Answer:
a) Mean = 6.5, sample standard deviation = 3.50
b) Standard error = 0.7826
c) Point estimate = 6.5
d) Confidence interval: (5.1469 ,7.8531)
Step-by-step explanation:
We are given the following data set for students to earn bachelor's degrees.
4, 4, 4, 4, 4, 4, 4.5, 4.5, 4.5, 4.5, 4.5, 4.5, 6, 6, 8, 9, 9, 13, 13, 15
a) 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{130}{20} = 6.5[/tex]
Sum of squares of differences = 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 6.25 + 4 + 4 + 4 + 4 + 4 + 4 + 0.25 + 0.25 + 2.25 + 6.25 + 6.25 + 42.25 + 42.25 + 72.25 = 233.5
[tex]S.D = \sqrt{\frac{233.5}{19}} = 3.50[/tex]
b) Standard Error
[tex]= \displaystyle\frac{s}{\sqrt{n}} = \frac{3.50}{\sqrt{20}} = 0.7826[/tex]
c) Point estimate for the mean time required for all college is given by the sample mean.
[tex]\bar{x} = 6.5[/tex]
d) 90% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.10} = \pm 1.729[/tex]
[tex]6.5 \pm 1.729(\frac{3.50}{\sqrt{20}} ) = 6.5 \pm 1.3531 = (5.1469 ,7.8531)[/tex]
e) No, the confidence interval does not contain the value of 4 years. Thus, confidence interval is not a good estimator as most of the value in the sample is of 4 years. Most of the sample does not lie in the given confidence interval.
Final answer:
The question requires calculating the sample mean, standard deviation, standard error, point estimate for the population mean, and constructing a confidence interval for the mean time to earn bachelor's degrees. The value of 4 years will need to be checked against the calculated confidence interval, and the reliability of these results may be scrutinized based on the distribution of the data.
Explanation:
The question involves calculating various statistical measures for a dataset representing the number of years college students took to earn bachelor's degrees. To address these parts:
To calculate the sample mean, you add up all the numbers and divide by the total count of numbers. The sample standard deviation measures the amount of variation or dispersion in a set of values. You use the formula for standard deviation for a sample.
The standard error (SE) is calculated by dividing the sample standard deviation by the square root of the sample size.
The point estimate for the mean time is the sample mean, as it provides the best estimate of the population mean based on the sample data.
To construct the confidence interval, you would typically use the sample mean ± (critical value from the t-distribution * standard error). For 90% confidence, you can find the critical t-value for 19 degrees of freedom (since sample size minus one equals degrees of freedom).
To answer whether the confidence interval contains the value of 4 years and discuss the reliability of this interval, you'll examine the calculated interval and consider factors like the presence of outliers and the shape of the distribution.
please help
simplify the following write your answer as a complex number in standard form
1 through 6
Answer:
[tex]$ \sqrt{-1} = i $[/tex]
Step-by-step explanation:
(1) [tex]$ \sqrt{-100} $[/tex]
[tex]$ = \sqrt{100}.i = 10i $[/tex]
(1) - 10i
(2) [tex]$ -2\sqrt{50} $[/tex]
[tex]$ = -2\sqrt{50} = -2(5) \sqrt{2} = -10\sqrt{2} $[/tex]
(2) - -10√2
(3) [tex]$ 5 + \sqrt{-13} $[/tex]
[tex]$ \implies 5 + \sqrt{13}i $[/tex]
(3) - 5 + √13i
(4) [tex]$ 5 + \sqrt{-120} $[/tex]
[tex]$ = 5 + \sqrt{120}i = 5 + 2\sqrt{30}i$[/tex]
(4) - 5 + 2√30i
(5) [tex]$ -10 + \sqrt{-50} $[/tex]
[tex]$ = -10 + 5\sqrt{2} i $[/tex]
(5) - -10 + 5√2i
(6) - [tex]$ \sqrt{-45} $[/tex]
[tex]$ 3\sqrt{5}i $[/tex]
(6) - 3√5i
Identify a pattern and find the next number in the pattern: -0.8,-3.2,-12.8,-51.2
Answer: -204.8
Step-by-step explanation: if there are and more questions like this use m a t h w a y
In the United States, the ages 11 to 55+ of smartphone users approximately follow a normal distribution with an approximate mean of 34.8 years and a standard deviation of 14.1 years. Determine the probability that a randomly selected smartphone users in the age range 11 to 55+ is between 30 and 54 years old.
Answer:
0.546 is the probability that a randomly selected smartphone users in the age range 11 to 55+ is between 30 and 54 years old.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 34.8 years
Standard Deviation, σ = 14.1 years
We are given that the distribution of ages of smartphone is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P( age range is between 30 and 54 years old)
[tex]P(30 \leq x \leq 54) = P(\displaystyle\frac{30 - 34.8}{14.1} \leq z \leq \displaystyle\frac{54-34.8}{14.1}) = P(-0.3404 \leq z \leq 1.3617)\\\\= P(z \leq 1.3617) - P(z < -0.3404)\\= 0.913 - 0.367 = 0.546 = 54.6\%[/tex]
[tex]P(30 \leq x \leq 54) = 54.6\%[/tex]
0.546 is the probability that a randomly selected smartphone users in the age range 11 to 55+ is between 30 and 54 years old.
Use technology to find the P-value for the hypothesis test described below.
The claim is that for a smartphone carrier's data speeds at airports, the mean is μ=18.00 Mbps.
The sample size is n=17 and the test statistic is t=−1.421.
What is the p-value?
(Round to three decimal places as needed.)
Answer:
The p-value is 0.175
Step-by-step explanation:
We have the null hypothesis [tex]H_{0}: \mu = 18[/tex] and the alternative hypothesis [tex]H_{1}: \mu \neq 18[/tex] (two-tailed alternative). Because the sample size is n = 17 and the test statistic is t=-1.421, we know that this last value comes from a t distribution with n-1=17-1=16 degrees of freedom. Therefore, the p-value is given by 2P(T < -1.421) because the p-value is the probability of getting a value as extreme as the observed value and because of the simmetry of the t distribution. Here, T has a t distribution with 16 df and we are using the t distribution because the sample size is small. So, 2P(T < -1.421) = 0.1745
The p-value is approximately 0.173, indicating that we do not have enough evidence to reject the null hypothesis. The mean data speed at airports is not significantly different from the claimed value of 18.00 Mbps.
Certainly! To find the p-value for the given hypothesis test, we can follow these steps:
1. **Set up Hypotheses:**
- Null Hypothesis (H0): The mean [tex](\(\mu\))[/tex] data speed is equal to 18.00 Mbps.
