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MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Mean - 8953 Standard Deviation 3273.46 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. A) Assume that you disagree with the claim. Do you think the average tuition cost is higher? Or lower? E) Using a significance level a = 0.05, find the critical value using the following steps. 1. What are the degrees of freedom? Average tuition cost is higher than $10230. B) Using your answer in part A, write a set of hypotheses (null hypothesis and alternative othesis) to represent that claim and its complement. Replace the ?s below with the appropriate inequality signs (<,<>,2). Highlight the hypothesis that represents your claim. d.f.= n- 1 (1)Here the sample size is 20 , we know degrees of freedom = n - 1 = 20 -1 =19 Ho:u? $10,230 Ha:p=10230 Hi:p>10230 2. Will this be a one-tailed or two tailed test? Ha:u?$10,230 A right-tailed test (one-tailed), for which a t-test for one mean will be used. C) Gather data for tuition cost of 20 different universities. Assume the data you gather is normally distributed. 3. Using table 5 from the back of your textbook (page A18 in Appendix B), what is the critical value? How to Create your data set: Informåtion provided, the significance level is a=0.05\alpha = 0.05, and the critical value for a right-tailed test is te=1.729 Follow the directions to fill in the table below with 20 college's tuition costs. • Navigate to https://www.collegetuitioncompare.com/compare/tables/ in the drop down menus, ONLY select School Type as "Public School" and School Level as 4 year or higher." Leave the rest of the drop down menus as is. Then select "update." Scroll down to the bottom of the list and for "Number of schools to show," select "All." The list given will be in alphabetical order. You will be selecting 20 college tuition costs from this list. Do NOT just use the first 20 entries or even 20 consecutive entries. You will need to randomly select 20 entries from the entire list. Try to make sure you have a good range of tuition costs as well. • Write the tuition cost from the first column, "Tuition & Fees/In-State" for that college in the table below. F) 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--1.745St=1.729t=-1.745 le t ce=1.729, it is then concluded that the null hypothesis is not rejected. 3565 7290 7409 11068 Using the P-value approach: The p-value is p-0.9514p = 0.9514, and since p=0.951420.05p= 0.9514 \ge 0.05, it is concluded that the null hypothesis is not rejected. %3D 10299 8315 3950 9672 %3D 14924 10958 7396 6946 6834 11108 14812 11068 8273 10904 3120 11149 D) Calculate the mean and standard deviation of your data set. You may use technology such as Google Sheets, Excel, or a T-83/TI-84. Round to the nearest hundredth.
Transcribed Image Text:Mean - 8953 Standard Deviation 3273.46 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. A) Assume that you disagree with the claim. Do you think the average tuition cost is higher? Or lower? E) Using a significance level a = 0.05, find the critical value using the following steps. 1. What are the degrees of freedom? Average tuition cost is higher than $10230. B) Using your answer in part A, write a set of hypotheses (null hypothesis and alternative othesis) to represent that claim and its complement. Replace the ?s below with the appropriate inequality signs (<,<>,2). Highlight the hypothesis that represents your claim. d.f.= n- 1 (1)Here the sample size is 20 , we know degrees of freedom = n - 1 = 20 -1 =19 Ho:u? $10,230 Ha:p=10230 Hi:p>10230 2. Will this be a one-tailed or two tailed test? Ha:u?$10,230 A right-tailed test (one-tailed), for which a t-test for one mean will be used. C) Gather data for tuition cost of 20 different universities. Assume the data you gather is normally distributed. 3. Using table 5 from the back of your textbook (page A18 in Appendix B), what is the critical value? How to Create your data set: Informåtion provided, the significance level is a=0.05\alpha = 0.05, and the critical value for a right-tailed test is te=1.729 Follow the directions to fill in the table below with 20 college's tuition costs. • Navigate to https://www.collegetuitioncompare.com/compare/tables/ in the drop down menus, ONLY select School Type as "Public School" and School Level as 4 year or higher." Leave the rest of the drop down menus as is. Then select "update." Scroll down to the bottom of the list and for "Number of schools to show," select "All." The list given will be in alphabetical order. You will be selecting 20 college tuition costs from this list. Do NOT just use the first 20 entries or even 20 consecutive entries. You will need to randomly select 20 entries from the entire list. Try to make sure you have a good range of tuition costs as well. • Write the tuition cost from the first column, "Tuition & Fees/In-State" for that college in the table below. F) 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--1.745St=1.729t=-1.745 le t ce=1.729, it is then concluded that the null hypothesis is not rejected. 3565 7290 7409 11068 Using the P-value approach: The p-value is p-0.9514p = 0.9514, and since p=0.951420.05p= 0.9514 \ge 0.05, it is concluded that the null hypothesis is not rejected. %3D 10299 8315 3950 9672 %3D 14924 10958 7396 6946 6834 11108 14812 11068 8273 10904 3120 11149 D) Calculate the mean and standard deviation of your data set. You may use technology such as Google Sheets, Excel, or a T-83/TI-84. Round to the nearest hundredth.
T-Test Results: tstats =-1.745, p-value 0.9514 I) What is the P-value? View the video Hypothesis Testing for the Mean (Siqma 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 (Siqma Unknown) Part II video). 40 30 p- value = 0.9514 J) 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. 20 Don't reject Ho 0.20 0.05 K) Interpret the decision in the context of the original claim. 0.00 40 35 3.0 25 20 15 10 -0.5 0.0 0.5 10 15 20 25 3.0 3.5 40 G) Calculate the test statistic. Replace the ?s in the formula with the appropriate values. Round to the nearest hundredth. L) If you lower the level of significance to a = 0.01, does your decision change? Explain your reasoning. ミー ナこ815310230 3273.46 こ - 12.77 731.97 M) Describe the type I and type Il errors that could occur in our test by completing each of the following statements. : .745 (type I, type I) error will occur when the actual mean of college (at most, at least) $10,230 but you reject the null tuition cost is t = ? hypothesis, Ho: µ? $10,230. H) Add the test statistic to your graph from part F, and insert the image below. (type I, type Il) error will occur when the actual mean of college (less than, greater than) $10,230 but you fail to • A tuition cost is reject the null hypothesis, Ho: u ? $10,230. $דר1 < %3D to
Transcribed Image Text:T-Test Results: tstats =-1.745, p-value 0.9514 I) What is the P-value? View the video Hypothesis Testing for the Mean (Siqma 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 (Siqma Unknown) Part II video). 40 30 p- value = 0.9514 J) 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. 20 Don't reject Ho 0.20 0.05 K) Interpret the decision in the context of the original claim. 0.00 40 35 3.0 25 20 15 10 -0.5 0.0 0.5 10 15 20 25 3.0 3.5 40 G) Calculate the test statistic. Replace the ?s in the formula with the appropriate values. Round to the nearest hundredth. L) If you lower the level of significance to a = 0.01, does your decision change? Explain your reasoning. ミー ナこ815310230 3273.46 こ - 12.77 731.97 M) Describe the type I and type Il errors that could occur in our test by completing each of the following statements. : .745 (type I, type I) error will occur when the actual mean of college (at most, at least) $10,230 but you reject the null tuition cost is t = ? hypothesis, Ho: µ? $10,230. H) Add the test statistic to your graph from part F, and insert the image below. (type I, type Il) error will occur when the actual mean of college (less than, greater than) $10,230 but you fail to • A tuition cost is reject the null hypothesis, Ho: u ? $10,230. $דר1 < %3D to
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