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 30.5. A random sample of 100 policy holders who had insured through the agent gave the following age distribution:
| Age | 16-20 | 21-25 | 26-30 | 31-35 | 36-40 | 41-45 | 46-50 |
|---|---|---|---|---|---|---|---|
| Number of persons | 12 | 16 | 18 | 20 | 14 | 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 = 31.9 .
Calculate the sample standard deviation: s = 9.117 .
Calculate the test statistic: t = 1.536 .
Compare the test statistic with the critical value: Since -1.660"> 1.536 > − 1.660 , fail to reject the null hypothesis. There is not enough evidence to support the claim. There is not enough evidence to support the claim.
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, which is 30.5 years. We have a sample of 100 policyholders with the given age distribution. Our goal is to test this claim at a 5% significance level.
Calculate the sample mean First, we need to calculate the sample mean age x ˉ from the given age distribution. We use the midpoint of each age range as the age for that group:
Age groups: 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, 46-50 Midpoints: 18, 23, 28, 33, 38, 43, 48 Number of persons: 12, 16, 18, 20, 14, 10, 10
The sample mean is calculated as: x ˉ = n ∑ ( mi d p o in t × n u mb er o f p erso n s ) x ˉ = 100 ( 18 × 12 ) + ( 23 × 16 ) + ( 28 × 18 ) + ( 33 × 20 ) + ( 38 × 14 ) + ( 43 × 10 ) + ( 48 × 10 ) x ˉ = 100 216 + 368 + 504 + 660 + 532 + 430 + 480 = 100 3190 = 31.9
Calculate the sample standard deviation Next, we calculate the sample standard deviation s :
s = n − 1 ∑ ( n u mb er o f p erso n s × ( mi d p o in t − x ˉ ) 2 ) s = 99 12 ( 18 − 31.9 ) 2 + 16 ( 23 − 31.9 ) 2 + 18 ( 28 − 31.9 ) 2 + 20 ( 33 − 31.9 ) 2 + 14 ( 38 − 31.9 ) 2 + 10 ( 43 − 31.9 ) 2 + 10 ( 48 − 31.9 ) 2 s = 99 12 ( − 13.9 ) 2 + 16 ( − 8.9 ) 2 + 18 ( − 3.9 ) 2 + 20 ( 1.1 ) 2 + 14 ( 6.1 ) 2 + 10 ( 11.1 ) 2 + 10 ( 16.1 ) 2 s = 99 12 ( 193.21 ) + 16 ( 79.21 ) + 18 ( 15.21 ) + 20 ( 1.21 ) + 14 ( 37.21 ) + 10 ( 123.21 ) + 10 ( 259.21 ) s = 99 2318.52 + 1267.36 + 273.78 + 24.2 + 520.94 + 1232.1 + 2592.1 s = 99 8229.0 = 83.1212 ≈ 9.117
State the null and alternative hypotheses Now, we state the null and alternative hypotheses:
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 test statistic We calculate the test statistic t :
t = s / n x ˉ − μ 0 = 9.117/ 100 31.9 − 30.5 = 9.117/10 1.4 = 0.9117 1.4 ≈ 1.536
Determine the degrees of freedom and critical value The degrees of freedom are df = n − 1 = 100 − 1 = 99 .
For a one-tailed t-test with α = 0.05 and df = 99 , the critical value t α is approximately -1.660.
Compare the test statistic with the critical value We compare the calculated test statistic t = 1.536 with the critical value t α = − 1.660 . Since -1.660"> 1.536 > − 1.660 , we fail to reject the null hypothesis.
State the conclusion Conclusion: There is not 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
In insurance, understanding the age distribution of policyholders is crucial for risk assessment and premium calculation. For instance, if an agent claims to attract younger clients, this could imply lower risk and potentially justify different premium rates. Hypothesis testing, as demonstrated here, helps validate such claims by comparing sample data against established norms, ensuring fair and accurate business practices.
