Question

A manager for an insurance company believes that customers have the following preferences for life insurance products: 40 % prefer Whole Life, 20 % prefer Universal Life, and 40 % prefer Life Annuities. The results of a survey of 212 212 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	70
universal	50
annuities	92

state the null and alternative hypothesis

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

State the null and alternative hypothesis in terms of the expected proportions for each category

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

Find the expected value for the number of customers who prefer Life Annuities. Round your answer to two decimal places.

Find the value of the test statistic. Round your answer to three decimal places.

Find the degrees of freedom associated with the test statistic for this problem.

Find the critical value of the test at the 0.01 level of significance. Round your answer to three decimal places.

Make the decision to reject or fail to reject the null hypothesis at the 0.01 level of significance

State the conclusion of the hypothesis test at the 0.01 level of significance

Answer 1

Null Hypothesis:

Ho: Proportion of customers who prefer each product do not differ from the Hypothetical frequencies.

Alternative Hypothesis:

Ha: Proportion of customers who prefer each product differ from the Hypothetical frequencies

Find the expected value for the number of customers who prefer Whole Life.

= 40% of 212 = 84.80

Find the expected value for the number of customers who prefer Universal.

= 20% of 212 = 42.40

Find the expected value for the number of customers who prefer Life Annuities.

= 40% of 212 = 84.80

Computational Table:

Test statistic:

Degrees of Freedom: k-1 = 3-1 = 2

Where, K = Number of Categories = 3

Critical value:

   ................Using Chi square table

Conclusion:

Test statistic < critical value, i.e 4.557 < 9.210, That is Fail to Reject Ho at 1% level of significance.

Fail to reject the null hypothesis at the 0.01 level of significance

Therefore, Proportion of customers who prefer each product do not differ from the Hypothetical frequencies.