Question

AVERAGE STUDENT LOAN DEBT FOR THE CLASS OF 2013 New Hampshire $32,795 Pennsylvania $32,528 10 00 Ohio $29,090 U.S. AVERAGE $28,400 100 to New York $26,381 BRE Tosition B01 Texas $25,244 California DO $20,340 DEBT New Mexico $18,656 Data source: The Project on Student Debt, The Institute for College Access & Success Data source: The Project on Student Debt, The Institute for College Access & Success According to estimates by the Project on Student Debt study, 69% of college students who graduated in 2013 had loans to pay. The accompanying chart lists the U.S. average and the averages of such debts for a few selected states for students in the class of 2013 who graduated with loans. Remember that each of these averages is based on a survey of students who graduated in 2013 with debt. For example, U.S. college graduates in 2013 owed an average of $28,400. Students who graduated with loans from colleges in New Hampshire in 2013 had the highest average debt of $32,795, and those who graduated with loans from colleges in New Mexico had the lowest average debt of $18,656. Note that these averages are on sample surveys. Suppose we want to find out if the average r stude: hio was higher than U.S. average of $28,400. Supposet the average debt for Ohio college students in the class of 2013 is based on a random sample of 900 students who graduated with loans. Assume that the standard deviation of the loans for all Ohio students in the class of 2013 was $4800 and the significance level is 1%. The test is right-tailed because we are testing the hypothesis that the average debt for the class of 2013 in Ohio (which was $29,090) was higher than $28,400. The null and alternative hypotheses are Họ: A = $28,400 H: >$28,400 Here, n=900, 7 = $29,090, 0 = $4800, and a=.01. The population standard deviation is known, and the sample is large. Hence, we can use the normal distribution to perform this test. Using the normal distribution to perform the test, we find that the critical value of z is 2.33 for 01 area in the right tail of the normal curve. We find the observed value of z as follows. 0 4800 $160 900 29,090-28,400 = 4.31 160 - The value of the test statistic z = 4.31 for î is larger than the critical value of z=2.33, and it falls in the rejection region. Consequently, we reject H, and conclude that the average debt of students who graduated in 2013 from colleges in Ohio is higher than $28,400, which is the average for the United States. To use the p-value approach, we find the area under the normal curve to the right of z = 4.31 from the normal distribution table. This area is (approximately).0000. Therefore, the p-value is .oooo. Since a .01 is larger than the p-value = .0000, we reject the null hypothesis.

Using Case-Study 9-1 that appears on Page 363 in Course Textbook(Image Shown Above), summarize in five steps the test of hypothesis being undertaken in Case Study 9-1 about the population mean μ. ?

Answer 1

Let u denotes the population mean. Hypotheses : To test null hypothesis Ho := 28400 against alternative hypothesis H1 :p> 28400 Let i denotes the sample mean, o denotes the sample standard deviation and n denotes the sample size. Here, Ẽ = 29090.000, = 23040000.0000,3 = 4800.00, n = 900 The test statistic can be written as: (3 – 28400.0) 0 which under H, follows a standard normal distribution. Decision rule / Rejection region : We reject H, at 0.01 level of significance if P-value < 0.01 or if |zobs > 20.005 = 2.326 0.4 - 0.3 0.2 0.1 - 0.0 - -2 Now, Value of test statistic: The value of the test statistic is Zobs = 900(29090.000 – 28400) 4800.00 4.312500 4.31 The critical value = 2.326348 2.326 P-value = P(Z > Zobs) = P(Z > 4.312500) =1-0.999992 = 0.000008 0.0000 Conclusion: Since p-value < 0.01 and Zobs > Zcritical = 2.326, so we reject H, at 0.01 level of significance Hence, we can conclude that the the population mean is greater than 28400.