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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 Mean StDev SE Mean P Students Score 100 75.4 12.99 ** *** i. Calculate the SEMean Construct a 98% confidence interval for the population scores of Mrs. Paul's students. i. iii. State the null and alternate hypothesis for this test.

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 Mean StDev SE Mean P Students Score 100 75.4 12.99 ** *** i. Calculate the SEMean Construct a 98% confidence interval for the population scores of Mrs. Paul's students. i. iii. State the null and alternate hypothesis for this test.

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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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 Students Score 100 75.4 12.99 ** *** i. Calculate the SEMean 11. Construct a 98% confidence interval for the population scores of Mrs. Paul’s students. iii. State the null and alternate hypothesis for this test.
Transcribed Image Text: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 Students Score 100 75.4 12.99 ** *** i. Calculate the SEMean 11. Construct a 98% confidence interval for the population scores of Mrs. Paul’s students. iii. State the null and alternate hypothesis for this test.
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