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 cudents 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. 111. 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
Z
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.
State the null and alternate hypothesis for this test.
111.
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 Z 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. State the null and alternate hypothesis for this test. 111.
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