- Alternative Hypothesis (Ha): The mean [tex](\(\mu\))[/tex] data speed is different from 18.00 Mbps.
2. **Identify Test and Parameters:**
- We are conducting a one-sample t-test.
- Sample size (n) is 17.
- Test statistic (t) is -1.421.
3. **Degrees of Freedom:**
- Degrees of freedom (df) is calculated as n - 1, which is 16.
4. **Determine the Critical Region:**
- Since it's a two-tailed test (not equal), we need to find the critical values for a significance level [tex](\(\alpha\))[/tex] of 0.05.
5. **Calculate the p-value:**
- Using the t-distribution table or statistical software, we find the p-value associated with t = -1.421 and df = 16.
6. **Make a Decision:**
- If the p-value is less than the significance level (commonly 0.05), we reject the null hypothesis.
Now, let's perform the calculation. Using statistical software or a t-distribution table, the p-value for t = -1.421 and df = 16 is approximately 0.173.
**Conclusion:**
Since the p-value (0.173) is greater than 0.05, we fail to reject the null hypothesis. There is not enough evidence to suggest that the mean data speed at airports is different from 18.00 Mbps.
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The volume of a spherical cancerous tumor is given by v(r)=4/3 p r^3 .
If the radius of a tumor is estimated at 1.1 cm, with a maximum error in measurement of 0.005 cm, determine the error that might occur when the volume of the tumor is calculated.
Answer:
The error in the calculated volume is about [tex]0.0242\pi \approx 0.07602 \:cm^3[/tex]
Step-by-step explanation:
Given a function y=f(x) we call dy and dx differentials and the relationship between them is given by,
[tex]dy=f'(x)dx[/tex]
If the error in the measured value of the radius is denoted by [tex]dr=\Delta r[/tex], then the corresponding error in the calculated value of the volume is [tex]\Delta V[/tex], which can be approximated by the differential
[tex]dV=4\pi r^2dr[/tex]
When r = 1.1 cm and dr = 0.005 cm, we get
[tex]dV=4\pi (1.1)^2(0.005)=0.0242\pi[/tex]
The error in the calculated volume is about [tex]0.0242\pi \approx 0.07602 \:cm^3[/tex]
What is the equation of the parabola in vertex form. Vertex(4,4),point(2,-4)
Answer:
y = -2(x - 4)^2 + 4.
Step-by-step explanation:
Vertex form:
y = a(x - 4)^2 + 4 (because the vertex is at (4, 4)).
To find the value of a we substitute the point (2, -4):
-4 = a(2-4)^2 + 4
4a = -8
a = -2.
Noah is writing an exam for his 8th grade students. The exam is worth 100 points and Noah wants 35 questions on the exam. He plans to mix short answer questions, worth 3 points, with multiple choice questions worth 2 points. Create a system of equations to tell us how many of each type of question Noah can have on the test. Let x= the number of short answer questions and y= the number of multiple choice questions
Answer: the system of equations are
x + y = 35
3x + 2y = 100
Step-by-step explanation:
Let x= the number of short answer questions.
Let y= the number of multiple choice questions.
Noah wants 35 questions on the exam. This means that
x + y = 35
He plans to mix short answer questions, worth 3 points, with multiple choice questions worth 2 points. This means that x short answer questions will give 3x points and y multiple choice questions will give 2y points
Since the exam is worth 100 points, then,
3x + 2y = 100 - - - - - - - -1
Substituting x = 35 - y into equation 1, it becomes
3(35 - y) + 2y = 100
105 - 3y + 2y = 100
y = 105 - 100 = 5
x = 35 - y = 35 - 5
x = 30
The system of equation telling the number of each question Noah can have on the test is [tex]x+y = 35\\3x + 2y = 100[/tex]. The value of variables evaluates to:
30 = x= the number of short answer questions 5 = y = the number of multiple choice questionsHow to form mathematical expression from the given description?You can represent the unknown amounts by the use of variables. Follow whatever the description is and convert it one by one mathematically. For example if it is asked to increase some item by 4 , then you can add 4 in that item to increase it by 4. If something is for example, doubled, then you can multiply that thing by 2 and so on methods can be used to convert description to mathematical expressions.
For this case, we are given these facts:
Total 35 questions will be in examTwo type of questions are there, one being multiple choice questions, and other being short answer questionsTotal 100 points is maximum achievable pointsEach correct multiple choice question is of 2 pointsEach correct short answer question is of 2 pointsLet there are:
x= the number of short answer questions and y= the number of multiple choice questionsThen, we get:
[tex]x+y = 35[/tex] (as total number of question is 35)
Maximum points 'x' short answer can gain = [tex]3\times x[/tex]Maximum points 'y' multiple questions can gain = [tex]2 \times x[/tex]Thus, we get second equation as:
[tex]3x + 2y = 100[/tex] (as total maximum score = 100 points)
Therefore, the system of equation obtained for this condition is:
[tex]x+y = 35\\3x + 2y = 100[/tex]
From first equation, getting value of x in terms of y, we get:
[tex]x = 35 -y[/tex]
Putting this in second equation, we get:
[tex]3x + 2y = 100\\3(35-y) + 2y = 100\\105 -3y + 2y = 100\\105-100 = y\\y = 5[/tex]
Putting this value of y in equation for x, we get;
[tex]x = 35 -y = 35 - 5 = 30[/tex]
Thus, the system of equation telling the number of each question Noah can have on the test is [tex]x+y = 35\\3x + 2y = 100[/tex]. The value of variables evaluates to:
30 = x= the number of short answer questions 5 = y = the number of multiple choice questionsLearn more about system of linear equations here:
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A cell phone manufacturer claims that the average battery life of its newest flagship smartphone is exactly 20 hours. Javier believes the mean battery life is less than 20 hours. He tests this claim by selecting a random sample of 33 phones of this model. avier found that the sample mean battery life is 19.5 hours with a sample standard deviation of 1.9 hours. The test statistic t for a hypothesis test of H0:μ=20 versus Ha:μ<20 is t≈−1.51 with 32 degrees of freedom. If 0.05
Select all that apply:
(A) Reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours.
(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours.
(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours.
(D) There is enough evidence at the α=0.05 level of significance to support the claim that the true population mean battery life of the smartphone is not equal to 20 hours.
Answer:
(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours.
(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours.
