*t ± a*t²/2
We are going to use a(t) = g = 32 ft/sec²
a) V₀ = ?? a = -g
Then as d = 580 ft and V(f) = 0
we have
V(f)² = ( V₀)² - 2*32*580 ⇒ ( V₀)² = 64*580 (ft²/sec²)
( V₀)² = 37120 V₀ = 192,666 ft/sec
a) t = ?? in this case V₀ = 0
d = V₀*t + gt²/2 ⇒ 32 = ( 32 t²) /2
t² = 2
t = 1.41 sec
How would you describe the difference between the graphs of f(x) = x ^ 2 + 4 and g(y) = y ^ 2 + 4 OA. g(y) is a reflection of f(x) ) over the line y = 1 . O O B. g(y) is a reflection of f(x) ) over the line y=x 1 g(y) is a reflection of f(x) over the x-axis OD. g(y) is a reflection of f(x) over the .
x is replaced with Y, so it is a reflection across the line y = x
Answer:
y=x
Step-by-step explanation:
In simple regression analysis, if the correlation coefficient is a positive value, then Select one:
A. the slope of the regression line must also be positive.
B. the least squares regression equation could have either a positive or a negative slope.
C. the y-intercept must also be a positive value.
D. the coefficient of determination can be either positive or negative, depending on the value of the slope.
E. the standard error of estimate can have either a positive or a negative value.
Final answer:
In simple regression analysis, a positive correlation coefficient means that the slope of the regression line must also be positive, as both the dependent and independent variables increase or decrease together.
Explanation:
In simple regression analysis, the correlation coefficient's sign (positive or negative) informs us about the direction of the relationship between the two variables.
A positive correlation coefficient indicates that the variables move in the same direction: as one increases, so does the other, and as one decreases, so does the other. This characteristic of correlation directs us to the correct answer to the student's question.
Considering the options provided:
A. The slope of the regression line must also be positive. B. The least squares regression equation could have either a positive or a negative slope. C. The y-intercept must also be a positive value. D. The coefficient of determination can be either positive or negative, depending on the value of the slope. E. The standard error of estimate can have either a positive or a negative value.The slope of the regression line is directly determined by the correlation coefficient in the context of simple linear regression. Given that the correlation coefficient is positive, the slope of the regression line (which represents the change in the dependent variable for a unit change in the independent variable) must also be positive.
Therefore, option A is correct.
A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.
Construct the 95 % confidence interval for the true proportion of all voters in the state who favor approval.
A) 0.435 < p < 0.508
B) 0.444 < p < 0.500
C) 0.471 < p < 0.472
D) 0.438 < p < 0.505
Answer: D) 0.438 < p < 0.505
Step-by-step explanation:
We know that the confidence interval for population proportion is given by :-
[tex]\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]
, where n= sample size
[tex]\hat{p}[/tex] = Sample proportion.
z* = critical value.
Given : A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature.
i.e. n= 865
[tex]\hat{p}=\dfrac{408}{865}\approx0.4717[/tex]
Two-tailed critical avlue for 95% confidence interval : z* = 1.96
Then, the 95 % confidence interval for the true proportion of all voters in the state who favor approval will be :-
[tex]0.4717\pm (1.96)\sqrt{\dfrac{0.4717(1-0.4717)}{865}}\\\\\approx0.4717\pm 0.03327\\\\=(0.4717-0.03327,\ 0.4717+0.03327)=(0.43843,\ 0.50497)\approx(0.438,\ 0.505)[/tex]
Thus, the required 95% confidence interval : (0.438, 0.505)
Hence, the correct answer is D) 0.438 < p < 0.505
A random sample of 10 shipments of stick-on labels showed the following order sizes.10,520 56,910 52,454 17,902 25,914 56,607 21,861 25,039 25,983 46,929PictureClick here for the Excel Data File(a) Construct a 95 percent confidence interval for the true mean order size. (Round your answers to the nearest whole number.) The 95 percent confidence interval to
Answer:
Confidence Interval: (21596,46428)
Step-by-step explanation:
We are given the following data set:
10520, 56910, 52454, 17902, 25914, 56607, 21861, 25039, 25983, 46929
Formula:
[tex]\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}[/tex]
where [tex]x_i[/tex] are data points, [tex]\bar{x}[/tex] is the mean and n is the number of observations.
