[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
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
JMP Pro 15 preferred but Excel is okay.
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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 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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fedd60f12-fe5d-409b-96fb-afd8188a6209%2F879403c1-b8aa-45b1-9a5f-da1014b4a3c0%2F8s2psyo_processed.png&w=3840&q=75)
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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