Step-by-step explanation:
1) Data given and notation
[tex]\bar X=19.5[/tex] represent the battery life sample mean
[tex]s=1.9[/tex] represent the sample standard deviation
[tex]n=33[/tex] sample size
[tex]\mu_o =20[/tex] represent the value that we want to test
[tex]\alpha=0.05[/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)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean battery life is less than 20 :
Null hypothesis:[tex]\mu \geq 20[/tex]
Alternative hypothesis:[tex]\mu < 20[/tex]
Since we don't know the population deviation, 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".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{19.5-20}{\frac{1.9}{\sqrt{33}}}=-1.51[/tex]
4) P-value
First we need to calculate the degrees of freedom given by:
[tex]df=n-1=33-1=32[/tex]
Since is a one-side lower test the p value would be:
[tex]p_v =P(t_{(32)}<-1.51)=0.0704[/tex]
5) Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the average battery life it's not significantly different less than 20 hours at 5% of signficance. If we analyze the options given we have:
(A) Reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. FALSE, we FAIL to reject the null hypothesis.
(B) Fail to reject the null hypothesis that the true population mean battery life of the smartphone is equal to 20 hours. TRUE, we fail to reject the null hypothesis that the mean would be 20 or higher .
(C) There is not enough evidence at the α=0.05 level of significance to suggest that the true population mean battery life of the smartphone is less than 20 hours. TRUE, we FAIL to reject the null hypothesis that the mean is greater or equal to 20 hours, so we reject the alternative hypothesis that the mean is less than 20 hours.
(D) There is enough evidence at the α=0.05 level of significance to support the claim that the true population mean battery life of the smartphone is not equal to 20 hours. FALSE the claim is not that the mean is different from 20. The real claim is: "Javier believes the mean battery life is less than 20 hours".
Javier's t-test with a test statistic of −1.51 and 32 degrees of freedom has a p-value greater than 0.05, hence we fail to reject the null hypothesis, indicating not enough evidence to suggest the true mean is less than the claimed 20 hours. Thus, the correct answer is option (C).
This problem involves conducting a one-sample t-test to determine whether the true population mean battery life is different from the claimed 20 hours. The null hypothesis here is H0: [tex](mu = 20)[/tex] hours, and the alternative hypothesis is Ha: [tex](mu < 20)[/tex] hours. With 32 degrees of freedom, Javier's calculated test statistic is t ≈ −1.51. To decide whether to reject or fail to reject the null hypothesis, we must compare the p-value to the significance level, [tex]\(\alpha = 0.05\)[/tex].
If the p-value is less than [tex]\(\alpha = 0.05\)[/tex], then we reject the null hypothesis. If the p-value is higher, we fail to reject. In this exercise, the p-value associated with Javier's test statistic of t ≈ −1.51 for a one-tailed test is greater than 0.05, thus we should fail to reject the null hypothesis (Option B). This implies that there is not enough evidence at the 0.05 level of significance to support the claim that the true population mean battery life of the smartphone is less than 20 hours (Option C).
Belinda placed stepping stones in the shape of the irregular polygon shown. She will plant thyme around the edge of each stepping stone. What is the total length of planting around each stepping stone ?
Answer:
The total length can be calculated by doing thesum of all sides
Step-by-step explanation:
The total length of planting around each stepping stone is 15x - 2.
To find the total length of planting around each stepping stone, we need to add up the lengths of all the sides of the irregular polygon.
The sides of the polygon are (3x-1), (2x+1), (4x-2), (4x-4), and (2x+4).
To find the total length, we can add up the lengths of all the sides:
(3x-1) + (2x+1) + (4x-2) + (4x-4) + (2x+4)
Simplifying the expression, we get:
15x - 2
Therefore, The total length of planting around each stepping stone is 15x - 2.
The probable question may be:
Belinda placed stepping stones in the shape of the irregular polygon shown. She will plant thyme around the edge of each stepping stone. What is the total length of planting around each stepping stone ?
The sides of the irregular polygon are (3x-1), (2x+1), (4x-2), (4x-4), (2x+4)
there are some pets in a pet shop. 36% of them are rabbits, 28% of them are dogs and the rest are hamsters. If there are 160 dogs and hamsters, how many pets are there in the pet shop altogether?
Answer:
250
Step-by-step explanation:
36% are rabbits, and 28% are dogs. That means the percent that are hamsters is:
100% − 36% − 28% = 36%
There are 160 dogs and hamsters, which are 28% + 36% = 64% of the pets. Therefore, we can write and solve a proportion:
160 / 64% = x / 100%
x = 250
There are a total of 250 pets.
Answer: 250 pets
Step-by-step explanation:
Let the total number of pets in the pet shop be x
36% of them are rabbits. This means that the number of pets that are rabbit 36/100 × x = 0.36×x = 0.36x
28% of them are dogs. This means that the number of pets that are dogs are 28/100 × x = 0.28×x = 0.28x
The rest are hamsters. This means that total number of hamsters is total number of pets minus sum of rabbits and dogs.
Number of hamsters = x -(0.28x + 0.36x) = 0.36x
If there are 160 dogs and hamsters,it means that
0.28x + 0.36x = 160
0.64x =160
x = 160/0.64
x = 250
Consider a collection of envelopes consisting of 3 red envelopes, 3 blue envelopes, 1 green envelope, and 3 yellow envelopes. If three envelopes are selected at random, without replacement, determine the probability that at least one envelope is a red envelope. The probability that at least one envelope is red is
Answer: Our required probability is [tex]\dfrac{31}{35}[/tex]
Step-by-step explanation:
Since we have given that
Number of red envelopes = 3
Number of blue envelopes = 3
Number of green envelopes = 1
We need to select 3 envelopes in such a way that at least one envelope is a red.
So, it becomes,
[tex]\dfrac{^3C_1\times ^4C_2}{^7C_3}+\dfrac{^3C_2\times ^4C_1}{^7C_3}+\dfrac{^3C_3}{7C_3}\\\\=\dfrac{18}{35}+\dfrac{12}{35}+\dfrac{1}{35}\\\\=\dfrac{18+12+1}{35}\\\\=\dfrac{31}{35}[/tex]
Hence, our required probability is [tex]\dfrac{31}{35}[/tex]
Final answer:
The probability of not selecting any red envelope is determined by choosing 3 envelopes without any red ones. The probability of selecting at least one red envelope is 17/24.
Explanation:
To find the probability of selecting at least one red envelope, we need to consider the complement of selecting no red envelopes.
The total number of ways to select 3 envelopes out of the given collection is (3+3+1+3)C3 = 10C3 = 120.
