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[2] A random sample of 69 health insurance firms, 47 did public relations in-house, as did 40 of an independent random sample of 69 casualty insurance firms. Conduct a hypothesis test of equality of t population proportions against a two-sided alternative. Use JMP to conduct the appropriate hypothes test. Use a 5% significance level.

[2] A random sample of 69 health insurance firms, 47 did public relations in-house, as did 40 of an independent random sample of 69 casualty insurance firms. Conduct a hypothesis test of equality of t population proportions against a two-sided alternative. Use JMP to conduct the appropriate hypothes test. Use a 5% significance level.

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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JMP Pro 15 preferred but Excel is okay.

STUDENT_PA COURS NO IR E COURSE 1 1 82 75 2 2 73 71 3 3 59 52 4 4 48 46 5 5 69 70 6 93 83 CO
Transcribed Image Text:STUDENT_PA COURS NO IR E COURSE 1 1 82 75 2 2 73 71 3 3 59 52 4 4 48 46 5 5 69 70 6 93 83 CO
[2] A random sample of 69 health insurance firms, 47 did public relations in-house, as did 40 of an independent random sample of 69 casualty insurance firms. Conduct a hypothesis test of equality of th population proportions against a two-sided alternative. Use JMP to conduct the appropriate hypothesis test. Use a 5% significance level. [2.1] State Assumptions: 1. The sample must be independent 2. The sample must be random 3. Normal distribution [2.2] State the Null and Alternative Hypothesis statements Ho: P1 = P2 Ha: P1 not equal to P2 P1 = 47/69 = 0.6812 P2 = 40/69 = 0.5797 Conduct hypothesis test using a 5% level of significance: a = 0.05 Test Statistic Phealth - Pcasualty z = where, Po(1– po) , Po(1 –- Po) + Neasualty Xhealth +Xcasualty nhealth +ncasualty Nhealth [2.3] Sample Data, Computation of Test Statistic and p-value. (JMP output with data input fields) (Paste JMP output with data input fields here.) Z-stats = 1.235 p-value = 0.2168 JMP Assignment 5 [2.4] State decision rules in terms of the p-value. What can we conclude based on the sample information? What does this mean in terms of the problem? p-value is greater than 0.05 so we do not reject the null hypothesis We can state that the proportion of health insurance firms is the same as the proportion of casualty insurance firms doing public relations i house.
Transcribed Image Text:[2] A random sample of 69 health insurance firms, 47 did public relations in-house, as did 40 of an independent random sample of 69 casualty insurance firms. Conduct a hypothesis test of equality of th population proportions against a two-sided alternative. Use JMP to conduct the appropriate hypothesis test. Use a 5% significance level. [2.1] State Assumptions: 1. The sample must be independent 2. The sample must be random 3. Normal distribution [2.2] State the Null and Alternative Hypothesis statements Ho: P1 = P2 Ha: P1 not equal to P2 P1 = 47/69 = 0.6812 P2 = 40/69 = 0.5797 Conduct hypothesis test using a 5% level of significance: a = 0.05 Test Statistic Phealth - Pcasualty z = where, Po(1– po) , Po(1 –- Po) + Neasualty Xhealth +Xcasualty nhealth +ncasualty Nhealth [2.3] Sample Data, Computation of Test Statistic and p-value. (JMP output with data input fields) (Paste JMP output with data input fields here.) Z-stats = 1.235 p-value = 0.2168 JMP Assignment 5 [2.4] State decision rules in terms of the p-value. What can we conclude based on the sample information? What does this mean in terms of the problem? p-value is greater than 0.05 so we do not reject the null hypothesis We can state that the proportion of health insurance firms is the same as the proportion of casualty insurance firms doing public relations i house.
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