Question 1

1. Mrs. Paul wants to test that claim that, on average, her students average mark is 85. She pulls the records of 100 of per past students over a three-year period. The grades inputted into MINITAB for analysis. Exhibit 1 below shows the results of an analysis carried out on the data obtained.

Exhibit 1 One-Sample Z: Level I/II

Test of mu = 85 vs mu not = 85 The assumed sigma = 15.0

Variable N Mean StDev SE Mean Z P

Students Score 100 75.4 12.99 * ** ***

i. Calculate the SEMean ii. Construct a 98% confidence interval for the population scores of Mrs. Paul's students. iii. State the null and alternate hypothesis for this test. iv. Calculate the missing value ** and use tables to approximate the value of ***. v. Using a 1% level of significance, state the decision criteria for this test vi. State the conclusion for the test and why

Question 2

a). The prime Minister of a small Caribbean Island stated that 95% of the population was vaccinated from the Covid-19 virus. The opposition believes that the Minister is overstating the proportion of vaccinated citizens. He randomly selects 300 citizens and found that 240 of them were fully vaccinated.

i. Calculate a 99% confidence interval for the true proportion of all citizens who were vaccinated. ii. Interpret you answer in i). iii. State the null and alternative hypothesis of this test. iv. Calculate the value of the test statistics. v. At the 5% level of significance, determine if the Politian overstated the proportion of vaccinated citizen. Use the classical approach.

b) The following data shows the number of hours slept by a patient in 1 weeks

12 18 09 16 11 10 08

i. Calculate the unbiased estimate for the population mean ii. Calculate the unbiased estimate for the population standard deviation iii. Calculate a 95% confidence interval for the mean number of hours

Question 3

In a business, the Chief Executive Officer wants to know if his workers are for or against mandatory vaccinations. A random sample of 200 workers was taken and the results are shown in the MINITAB output below

Expected counts are printed below observed counts

Managers Team leaders Employees Total

17 22 15

Approve 21.330 17.820 14.850 54

34 25 14

Against 28.835 24.090 20.075 73

28 19 26

No Opinion 28.835 24.090 20.075 73

Total 79 66 55 200

Chi-sq = 0.878992 + 0.980494 + 0.001515 +

0.925168+ 0.034375 + 1.838387 +

0.02418 + 1.075471 + 1.74874 =7.507306

i. What is the value of the degree of freedom for this test? ii. What is the p-value for this test? iii. Carefully define the null and alternative hypotheses of the ² test underlying the generation of the above table. iv. Using = 10%, what is the conclusion of this test? Give reasons for your answer.