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?

Product	Number
Whole	189189
Universal	116116
Annuities	4646

Copy Data

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.

Answer 1

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)

H_a:Data does not fit the specified distribution.

Expected count(E)=351*p_i

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

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

6)

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

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.