The number of ways to select 3 envelopes without selecting any red envelope is (3+1+3)C3 = 7C3 = 35.
Therefore, the probability of selecting at least one red envelope is 1 - P(no red envelope) = 1 - (35/120) = 85/120 = 17/24.
Match the integrals with the type of coordinates which make them the easiest to do. Put the letter of the coordinate system to the left of the number of the integral. 1. ∫10∫y20 1x dx dy 2. ∫∫D 1x2+y2 dA where D is: x2+y2≤4 3. ∫∫∫E z2 dV where E is: −2≤z≤2, 1≤x2+y2≤2 4. ∫∫∫E dV where E is: x2+y2+z2≤4, x≥0, y≥0, z≥0 5. ∫∫∫E z dV where E is: 1≤x≤2, 3≤y≤4, 5≤z≤6
Answer:
for 1 ) Normal (rectangular) coordinates
for 2) Polar coordinates
for 3) Cylindrical coordinates
for 4) Spherical coordinates
for 5) Normal (rectangular) coordinates
Step-by-step explanation:
1. ∫10∫y20 1x dx dy 2. → Normal (rectangular) coordinates x=x , y=y → integration limits ∫ [20,1] and ∫ [10,2]
2. ∫∫D 1x2+y2 dA. , D is: x2+y2≤4 → Polar coordinates x=rcosθ , y=rsinθ → integration limits ∫ [2,0] for dr and ∫ [2π,0] for dθ
3. ∫∫∫E z2 dV , E is: −2≤z≤2, 1≤x2+y2≤2 → Cylindrical coordinates x=rcosθ , y=rsinθ , z=z → integration limits ∫ [2,-2] for dz , ∫ [√2,1] for dr and ∫ [2π,0] for dθ
4. ∫∫∫E dV where E is: x2+y2+z2≤4, x≥0, y≥0, z≥0 → Spherical coordinates x=rcosθcosФ y=rsinθcosФ , z=rsinФ → integration limits ∫ [2,0] for dr ,∫ [-π/2,π/2] for dθ , ∫ [π/2,0] for dθ
5. ∫∫∫E z dV where E is: 1≤x≤2, 3≤y≤4, 5≤z≤6 → Normal (rectangular) coordinates x=x , y=y , z=z → integration limits ∫ [2,1] for dx ,∫ [4,3] for dy and ∫ [6,5] for dz
The type of coordinates easiest to use when solving integrals depends on the nature of the integral and its bounds. In cases where circular symmetry is present, Polar, Cylindrical, or Spherical coordinates may be used instead of Cartesian Coordinates.
Explanation:The integrals listed in your question can be best solved depending on the coordinate system expressed in their limits of integration, or the area or volume they represent. Let's match them below:
Polar Coordinates: ∫∫D 1/(x²+y²) dA where D is: x²+y²≤4 Cartesian Coordinates: ∫10∫y20 1/x dx dy Cylinder Coordinates: ∫∫∫E z² dV where E is: −2≤z≤2, 1≤x²+y²≤2 Spherical Coordinates: ∫∫∫E dV where E is: x²+y²+z²≤4, x≥0, y≥0, z≥0 Cartesian Coordinates: ∫∫∫E z dV where E is: 1≤x≤2, 3≤y≤4, 5≤z≤6
Please note, the type of coordinates that are the easiest to use often depends heavily on the specific integrals and their bounds. For instance, in situations where circular symmetry is present, it's preferable to use Polar, Cylindrical or Spherical coordinates as compared to Cartesian coordinates.
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Using L' Hopital's rule, find the limit of
[tex]\lim_{x \to \frac{\pi }{2} } 3secx-3tanx[/tex]
Answer:
L=0
Step-by-step explanation:
[tex]L=\lim\limits_{x \rightarrow \frac{\pi}{2}}3secx-3tanx[/tex]
Replacing the value of x we get ∞ - ∞ which is an indetermined expression
We must transform the limit so it can be shown as a fraction and the L'Hopital's rule can be applied:
[tex]L=\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{3-3sinx}{cosx}=\frac{0}{0}[/tex]
Now we can take the derivative in both parts of the fraction
[tex]L=\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{-3cosx}{-sinx}=3\lim\limits_{x \rightarrow \frac{\pi}{2}}\frac{cosx}{sinx}=3\times 0=0[/tex]
Julian is packaging cold cuts. He prepares 9 bags that each contain 3 tenths of a pound of turkey. He calculates that the total weight of the bags combined is 0.27 pounds. What error did Julian make?
He placed the decimal point incorrectly after he multiplied.
He divided the weight of one bag by 9 instead of multiplying.
He multiplied the number of bags by 3 instead of 0.3.
He forgot to multiply and found the weight of one bag of turkey.
Answer: the error is he placed the decimal point incorrectly after he multiplied. the answer is actually 2.7, as 0.3 times 9 is 2.7. Hope this helps.
Answer:
the answer is a
Step-by-step explanation:
i just took the test
A CI is desired for the true average stray-load loss ? (watts) for a certain type of induction motor when the line current is held at 10 amps for a speed of 1500 rpm. Assume that stray-load loss is normally distributed with σ = 2.8. (Round your answers to two decimal places.)
(a)Compute a 95% CI for ? when n = 25 and x = 51.4.
(_________________, __________________) Watts
(b) Compute a 95% CI for ? when n = 100 and x = 51.4.
(________________________ , _____________________) watts
(c) Compute a 99% CI for ? when n = 100 and x = 51.4.
(___________________________, _______________________) watts
(d) Compute an 82% CI for ? when n = 100 and x = 51.4.
(_________________________, ___________________________) watts
(e) How large must n be if the width of the 99% interval for ? is to be 1.0? (Round your answer up to the nearest whole number.)
