According to the College Board’s website, the average tuition cost per year for a public four-year college (in-state cost) is $10,230. Assume that you disagree with the claim. Do you think the average tuition cost is higher? Or lower? Average tuition cost is higher than $10230. Using your answer in part A, write a set of hypotheses (null hypothesis and alternative hypothesis) to represent that claim and its complement. Replace the ?s below with the appropriate inequality signs (). Highlight the hypothesis that represents your claim. H0:μ=10230 H1:μ>10230 Gather data for tuition cost of 20 different universities. Assume the data you gather is normally distributed. 3565 7290 7409 11068 10299 8315 3950 9672 14924 10958 7396 6946 6834 11108 14812 11068 8273 10904 3120 11149 Calculate the mean and standard deviation of your data set. You may use technology such as Google Sheets, Excel, or a TI-83/TI-84. Round to the nearest hundredth. Mean 8953 Standard Deviation 3273.46 Using a significance level , find the critical value using the following steps. What are the degrees of freedom? (1)Here the sample size is 20 , we know degrees of freedom = n - 1 = 20 -1 =19 Will this be a one-tailed or two tailed test? A right-tailed test (one-tailed), for which a t-test for one mean will be used. Using table 5 from the back of your textbook (page A18 in Appendix B), what is the critical value? Information provided, the significance level is α=0.05\alpha = 0.05, and the critical value for a right-tailed test is tc=1.729 Add the critical values to the graph below and shade the rejection region. (Special Instruction: For this part, it is recommended that you do this by hand. You can print out this page and use the image below as a template to write on or you can create a handwritten normal curve. Take a photo of your graph and replace the image below with it. If you need any technical assistance with this process. Since it is observed that t=−1.745≤tc=1.729t = -1.745 \le t_c = 1.729, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.9514p = 0.9514, and since p=0.9514≥0.05p = 0.9514 \ge 0.05, it is concluded that the null hypothesis is not rejected. Calculate the test statistic. Replace the ?s in the formula with the appropriate values. Round to the nearest hundredth. Add the test statistic to your graph from part F, and insert the image below. What is the P-value? View the video Hypothesis Testing for the Mean (Sigma Unknown) Part II to learn how to use Google Sheets to calculate the P-value of a t-test (transcript for Hypothesis Testing for the Mean (Sigma Unknown) Part II video). Make a decision to reject or fail to reject the null hypothesis using either the test statistic or the P-value. Note that the same conclusion will be reached using either method. Interpret the decision in the context of the original claim. If you lower the level of significance to , does your decision change? Explain your reasoning. Describe the type I and type II errors that could occur in our test by completing each of the following statements. A ____________ (type I, type II) error will occur when the actual mean of college tuition cost is ______________ (at most, at least) $10,230 but you reject the null hypothesis, . A ____________ (type I, type II) error will occur when the actual mean of college tuition cost is ______________ (less than, greater than) $10,230 but you fail to reject the null hypothesis, .

According to the College Board’s website, the average tuition cost per year for a public four-year college (in-state cost) is $10,230. Assume that you disagree with the claim. Do you think the average tuition cost is higher? Or lower? Average tuition cost is higher than $10230. Using your answer in part A, write a set of hypotheses (null hypothesis and alternative hypothesis) to represent that claim and its complement. Replace the ?s below with the appropriate inequality signs (). Highlight the hypothesis that represents your claim. H0:μ=10230 H1:μ>10230 Gather data for tuition cost of 20 different universities. Assume the data you gather is normally distributed. 3565 7290 7409 11068 10299 8315 3950 9672 14924 10958 7396 6946 6834 11108 14812 11068 8273 10904 3120 11149 Calculate the mean and standard deviation of your data set. You may use technology such as Google Sheets, Excel, or a TI-83/TI-84. Round to the nearest hundredth. Mean 8953 Standard Deviation 3273.46 Using a significance level , find the critical value using the following steps. What are the degrees of freedom? (1)Here the sample size is 20 , we know degrees of freedom = n - 1 = 20 -1 =19 Will this be a one-tailed or two tailed test? A right-tailed test (one-tailed), for which a t-test for one mean will be used. Using table 5 from the back of your textbook (page A18 in Appendix B), what is the critical value? Information provided, the significance level is α=0.05\alpha = 0.05, and the critical value for a right-tailed test is tc=1.729 Add the critical values to the graph below and shade the rejection region. (Special Instruction: For this part, it is recommended that you do this by hand. You can print out this page and use the image below as a template to write on or you can create a handwritten normal curve. Take a photo of your graph and replace the image below with it. If you need any technical assistance with this process. Since it is observed that t=−1.745≤tc=1.729t = -1.745 \le t_c = 1.729, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.9514p = 0.9514, and since p=0.9514≥0.05p = 0.9514 \ge 0.05, it is concluded that the null hypothesis is not rejected. Calculate the test statistic. Replace the ?s in the formula with the appropriate values. Round to the nearest hundredth. Add the test statistic to your graph from part F, and insert the image below. What is the P-value? View the video Hypothesis Testing for the Mean (Sigma Unknown) Part II to learn how to use Google Sheets to calculate the P-value of a t-test (transcript for Hypothesis Testing for the Mean (Sigma Unknown) Part II video). Make a decision to reject or fail to reject the null hypothesis using either the test statistic or the P-value. Note that the same conclusion will be reached using either method. Interpret the decision in the context of the original claim. If you lower the level of significance to , does your decision change? Explain your reasoning. Describe the type I and type II errors that could occur in our test by completing each of the following statements. A (type I, type II) error will occur when the actual mean of college tuition cost is (at most, at least) $10,230 but you reject the null hypothesis, . A (type I, type II) error will occur when the actual mean of college tuition cost is ____ (less than, greater than) $10,230 but you fail to reject the null hypothesis, .

