A manager for an insurance company believes that customers have the following pr

Question

A manager for an insurance company believes that customers have the following preferences for life insurance products: 10%10% prefer Whole Life, 40%40% prefer Universal Life, and 50%50% prefer Life Annuities. The results of a survey of 351351 customers were tabulated. Is it possible to refute the sales manager's claimed proportions of customers who prefer each product using the data?

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Step 1 of 10:

State the null and alternative hypothesis.

Step 2 of 10: What does the null hypothesis indicate about the proportions of customers who prefer each insurance product?

Step 3 of 10: State the null and alternative hypothesis in terms of the expected proportions for each category.

Whole= Universal= Annual =

Step 4 of 10: Find the expected value for the number of customers who prefer Whole Life. Round your answer to two decimal places.

Step 5 of 10: Find the expected value for the number of customers who prefer Universal Life. Round your answer to two decimal places.

Step 6 of 10: Find the value of the test statistic. Round your answer to three decimal places.

Step 7 of 10: Find the degrees of freedom associated with the test statistic for this problem.

Step 8 of 10: Find the critical value of the test at the

0.1

0.1

level of significance. Round your answer to three decimal places.

Step 9 of 10: Make the decision to reject or fail to reject the null hypothesis at the

0.1

0.1

level of significance.

Step 10 of 10: State the conclusion of the hypothesis test at the

0.1

0.1

level of significance.

Product Number Whole 189189 Universal 116116 Annuities 4646

Explanation / Answer

Ans:

2)Null hypothesis indicates that given data fits the specified distribution(i.e.p1=0.1,p2=0.4,p3=0.5)

1,3)H0:p1=0.1,p2=0.4,p3=0.5

(where 1=whole,2=universal,3=annual)

Ha:Data does not fit the specified distribution.

Expected count(E)=351*pi

4)E(whole)=351*0.1=35.10

5)E(universal)=351*0.4=140.40

6)

Chi square test statistic=774.590

7)df=3-1=2

8)alhpa=0.1

Critical chi square value=CHIINV(0.1,2)=4.605

9)Reject the null hypothesis.

10)There is sufficient evidence to refute the sales manager's claimed proportions of customers who prefer each product using the data.

Product Observed(O) pi Expected(E)   (O-E)^2/E Whole 189 0.1 35.1 674.792 Universal 116 0.4 140.4 4.240 Annuities 46 0.5 175.5 95.557 Total 351 1 351 774.590

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