Answer:
a) (50.30 , 52.50)
b) (50.85 , 51.95)
c) (50.68 , 52.12)
d) (51.02 , 51.78)
e) 209
Step-by-step explanation:
(a) Sample Mean = 51.4
σ = 2.8
Sample Size, n = 25
Standard Error, E = [tex]\frac{\sigma}{\sqrt{n}}[/tex] = 0.56
z critical value for 95% confidence interval
z = 1.96
Margin of Error (ME) = z × E = 1.0976
95% confidence interval is given as
⇒ Mean ± ME
= 51.4 ± 1.0976
or
= (50.30 , 52.50)
b) Sample Mean = 51.4
σ = 2.8
Sample Size, n = 100
Standard Error, E = [tex]\frac{2.8}{\sqrt{100}}[/tex] = 0.28
z critical value for 95% confidence interval
z = 1.96
Margin of Error (ME) = z × E = 0.5488
95% confidence interval is given as
⇒ Mean ± ME
= 51.4 ± 0.5488
or
= (50.85 , 51.95)
c) Sample Mean = 51.4
σ = 2.8
Sample Size, n = 100
Standard Error, E = [tex]\frac{2.8}{\sqrt{100}}[/tex] = 0.28
z critical value for 99% confidence interval
z = 2.5758
Margin of Error (ME) = z × E = 0.7212
99% confidence interval is given as
⇒ Mean ± ME
= 51.4 ± 0.7212
or
= (50.68 , 52.12)
d) Sample Mean = 51.4
σ = 2.8
Sample Size, n = 100
Standard Error, E = [tex]\frac{2.8}{\sqrt{100}}[/tex] = 0.28
z critical value for 82% confidence interval
z = 1.3408
Margin of Error (ME) = z × E = 0.3754
82% confidence interval is given as
⇒ Mean ± ME
= 51.4 ± 0.3754
or
= (51.02 , 51.78)
e) Margin of Error, ME = (width of interval) ÷ 2 = 0.5
Now,
σ = 2.8
as ME = z × Standard Error,
z = 2.5758 for 99% confidence level
For ME = 0.5,
i,e
[tex]\frac{z\times\sigma}{\sqrt{n}}[/tex] = 0.5
or
[tex]\frac{2.5758 \times2.8}{\sqrt{n}}[/tex] = 0.5
or
n = [tex](\frac{2.5758 \times2.8}{0.5})^2[/tex]
or
n = 208.06
or
n ≈ 209
The question involves calculating different confidence intervals for an unknown population parameter. The computations require using the given standard deviation, sample size, and sample mean, along with appropriate Z-scores. The calculated intervals range between about 49.896 and 52.904 Watts for a 95% CI with a sample size of 25, and as narrow as between 50.626 and 52.174 Watts for an 82% CI with a sample size of 100. Using a desire for a 99% CI width of 1.0, a necessary sample size of about 43 is computed.
Explanation:In statistics, confidence intervals (CI) provide an estimated range of values which is likely to include an unknown population parameter. Given the parameters of standard deviation (σ), sample size (n), and the sample mean (x), we can compute the confidence intervals. This involves finding the standard error of the mean (σ/√n), and using Z-scores depending on the percentage of the confidence interval. For 95%, 99%, and 82% CIs, the Z-scores are approximately 1.96, 2.58, and 1.34 respectively.
(a) 95% CI: 51.4 ± 1.96 * (2.8/√25) = (49.896, 52.904) Watts(b) 95% CI: 51.4 ± 1.96 * (2.8/√100) = (50.448, 52.352) Watts(c) 99% CI: 51.4 ± 2.58 * (2.8/√100) = (50.278, 52.522) Watts(d) 82% CI: 51.4 ± 1.34 * (2.8/√100) = (50.626, 52.174) WattsFor part (e), we want the width of the 99% interval to be 1.0. This involves setting the equation for the interval to 1.0 and solving for n. This results in n being approximately 42.64, but since we can't have a fractional part of an individual, we round up to 43.
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Suppose x has a distribution with a mean of 80 and a standard deviation of 3. Random samples of size n = 36 are drawn. (a) Describe the x distribution and compute the mean and standard deviation of the distribution. x has distribution with mean μx = and standard deviation σx = . (b) Find the z value corresponding to x = 81. z = (c) Find P(x < 81). (Round your answer to four decimal places.) P(x < 81) = (d) Would it be unusual for a random sample of size 36 from the x distribution to have a sample mean less than 81? Explain. Yes, it would be unusual because less than 5% of all such samples have means less than 81. No, it would not be unusual because more than 5% of all such samples have means less than 81. No, it would not be unusual because less than 5% of all such samples have means less than 81. Yes, it would be unusual because more than 5% of all such samples have means less than 81.
Final answer:
The sampling distribution of the sample mean has the same mean as the original distribution (80) and a calculated standard error of 0.5. The z-value for x = 81 is 2, corresponding to a probability of P(x < 81) = 0.9772. It would not be unusual for a sample mean to be less than 81 based on this z-score.
Explanation:
If we have a variable x that follows a normal distribution with a known mean (μ) and standard deviation (σ), and we draw random samples of size n from it, we can describe the distribution of the sample means. The distribution of sample means will also be normally distributed, known as the sampling distribution of the sample mean, thanks to the Central Limit Theorem. For a sufficiently large sample size, this holds true regardless of the shape of the original distribution.
(a) Description of x Distribution and Calculation of Mean and Standard Deviation
The variable x has a distribution with a mean (μx) of 80 and a standard deviation (σx) of 3. When taking a sample size n = 36, by the Central Limit Theorem, the mean of the sampling distribution will remain the same (μx = 80), but the standard deviation will be the original standard deviation divided by the square root of the sample size (n), known as the standard error (SE). Hence, the standard error will be σx/√n = 3/√36 = 3/6 = 0.5.
(b) Finding the z-value for x = 81
To find the z-value for x = 81, we use the formula:
z = (x - μx) / SE
So, z = (81 - 80) / 0.5 = 1/0.5 = 2.
(c) Computing P(x < 81) - Probability Calculation
To find the probability P(x < 81), we would look up the z-value we just calculated in a standard normal distribution table. Let's assume it corresponds to a probability of 0.9772. Thus, P(x < 81) = 0.9772.
(d) Unusualness of a Sample Mean Less Than 81
To determine if it would be unusual for a random sample of size 36 from the x distribution to have a sample mean less than 81, we consider the z-value and the empirical rule. Since our z-score of 2 corresponds to a percentage greater than 5% of the tail (assuming the previously stated probability is correct), it is not unusual for a sample mean to be less than 81 because more than 5% of samples would have means less than this.
Kristin wants to spend no more than $50 for a birthday dinner at a restaurant, including a 15% tip. What is the most that the meal can cost before tip
Answer:
The most that a meal can cost before tip = $43.47
Step-by-step explanation:
Kristin's maximum budget for a birthday dinner = $50 inclusive of 15% tip.
Let the cost of meal that Kristin orders in dollars be [tex]=x[/tex]
15% of the cost of meal is tip which would be in dollars = 15% of [tex]x=0.15\ x[/tex]
So total cost of dinner would be [tex]=x+0.15x[/tex]
We know that the total should be no more that $50. So, we have
[tex]x+0.15x\leq50[/tex]
⇒ [tex]1.15x\leq50[/tex]
Dividing both sides by 1.15.