[tex]Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}[/tex]
[tex]Mean =\displaystyle\frac{340119}{10} = 34011.9[/tex]
Sum of squares of differences = 551869365.6 + 524322983.6 + 340111052.4 + 259528878 + 65575984.41 + 510538544 + 147644370.8 + 80512934.41 + 64463235.21 + 166851472.4 = 2711418821
[tex]S.D = \sqrt{\frac{2711418821}{9}} = 17357.09[/tex]
Confidence interval:
[tex]\mu \pm t_{critical}\frac{\sigma}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 9 and}~\alpha_{0.05} = \pm 2.2621[/tex]
[tex]34011.9 \pm 2.2621(\frac{17357.09}{\sqrt{10}} ) = 34011.9 \pm 12416.20 = (21595.7,46428.1) \approx (21596,46428)[/tex]
f(1)=-3
f(n)=-5*f(n-1)-7
f(2)=?
pls help lol.
Answer:
f(2)=8
Step-by-step explanation:
f(2)=(-5)*f(1)-7=
(-5)*(-3)-7=8
f(2)=8
If all the beads in a jar are either orange or purple and the ratio of orange to purple beads is 6 to 5, what is the ratio of the following:
a. purple to orange
b. purple to all beads
c. all beads to orange
Please help ASAP on all 3 parts!!! :(
Answer:
a. purple to orange =[tex]\frac{5}{6}[/tex]
b. purple to all beads=[tex]\frac{5){11}[/tex]
c. all beads to orange=[tex]\frac{11}{6}[/tex]
Step-by-step explanation:
Given:
Colour of beads in jar = orange or purple
The ratio of orange to purple beads = 6 : 5
Solution:
Let number of orange beads be= 6x
Number of purple beads=5x
Total no of beads= 6x+5x=11x
a. Ratio of purple to orange.
Ratio of purple to orange=[tex]\frac{\text{number of purple boads}}{\text{number of orange beads}}[/tex]
Ratio of purple to orange=[tex]\frac{5x}{6x}[/tex]
Ratio of purple to orange=[tex]\frac{5}{6}[/tex]
b. Ratio of purple to all beads
Ratio of purple to all beads=[tex]\frac{\text{number of purple beads}}{\text{Total number of beads}}[/tex]
Ratio of purple to all beads=[tex]\frac{5x}{11x}[/tex]
Ratio of purple to all beads=[tex]\frac{5}{11}[/tex]
c. Ratio of all beads to orange
Ratio of purple to all beads=[tex]\frac{\text{Total number of beads}}{\text{number of orange beads}}[/tex]
Ratio of purple to all beads=[tex]\frac{11x}{6x}[/tex]
Ratio of purple to all beads=[tex]\frac{11}{6}[/tex]
For each of the following scenarios state whether H0 should be rejected or not. State any assumptions that you make beyond the information that is given.
(a) H0 : µ = 4, H1 : µ 6= 4, n = 15, X = 3.4, S = 1.5, α = .05.
(b) H0 : µ = 21, H1 : µ < 21, n = 75, X = 20.12, S = 2.1, α = .10.
(c) H0 : µ = 10, H1 : µ 6= 10, n = 36, p-value = 0.061.
Answer:
a)[tex]H_0 :\mu = 4\\ H_1 : \mu \neq 4[/tex] , n = 15 , X=3.4 , S=1.5 , α = .05
Formula : [tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{3.4-4}{\frac{1.5}{\sqrt{15}}}[/tex]
[tex]t =-1.549[/tex]
p- value = 0.607(using calculator)
α = .05
p- value > α
So, we failed to reject null hypothesis
b)[tex]H_0 :\mu = 21\\ H_1 : \mu < 21[/tex] , n =75 , X=20.12 , S=2.1 , α = .10
Formula : [tex]t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}[/tex]
[tex]t = \frac{20.12-21}{\frac{2.1}{\sqrt{75}}}[/tex]
[tex]t =-3.6290[/tex]
p- value = 0.000412(using calculator)
α = .1
p- value< α
So, we reject null hypothesis
(c) [tex]H_0 :\mu = 10\\ H_1 : \mu \neq 10[/tex], n = 36, p-value = 0.061.
Assume α = .05
p-value = 0.061.
p- value > α
So, we failed to reject null hypothesis
The decision to reject the null hypothesis in each scenario depends on comparing calculated test statistics (or given p-value) to critical values associated with the significance level. In (a) and (c), we may not reject H0, and in (b) we would reject H0 if our test statistic is larger than the critical value.
