An Insurance agent claims that the average age of policy holders who insure through him is less than the average age of all other agents, which is 30 years. 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 |
|---|---|---|---|---|---|---|
| Number of persons | 12 | 16 | 18 | 20 | 10 | 10 |
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 x ˉ = 25.58 and sample standard deviation s = 8.164 .
State the null hypothesis H 0 : μ = 30 and the alternative hypothesis H 1 : μ < 30 .
Calculate the test statistic t = s / n x ˉ − μ 0 = − 5.413 .
Since t = − 5.413 < t 0.05 = − 1.660 , reject the null hypothesis and conclude that the average age of the agent's policyholders is less than 30 years at a 5% significance level. Average age is less than 30 .
Explanation
Understand the problem and provided data We are given a problem where an insurance agent claims that the average age of their policyholders is less than 30 years, which is the average age of policyholders of all other agents. We have a sample of 100 policyholders with their age distribution, and we need to test this claim 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 midpoints are 18, 23, 28, 33, 38, and 43, with corresponding frequencies 12, 16, 18, 20, 10, and 10.
Calculate the sample mean The sample mean is calculated as follows: x ˉ = n ∑ ( a g e mi d p o in t × f re q u e n cy ) x ˉ = 100 ( 18 × 12 ) + ( 23 × 16 ) + ( 28 × 18 ) + ( 33 × 20 ) + ( 38 × 10 ) + ( 43 × 10 ) x ˉ = 100 216 + 368 + 504 + 660 + 380 + 430 = 100 2558 = 25.58 The sample mean is 25.58.
Calculate the sample standard deviation The sample standard deviation is calculated as follows: s = n − 1 ∑ (( a g e mi d p o in t − x ˉ ) 2 × f re q u e n cy ) s = 99 ( 18 − 25.58 ) 2 × 12 + ( 23 − 25.58 ) 2 × 16 + ( 28 − 25.58 ) 2 × 18 + ( 33 − 25.58 ) 2 × 20 + ( 38 − 25.58 ) 2 × 10 + ( 43 − 25.58 ) 2 × 10 s = 99 667.8768 + 107.3776 + 116.9592 + 1129.488 + 1541.476 + 3034.756 s = 99 6598 = 66.646 = 8.164 The sample standard deviation is approximately 8.164.
State the null and alternative hypotheses Now, we state the null and alternative hypotheses: H 0 : μ = 30 (The average age of the agent's policyholders is equal to 30 years) H 1 : μ < 30 (The average age of the agent's policyholders is less than 30 years)
Calculate the test statistic Next, we calculate the test statistic t :
t = s / n x ˉ − μ 0 = 8.164/ 100 25.58 − 30 = 0.8164 − 4.42 = − 5.413 The test statistic t is -5.413.
Determine the degrees of freedom The degrees of freedom are df = n − 1 = 100 − 1 = 99 .
Find the critical value We find the critical value t α for a one-tailed t-test with df = 99 and α = 0.05 . Using a t-table or calculator, we find that t 0.05 = − 1.660 .
Compare the test statistic with the critical value We compare the calculated test statistic t = − 5.413 with the critical value t α = − 1.660 . Since t < t α (-5.413 < -1.660), we reject the null hypothesis H 0 .
State the conclusion Conclusion: Since we rejected the null hypothesis, there is sufficient evidence to support the claim that the average age of the agent's policyholders is less than 30 years at a 5% significance level.
Examples
In marketing, an insurance company might want to target specific age groups with tailored products. By analyzing the age distribution of their policyholders, they can test whether their customer base is younger than the general population. This information helps them refine their marketing strategies and product offerings to better suit their target demographic.
