Question

Isabel Myers was a pioneer in the study of personality types. She identified four basic personality preferences that are described at length in the book Manual: A Guide to the Development and Use of the Myers-Briggs Type Indicator, by Myers and McCaulley.† Marriage counselors know that couples who have none of the four preferences in common may have a stormy marriage. A random sample of 375 married couples found that 295 had two or more personality preferences in common. In another random sample of 577 married couples, it was found that only 23 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. (a) Find a 99% confidence interval for p1 – p2. (Round your answers to three decimal places.) lower limit upper limit

Answer 1

i am using minitab to solve the problem.

steps :-

stat basic statistics2 proportions select summarized data in sample 1,type 295 in number of events and 375 in number of trials,in sample 2 ,type 23 in number of events and 577 in number of trialsoptions in confidence level type 99in hypothesized difference type 0select alternative hypothesis as difference hypothesized differenceokok.

needed part of minitab output be:-

Test and CI for Two Proportions

Method

p1: proportion where Sample 1 = Event

p2: proportion where Sample 2 = Event

Difference: p1-p2

Descriptive Statistics

Sample	N	Event	Sample p
Sample 1	375	295	0.786667
Sample 2	577	23	0.039861

Estimation for Difference

Difference	99% CI for Difference
0.746805	(0.688415, 0.805195)

CI based on normal approximation ]

**SOLUTION**

the 99% confidence interval for (p₁ – p₂) is:-

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