Explanation:In hypothesis testing, we compare the test statistic, often a t-value, to the critical value determined by our chosen significance level, α. If the test statistic is greater, we reject the null hypothesis (H0).
(a) In the first scenario, we must calculate the test statistic: t = (X - µ) / (S/√n) = (3.4 - 4) / (1.5/√15); if this absolute value is less than our critical value associated with α = .05 and degrees of freedom = 14, we do not reject H0.
(b) For the second, the test statistic would be (20.12 - 21) / (2.1/√75); compare its value to the critical value associated with α = .10 and degrees of freedom = 74. If it is larger, we reject H0 due to the lower tail test in this scenario.
(c) In the third scenario, the p-value is given. Since our p-value = 0.061 is larger than our α = .05, we would not reject H0.
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The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.
a) Find the probability of a pregnancy lasting 308 days or longer.
b) If the length of pregnancy is in the lowest 22?%, then the baby is premature.
Find the length that separates premature babies from those who are not premature.
a) the probability that a pregnancy will last 308 days or longer is?
b) babies who are born on or before ___ days are considered premature
Answer:
a) The probability that a pregnancy will last 308 days or longer is 0.0038
b) Babies who are born on or before 256 days are considered prematures.
Step-by-step explanation:
Let X be the random variable that represents the length of a pregnancy. Then, X is normally distributed with a mean of 268 days and a standard deviation of 15 days.
a) The z-score related to 308 days is z = (308-268)/15 = 2.6667, so, the probability of a pregnancy lasting 308 days or longer is P(Z > 2.6667) = 0.0038
b) We are looking for a value q such that P(X < q) = 0.22, i.e., P((X-268)/15 < (q-268)/15) = 0.22, here, Z = (X-268)/15 is a standard normal random variable and z = (q-268)/15 is the 22nd quantile of the standard normal distribution, i.e., z = -0.7722 = (q-268)/15 and (-0.7722)(15) + 268 = q, i.e., q = 256.417, so, babies who are born on or before 256 days are considered premature.
Answer:
a) P(X>308) = 0.00383
b) 256.42 days
Step-by-step explanation:
Population mean (μ) = 268 days
Standard deviation (σ) = 15 days
a) P(X>308)?
The z-score for any length of pregnancy 'X' is given by:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
The z-score for X=308 is:
[tex]z=\frac{308-268}{15}\\z=2.6667[/tex]
A z-score of 2.6667 is equivalent to the 99.617-th percentile of a normal distribution. Thus, the probability of a pregnancy lasting 208 days or longer is:
[tex]P(X>308)=1-0.99617\\P(X>308) = 0.00383[/tex]
b) X at which P(X) < 0.22?
At the 22-nd percentile, a normal distribution has an equivalent z-score of -0.772. Therefore, the length of pregnancy, X, that separates premature babies from those who are not premature is:
[tex]-0.772=\frac{X-268}{15}\\X= 256.42[/tex]
About 20,000 steel cans are recycled every minute in the United States. Expressed in scientific notation, about how many cans are recycled in 48 hours?
Answer:
57,600,000
Step-by-step explanation:
There are 25 dogs playing in the dog park. You are a "dog" person and wonder which dog will be the first to come up to you when you enter the park. There are 15 golden retrievers, 5 pugs, 4 pomeranians, and 1 terrier. What is the probability that the first dog to come over to you will be a pug?
I chose 5/25
2. One Psyc 317 class has 10 people in it; another has 50 people in it. The average grade in the 10-person class is 72% (high probability of getting a C); the average grade in the 50-person class is 85% (high probability of getting a B). Which class average is the most reliable?
a. the 10 person class
b. 50 person class
c. cannot determine with this information
d. both equally
I chose the 50 person class.
The probability of a pug approaching first in a park of 25 dogs is 5/25 or 1/5. The average grade of a larger class is typically a more reliable representation of the group's performance than that of a smaller one.
Explanation:For the first question, the probability of a pug being the first dog to approach you in a park with 25 dogs, 5 of them being pugs, is indeed 5/25, or 1/5. This fraction represents the ratio of the desired outcome (a pug approaching you first) to the total number of possible outcomes (any of the 25 dogs in the park could potentially be the first to approach).
For the second part of the question - determining the more reliable average grade between two classes - the class with 50 students is the better choice. This conclusion is based on statistical reasoning. A larger sample size (in this case, the larger class) generally provides a more reliable average than a smaller sample size.
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