18. An insurance agent has claimed that the average age of policy-holders who insure through him is-less than the average for all agents, which is 30-5 years. A random sample of 100 policy-holders who had insured through him gave the following age distribution : Age last birthday 16-20 21-25 26-30 31-35 36- 40 No. of persons 12 22 20 30 16 Calculate the arithmetic mean and standard deviation of this distribution and use these values to test his claim at the 5% level of significance. You are given that Z (1-645) = (0-95.
18. An insurance agent has claimed that the average age of policy-holders who insure through him is-less than the average for all agents, which is 30-5 years. A random sample of 100 policy-holders who had insured through him gave the following age distribution : Age last birthday 16-20 21-25 26-30 31-35 36- 40 No. of persons 12 22 20 30 16 Calculate the arithmetic mean and standard deviation of this distribution and use these values to test his claim at the 5% level of significance. You are given that Z (1-645) = (0-95.
Chapter1: Making Economics Decisions
Section: Chapter Questions
Problem 1QTC
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
Transcribed Image Text:18. An insurance agent has claimed that the average age of policy-holders
who insure through him is less than the average for all agents, which is 30-5 years.
A random sample of 100 policy-holders who had insured through him gave the following
age distribution :
Age last birthday
16-20
21-25
26-30
31-35
36-40
No. of persons
12
22
20
30
16
:
Calculate the arithmetic mean and standard deviation of this distribution and use these
values to test his claim at the 5% level of significance. You are given that Z (1-645) = (0-95.
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