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.

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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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