An Insurance agent claims that the average age of policy holders who insure through him is less than the average of all other agents, which is 30. A random sample of 100 policy holders who had insured through the agent gave the adjoining age distribution:
Age: 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, 46-50
Number of persons: 12, 16, 18, 20, 10, 10, 14
Test the claim that the average age is less than the figure for all the other agents at a 5% level of significance.
Answer (1)
Calculate the sample mean age from the given distribution: x ˉ = 30.3 .
Calculate the sample standard deviation: s = 9.588 .
Calculate the t-test statistic: t = − 0.2086 .
Compare the calculated t-value with the critical t-value and conclude: Fail to reject the null hypothesis. The average age is not less than 30.5. Fail to reject the null hypothesis
Explanation
Understand the problem and provided data The insurance agent claims that the average age of their policyholders is less than the average age of all other agents' policyholders, which is 30.5 years. We have a sample of 100 policyholders with the given age distribution. We need to test if the agent's claim is valid at a 5% significance level.
Calculate sample mean and standard deviation First, we need to calculate the sample mean ( x ˉ ) and sample standard deviation (s) from the given age distribution. The age groups are 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, and 46-50. We'll use the midpoints of these intervals as the ages: 18, 23, 28, 33, 38, 43, and 48. The number of policyholders in each group is 12, 16, 18, 20, 10, 10, and 14, respectively.
Calculate the sample mean The sample mean is calculated as follows: x ˉ = ∑ i = 1 n f i ∑ i = 1 n f i x i where f i is the frequency (number of policyholders) and x i is the midpoint of the age group. The calculation yields x ˉ = 30.3 .
Calculate the sample standard deviation The sample standard deviation is calculated as follows: s = n − 1 ∑ i = 1 n f i ( x i − x ˉ ) 2 where n is the total number of policyholders. The calculation yields s = 9.588 .
State the null and alternative hypotheses Now, we state the null hypothesis ( H 0 ) and the alternative hypothesis ( H 1 ). H 0 : μ = 30.5 (The average age of the agent's policyholders is equal to 30.5 years) H 1 : μ < 30.5 (The average age of the agent's policyholders is less than 30.5 years)
Calculate the t-test statistic We will use a t-test statistic since the population standard deviation is unknown: t = s / n x ˉ − μ 0 where μ 0 = 30.5 and n = 100 . Substituting the values, we get: t = 9.588/ 100 30.3 − 30.5 = 0.9588 − 0.2 = − 0.2086
Determine the critical t-value The degrees of freedom are df = n − 1 = 100 − 1 = 99 . The significance level is α = 0.05 . We need to find the critical t-value ( t α ) for a one-tailed test with df = 99 and α = 0.05 . The critical t-value is approximately -1.660.
Compare the calculated t-value with the critical t-value Now, we compare the calculated t-value (-0.2086) with the critical t-value (-1.660). Since -0.2086 > -1.660, we fail to reject the null hypothesis.
Conclusion Since we fail to reject the null hypothesis, we do not have enough evidence to support the claim that the average age of the agent's policyholders is less than 30.5 years at a 5% significance level.
Examples
Imagine you're a marketing analyst trying to determine if a new ad campaign is reaching a younger demographic than your current customer base. By comparing the average age of customers exposed to the new campaign with the average age of your existing customers, you can use a hypothesis test to see if the campaign is indeed attracting a younger audience. This helps you make informed decisions about your marketing strategies and budget allocation. Hypothesis testing is a fundamental tool in data-driven decision-making across various fields.
