A researcher for a car insurance company wishes to estimate themean annual premium that men...

A researcher for a car insurance company wishes to estimate the mean annual premium that men aged 20-24 years pay for their car insurance. A random sample of 50 men aged 20-24 years shows a mean of $711. Assume that the distribution of annual premiums has a standard deviation of $185.

a) Find a 94% confidence interval for the mean annual premium for all men aged 20-24 years.

b) What is the error of the estimate?

c) Find the 99% confidence interval for the mean annual premium for all men aged 20-24 years. Which confidence interval is wider?

d) How large a sample is necessary if we wish to be 94% confident that the sample mean will be within $20 of the true mean?

Solutions

Expert Solution

By comparing 94 % and 99 % confidence interval we know that 99% confidence interval is wider.

d) C = 0.94 , = 185 , E = 20

We have asked to find the sample size n.

We know that-

Sample size n = [(Z_0.94 * ) / E]²

= [ (1.88 * 185) / 20 ]²

= 302.67

303 (Rounded to the next integer)

Hence the required sample size n = 303

orchestra answered 1 year ago

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