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 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 (2)
Calculate the sample mean age: x ˉ = 29.744 .
Calculate the sample standard deviation: s = 7.737 .
Calculate the test statistic: t = − 0.977 .
Compare the test statistic to the critical value and conclude: Fail to reject the null hypothesis. Fail to reject the null hypothesis .
Explanation
Calculate the sample mean First, we need to calculate the sample mean age from the given age distribution. We take the midpoint of each age group as the representative age for that group: 16-20 (18), 21-25 (23), 26-30 (28), 31-35 (33), 36-40 (38), 41-45 (43). Then we calculate the weighted average:
Calculate the sample mean (corrected) The sample mean x ˉ is calculated as follows: 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 Oops! There was a calculation error in the previous step. Let's recalculate the mean correctly: x ˉ = 100 ( 18 × 12 ) + ( 23 × 16 ) + ( 28 × 18 ) + ( 33 × 20 ) + ( 38 × 14 ) + ( 43 × 20 ) x ˉ = 100 216 + 368 + 504 + 660 + 380 + 430 = 100 2558 = 25.58
Calculate the sample standard deviation Next, we calculate the sample standard deviation s . First, we calculate the variance: s 2 = n − 1 ∑ i = 1 n f i ( x i − x ˉ ) 2 s 2 = 99 12 ( 18 − 29.744 ) 2 + 16 ( 23 − 29.744 ) 2 + 18 ( 28 − 29.744 ) 2 + 20 ( 33 − 29.744 ) 2 + 14 ( 38 − 29.744 ) 2 + 20 ( 43 − 29.744 ) 2 s 2 = 99 12 ( 137.92 ) + 16 ( 45.42 ) + 18 ( 3.04 ) + 20 ( 10.60 ) + 14 ( 68.13 ) + 20 ( 175.79 ) s 2 = 99 1655.04 + 726.72 + 54.72 + 212 + 953.82 + 3515.8 = 99 7118.1 = 71.899 Then, the standard deviation is the square root of the variance: s = 71.899 = 8.479
State the null and alternative hypotheses Now, we state the null hypothesis H 0 : μ = 30.5 and the alternative hypothesis H 1 : μ < 30.5 . We are testing if the average age of the agent's policyholders is less than 30.5.
Calculate the test statistic We calculate the test statistic t :
t = s / n x ˉ − μ 0 = 7.737/ 100 29.744 − 30.5 = 0.7737 − 0.756 = − 0.977
Determine the degrees of freedom The degrees of freedom df = n − 1 = 100 − 1 = 99 .
Find the critical value We find the critical value t α for a one-tailed t-test with α = 0.05 and df = 99 . Using a t-table or calculator, we find that t 0.05 , 99 ≈ − 1.660 .
Compare the test statistic with the critical value We compare the calculated test statistic t = − 0.977 with the critical value t α = − 1.660 . Since -1.660"> − 0.977 > − 1.660 , we fail to reject the null hypothesis H 0 .
State the conclusion 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 at a 5% significance level.
Examples
In marketing, hypothesis testing can determine if a new advertising campaign attracts a younger demographic compared to the existing customer base. For example, an insurance company might want to know if their online ads are reaching a younger audience than their traditional marketing methods. By comparing the average age of customers acquired through each method, they can decide where to focus their marketing efforts to attract a specific age group.
We calculated the sample mean and standard deviation of the insurance agent's policyholders' ages. We found the test statistic and compared it to the critical value, leading us to reject the null hypothesis. Thus, we conclude that the average age of the agent's policyholders is indeed less than 30.5 years at a 5% significance level.
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