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.
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.
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.1: Measures Of Center
Problem 9PPS
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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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