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Description: According to the College Board’s website, the average tuition cost per year for a public four-year college (in-state cost) is $10,230. Assume that you disagree with the claim. Do you think the average tuition cost is higher? Or lower? Average tuition

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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According to the College Board’s website, the average tuition cost per year for a public four-year college (in-state cost) is $10,230.

Assume that you disagree with the claim. Do you think the average tuition cost is higher? Or lower?

Average tuition cost is higher than $10230.

Using your answer in part A, write a set of hypotheses (null hypothesis and alternative hypothesis) to represent that claim and its complement. Replace the ?s below with the appropriate inequality signs (). Highlight the hypothesis that represents your claim.

H0:μ=10230 H1:μ>10230

Gather data for tuition cost of 20 different universities. Assume the data you gather is normally distributed.

3565	7290	7409	11068
10299	8315	3950	9672
14924	10958	7396	6946
6834	11108	14812	11068
8273	10904	3120	11149

Calculate the mean and standard deviation of your data set. You may use technology such as Google Sheets, Excel, or a TI-83/TI-84. Round to the nearest hundredth.

Mean 8953 Standard Deviation 3273.46

Using a significance level , find the critical value using the following steps.

What are the degrees of freedom?

(1)Here the sample size is 20 , we know degrees of freedom = n - 1 = 20 -1 =19

Will this be a one-tailed or two tailed test?

A right-tailed test (one-tailed), for which a t-test for one mean will be used.

Using table 5 from the back of your textbook (page A18 in Appendix B), what is the critical value?

Information provided, the significance level is α=0.05\alpha = 0.05, and the critical value for a right-tailed test is tc=1.729

Add the critical values to the graph below and shade the rejection region.

(Special Instruction: For this part, it is recommended that you do this by hand. You can print out this page and use the image below as a template to write on or you can create a handwritten normal curve. Take a photo of your graph and replace the image below with it. If you need any technical assistance with this process.

Since it is observed that t=−1.745≤tc=1.729t = -1.745 \le t_c = 1.729, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.9514p = 0.9514, and since p=0.9514≥0.05p = 0.9514 \ge 0.05, it is concluded that the null hypothesis is not rejected.

Calculate the test statistic. Replace the ?s in the formula with the appropriate values. Round to the nearest hundredth.

Add the test statistic to your graph from part F, and insert the image below.

What is the P-value? View the video Hypothesis Testing for the Mean (Sigma Unknown) Part II to learn how to use Google Sheets to calculate the P-value of a t-test (transcript for Hypothesis Testing for the Mean (Sigma Unknown) Part II video).

Make a decision to reject or fail to reject the null hypothesis using either the test statistic or the P-value. Note that the same conclusion will be reached using either method.

Interpret the decision in the context of the original claim.

If you lower the level of significance to , does your decision change? Explain your reasoning.

Describe the type I and type II errors that could occur in our test by completing each of the following statements.

A ____________ (type I, type II) error will occur when the actual mean of college tuition cost is ______________ (at most, at least) $10,230 but you reject the null hypothesis, .

A ____________ (type I, type II) error will occur when the actual mean of college tuition cost is ______________ (less than, greater than) $10,230 but you fail to reject the null hypothesis, .

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Expert Solution

Step 1

Given,

Population mean = $μ$ = 10230

Sample size = n = 20

Level of significance = $α$ = 0.05

The null and alternative hypothesis are,

$H_{0} : μ = 10230 H_{1} : μ > 10230$

Here we use one-sample t-test.

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