⇒ [tex]\frac{1.15x}{1.15}\leq\frac{50}{1.15}[/tex]
∴ [tex]x\leq43.47[/tex]
So, the most that a meal can cost before tip = $43.47
It is believed that the average amount of money spent per U.S. household per week on food is about $99, with standard deviation $8. A random sample of 100 households in a certain affluent community yields a mean weekly food budget of $100. We want to test the hypothesis that the mean weekly food budget for all households in this community is higher than the national average. State the null and alternative hypotheses for this test, the test statistic and determine if the results significant at the 5% level.
Answer:
Step-by-step explanation:
The null hypothesis is that the mean weekly food budget for all households in the community is equal to the national average, while the alternative hypothesis is that it is higher. The test statistic is calculated using the sample mean and the standard deviation of the population. To determine if the results are significant, we compare the test statistic to the critical value.
Explanation:The null hypothesis for this test is that the mean weekly food budget for all households in the community is equal to the national average, which is $99. The alternative hypothesis is that the mean weekly food budget for all households in the community is higher than the national average.
The test statistic in this case is calculated using the sample mean and the standard deviation of the population. It is equal to (sample mean - population mean) / (standard deviation / sqrt(sample size)).
To determine if the results are significant at the 5% level, we compare the test statistic to the critical value for a one-tailed test with a significance level of 0.05. If the test statistic is greater than the critical value, we reject the null hypothesis and conclude that the mean weekly food budget for all households in the community is higher than the national average.
A random sample of 100 people was taken. Eighty of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 75%. At a .05 level of significance, it can be concluded that the proportion of the population in favor of candidate A is _____.
a. significantly greater than 75%b.not significantly greater than 75%c. significantly greater than 80%d.not significantly greater than 80%
Answer:
Option b - not significantly greater than 75%.
Step-by-step explanation:
A random sample of 100 people was taken i.e. n=100
Eighty of the people in the sample favored Candidate i.e. x=80
We have used single sample proportion test,
[tex]p=\frac{x}{n}[/tex]
[tex]p=\frac{80}{100}[/tex]
[tex]p=0.8[/tex]
Now we define hypothesis,
Null hypothesis [tex]H_0[/tex] : candidate A is significantly greater than 75%.
Alternative hypothesis [tex]H_1[/tex] : candidate A is not significantly greater than 75%.
Level of significance [tex]\alpha=0.05[/tex]
Applying test statistic Z -proportion,
[tex]Z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}[/tex]
Where, [tex]\widehat{p}=80\%=0.80[/tex] and [tex]p=75%=0.75[/tex]
Substitute the values,
[tex]Z=\frac{0.80-0.75}{\sqrt{\frac{0.75(1-0.75)}{100}}}[/tex]
[tex]Z=\frac{0.80-0.75}{\sqrt{\frac{0.1875}{100}}}[/tex]
[tex]Z=\frac{0.05}{0.0433}[/tex]
[tex]Z=1.1547[/tex]
The p-value is
[tex]P(Z>1.1547)=1-P(Z<1.1547)[/tex]
[tex]P(Z>1.1547)=1-0.8789[/tex]
[tex]P(Z>1.1547)=0.1241[/tex]
Now, the p-value is greater than the 0.05.
So we fail to reject the null hypothesis and conclude that the A is not significantly greater than 75%.
Therefore, Option b is correct.
To answer if the proportion of the population in favor of Candidate A is significantly more than 75% at a .05 level of significance, we'd need to perform a statistical test. If the p-value from this test is less than .05, we can say the proportion is significantly more than 75%. However, we haven't been given a statistical result so we can't definitively select between options a. and b.
Explanation:The problem is about determining the significance of a proportion in a population. In this case, the proportion represents the people who favor Candidate A. The question is whether this proportion is significantly more than 75% at a .05 level of significance. Eighty people out of the sample of hundred favor Candidate A, which is 80% of the population sample.
The next step is to set up the null hypothesis (H0) and the alternative hypothesis (Ha). The null hypothesis claims that the proportion of people in favor of Candidate A is 75%. The alternative hypothesis states that the proportion of people in favor of Candidate A is significantly more than 75%.
Next, we test the hypothesis using statistical analysis. As we are using a .05 level of significance, if the p-value is less than .05, we reject the null hypothesis in favor of the alternative hypothesis.
In this case, since our percentage in the sample (80%) is greater than the claim that we are testing against (75%), we can suggest that the proportion of the population in favor of Candidate A is significantly greater than 75% if our p-value is less than .05. However, without performing the statistical test or being given the resultant p-value we cannot choose between options a. and b.
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How do I do number 12
Answer:
d = √(k/R)
Step-by-step explanation:
Eliminate fractions by multiplying by the denominator, then divide by the coefficient of d². Finally, take the square root.
[tex]R=\dfrac{k}{d^2}\\\\Rd^2=k \quad\text{multiply by $d^2$}\\\\d^2=\dfrac{k}{R} \quad\text{divide by $R$}\\\\d=\sqrt{\dfrac{k}{R}} \quad\text{take the square root}[/tex]
Say a business found that 29.5% of customers in Washington prefer grey suits. The company chooses 8 customers in Washington and asks them if they prefer grey suits. What assumption must be made for this study to follow the probabilities of a binomial experiment?
Answer:
The assumption that must be made for this study to follow the probabilities of a binomial experiment is that there must be only two outcomes of each trail in this study (meaning that it is either they prefer grey suits or they do not prefer grey suits). There must be no other option apart from those two options and each of the independent trails must be mutually exclusive, meaning that the two required options cannot occur together. It is either the first option (prefer grey suits) or the second option (do not prefer grey suits).
Step-by-step explanation:
Consider a simple example of moral hazard. Suppose that Woodrow goes into a casino to make one bet a day. The casino is very basic; it has two bets: a safe bet and a risky bet. In the safe bet, a nickel is flipped. If the nickel lands on heads, Woodrow wins $ 100 . If it lands on tails, Woodrow loses $ 100 . The risky bet is similar: a silver dollar is flipped. If the silver dollar lands on heads, Woodrow wins $ 5,000 . If it lands on tails, Woodrow loses $ 10,000 . Each coin has a 50 % chance of landing on each side. What is the expected value of the safe bet?
Answer:
The expected value of the safe bet equal $0
Step-by-step explanation:
If
[tex]S=\left\{s_1,s_2,...,s_n\right\}[/tex]
is a finite numeric sample space and
[tex]P(X=s_k)=p_k[/tex] for k=1, 2,..., n
is its probability distribution, then the expected value of the distribution is defined as
[tex]E(X)=s_1P(X=s_1)+s_2P(X=s_2)+...+s_nP(X=s_n)X) [/tex]
What is the expected value of the safe bet?
In the safe bet we have only two possible outcomes: head or tail. Woodrow wins $100 with head and “wins” $-100 with tail So the sample space of incomes in one bet is
S = {100,-100}
Since the coin is supposed to be fair,
P(X=100)=0.5
P(X=-100)=0.5
and the expected value is
E(X) = 100*0.5 - 100*0.5 = 0
The expected value of the safe bet is $0, which means that you would neither gain nor lose money on average if you played this game repeatedly.
Explanation:The expected value of the safe bet can be calculated by multiplying the possible outcomes by their respective probabilities and summing them up. In this case, the safe bet has two possible outcomes: winning $100 with a 50% probability and losing $100 with a 50% probability. So, the expected value can be calculated as follows:
Expected value = (100 * 0.5) + (-100 * 0.5) = $0
Therefore, the expected value of the safe bet is $0, which means that you would neither gain nor lose money on average if you played this game repeatedly.
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Greenville County, South Carolina, has 461,299 adult residents, of which 59,969 are 65 years or older. A survey wants to contact residents.(a) Find the proportion of Greenville county adult residents who are 65 years or older. (Enter your answer rounded to two decimal places.)
The proportion of Greenville County adult residents who are 65 years or older is approximately 13%.
Explanation:To find the proportion of Greenville County adult residents who are 65 years or older, we need to divide the number of residents who are 65 years or older by the total number of adult residents in the county.
The proportion can be found using the formula: Proportion = Number of residents who are 65 years or older / Total number of adult residents.
Substituting the given values, we have:
Proportion = 59,969 / 461,299 = 0.1299
Therefore, the proportion of Greenville County adult residents who are 65 years or older is approximately 0.13, or 13%.
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A consumer protection group is concerned that a ketchup manufacturer is filling its 20-ounce family-size containers with less than 20 ounces of ketchup. The group purchases 10 family-size bottles of this ketchup, weighs the contents of each, and finds that the mean weight is equal to 19.86 ounces, and the standard deviation is equal to 0.22 ounce. Do the data provide sufficient evidence for the consumer group to conclude that the mean fill per family-size bottle is less than 20 ounces? Use α = 0.01
Answer:
We accept H₀, we dont have evidence to say that the family size container
of ketchup has smaller quantity
Step-by-step explanation:
Population mean μ₀ = 20 ounces
sample size n = 10 df = n -1 df = 10-1 df= 9
n < 30 use of t-student distribution
sample mean μ = 19.86
sample standard deviation s = 0,22
One tail-test ( left tail)
1.-Test Hypothesis
H₀ null hypothesis μ₀ = 20
Hₐ Alternative hypothesis μ₀ < 20
2.- α = 0,01 and one test tail
3.- Compute
t(s) = [ ( μ - μ₀ ) ] / s/√n t(s) = [( 19.86 - 20 )* √10 ] / 0.22
t(s) = - ( 0,14 * 3.16) / 0,22
t(s) = - 2.01
4.- We go to t-student table t(c) for df = 9 and 0,01 = α
and find t(c) = - 2.821
5.-Compare t (s) and t (c)
t (c) < t (s) - 2.821 < - 2.01
6. t(s) is inside de acceptance region we accept H₀
According to a past survey, 23% of Americans have hypertension. After a stringent regimen of diet and exercise, 75 people were then tested and 18 were found to have hypertension. Based on this sample, does diet and exercise reduce hypertension? Use a significance level of 0.05.a) yesb) no
Answer:No, Diet and exercise does not reduce hypertension.
Step-by-step explanation:
Since we have given that
p = 0.23
n = 75
x = 18
So, [tex]\hat{p}=\dfrac{x}{n}=\dfrac{18}{75}=0.24[/tex]
So, hypothesis would be
[tex]H_0:p=\hat{p}\\\\H_a:\hat{p}<p[/tex]
So, the test statistic value would be
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\\\\z=\dfrac{0.24-0.23}{\sqrt{\dfrac{0.23\times 0.77}{75}}}\\\\z=\dfrac{0.01}{0.049}\\\\z=0.204[/tex]
At α = 0.05 level of significance, we get
critical value = 1.96
and 1.96>0.204.
so, we will accept the null hypothesis.
Hence, No, Diet and exercise does not reduce hypertension.
Evaluate the line integral, where C is the given curve sin(x)dx + cos(y)dy C consists of the top half of the circle x2 + y2 = 25 from (5, 0) to (-5, 0) and the line segment from (-5, 0) to (-6, 3).
Answer:
-cos(6)+sin(3)+cos(5)
=-0.53538809312 (using calculator)
Step-by-step explanation:
if F is a vector field , if it is defined on a domain D and F=[tex]\nabla f[/tex] for any scalar function f on D then f is known as potential function of Fhere, F = sin(x)dx + cos(y)dy
then f = -cos(x) + sin(y)
( because, [tex]\nabla f[/tex] should be F. by applying [tex]\nabla[/tex] operator on f we must obtain F. so to satisfy this condition f must be -cosx + siny.
where, [tex]\nabla g(x,y) =[/tex] partial derive of g(x,y) with respect to x + partial derive of g(x,y) with respect to y )
therefore, line integral of F over the given curve C = f(-6,3)-f(-5,0)={-cos(-6)}+sin(3) -[{-cos(-5)}+sin(0)]
=-cos(6)+sin(3)-[-cos(5)] [since, sin(0)=0, cos(-a)=cos(a) where, a>0]
= -cos(6)+sin(3)+cos(5)
=-0.53538809312 (using calculator)
The evaluation of the line integral is: -0.53538809312 (using a calculator) or -cos(6)+sin(3)+cos(5)
Calculations and Parameters:If F is a vector field and if it is defined on a domain D and F
= ∇f for any scalar function f on D
Then f is known as the potential function of F
where, F = sin(x)dx + cos(y)dy
then f = -cos(x) + sin(y)
( because ∇f should be F. by applying ∇ operator on f we must obtain F. so to satisfy this condition f must be -cosx + siny.
where, ∇g(x,y) partial derive of g(x,y) with respect to x + partial derive of g(x,y) with respect to y )
Therefore, the line integral of F over the given curve C
= f(-6,3)-f(-5,0)={-cos(-6)}+sin(3) -[{-cos(-5)}+sin(0)]=-cos(6)+sin(3)-[-cos(5)] [since, sin(0)=0, cos(-a)=cos(a) where, a>0]= -cos(6)+sin(3)+cos(5)=-0.53538809312 (using calculator)Read more about line segments here:
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A journal article reports that a sample of size 5 was used as a basis for calculating a 95% CI for the true average natural frequency (Hz) of delaminated beams of a certain type. The resulting interval was (230.061, 233.807). You decide that a confidence level of 99% is more appropriate than the 95% level used. What are the limits of the 99% interval?
the limits of the 99% confidence interval are approximately [tex]\( (230.068, 233.799) \).[/tex]
To find the limits of the 99% confidence interval (CI) for the true average natural frequency (Hz), we'll use the formula for confidence intervals for the mean:
[tex]\[ \text{CI} = \bar{x} \pm z \frac{s}{\sqrt{n}} \][/tex]
Where:
- [tex]\( \bar{x} \)[/tex] is the sample mean,
- [tex]\( z \)[/tex] is the z-score corresponding to the desired confidence level,
- [tex]\( s \)[/tex] is the sample standard deviation,
- [tex]\( n \)[/tex] is the sample size.
Given:
- Sample size (n ) = 5,
- 95% CI: Lower limit = 230.061, Upper limit = 233.807.
1. Calculate the sample mean [tex](\( \bar{x} \)):[/tex]
[tex]\[ \bar{x} = \frac{\text{Lower limit} + \text{Upper limit}}{2} \][/tex]
[tex]\[ \bar{x} = \frac{230.061 + 233.807}{2} \][/tex]
[tex]\[ \bar{x} = \frac{463.868}{2} \][/tex]
[tex]\[ \bar{x} = 231.934 \][/tex]
2. Calculate the sample standard deviation [tex](\( s \)):[/tex]
To calculate the sample standard deviation, we need to know the z-score corresponding to the desired confidence level.
3. Find the z-score for a 99% confidence level:
At 99% confidence level, the critical z-value is approximately 2.576 (you can find this value using a standard normal distribution table or a calculator).
4. Substitute the values into the confidence interval formula:
[tex]\[ 231.934 \pm 2.576 \frac{s}{\sqrt{5}} \][/tex]
We know that the width of the interval at 95% confidence level is[tex]\( 233.807 - 230.061 = 3.746 \).[/tex]
So, we can set up an equation to find \( s \):
[tex]\[ 2.576 \frac{s}{\sqrt{5}} = \frac{3.746}{2} \][/tex]
[tex]\[ \frac{s}{\sqrt{5}} = \frac{3.746}{2.576 \times 2} \][/tex]
[tex]\[ \frac{s}{\sqrt{5}} \approx \frac{3.746}{5.152} \][/tex]
[tex]\[ s \approx \frac{3.746 \times \sqrt{5}}{5.152} \][/tex]
[tex]\[ s \approx \frac{3.746 \times 2.236}{5.152} \][/tex]
[tex]\[ s \approx \frac{8.3753}{5.152} \][/tex]
[tex]\[ s \approx 1.6247 \][/tex]
Now, we can substitute [tex]\( s = 1.6247 \)[/tex] into the CI formula to find the limits of the 99% interval:
[tex]\[ 231.934 \pm 2.576 \frac{1.6247}{\sqrt{5}} \][/tex]
[tex]\[ 231.934 \pm 2.576 \times \frac{1.6247}{\sqrt{5}} \][/tex]
[tex]\[ 231.934 \pm 2.576 \times 0.7241 \][/tex]
[tex]\[ 231.934 \pm 1.8658 \][/tex]
Therefore, the limits of the 99% confidence interval are approximately [tex]\( (230.068, 233.799) \).[/tex]
Select all expressions that equal -6-(-2)
Answer:
-6-(-2) is equivalent to
-6 +2
and 2-6
a) and b) are correct options
Answer:
a and b.
Step-by-step explanation:
-6-(-2) = -6 + 2 = -4.
2 - 6 = -4.
Mr. Taylor's 4th grade class uses Skittles to learn about probability. They open several randomly selected bags of Skittles and sort and count the different colors and want to determine if Skittles are evenly distributed by color.
Col1 Color Red Orange Yellow Purple Green
Col2 Count 107 101 87 115 10
A) Choose the appropriate null and alternate hypotheses.B)What is the p-value?C)What is your conclusion?
Answer:
The skitties are not evenly distributed by colour
Step-by-step explanation:
Given that Mr. T aylor's 4th grade class uses Skittles to learn about probability. They open several randomly selected bags of Skittles and sort and count the different colors and want to determine if Skittles are evenly distributed by color.
[tex]H_0: Skitties are equally distributed\\H_a: atleast two are not equally distributed[/tex]
(Two tailed chi square test)
If all are equally distributed then expected values would be equal to 420/5 =104
Observed Red Orange Yellow Purple Green Total
107 101 87 115 10 420
Expected 104 104 104 104 104 420
Chi square 0.0865 0.0865 2.7788 1.16345 84.9615 89.0769
Chi square is calculated as (obs-exp)^2/exp
Total chi square = 89.0769
df = 4
p value = <0.00001
Reject null hypothesis
The skitties are not evenly distributed by colour
The Ishares Bond Index fund (TLT) has a mean and annual standard deviation of returns of 5%
and 10%, respectively. What is the 66% confidence interval for the returns on TLT?
A) -7%, 10%
B) 5%, 10%
C) -5%, 15%
D) -10%, 10%
Answer:
C) (-5%.15%)
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".
[tex]\bar X =10\%=0.1 [/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma[/tex]=5% =0.05 represent the population standard deviation
2) Confidence interval
We assume that the random variable X who represent The Ishares Bond Index fund (TLT) follows this distribution:
[tex]X \sim N(\mu, \sigma=10\%=0.1)[/tex]
The confidence interval for the returns on TLT is given by the following formula:
[tex]\bar X \pm z_{\alpha/2}\sigma[/tex] (1)
In order to calculate the critical value [tex]z_{\alpha/2}[/tex]. Since the Confidence is 0.66 or 66%, the value of [tex]\alpha=0.34[/tex] and [tex]\alpha/2 =0.17[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-NORM.INV(0.17,0,1)".And we see that [tex]z_{\alpha/2}=0.95[/tex]
Now we have everything in order to replace into formula (1):
[tex]0.05-0.95(0.1)=-0.05[/tex]
[tex]0.05+0.95(0.1)=0.15[/tex]
So on this case the 66% confidence interval would be given by (-0.05;0.15) and we can convert this into % and wr got (-5%; 15%).