Question

Families USA, a monthly magazine that discusses issues related to health and health costs, survey 19 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,800. The standard deviation of the sample was $1095.

A. Based on the sample information, develop a 99% confidence interval for the population mean yearly premium

B. How large a sample is needed to find the population mean within $225 at 90% confidence? (Round up your answer to the next whole number.)

Answer 1

a

$10,151
`< \mu <` $11448.12

b

`n = 158`

Explanation:

From the question we are told that

The sample size is n = 19

The sample mean is
`\= x =`$10,800

The standard deviation is
`\sigma =`$1095

The population mean is
`\mu =`$225

Given that the confidence level is 99% the level of significance is mathematically represented as

`\alpha = 100 -99`

`\alpha = 1`%

=>
`\alpha = 0.01`

Now the critical values of
`\alpha = Z_{(\alpha )/(2) }` is obtained from the normal distribution table as

`Z_{(0.01)/(2) } = 2.58`

The reason we are obtaining values for
`(\alpha )/(2)` is because
`\alpha` is the area under the normal distribution curve for both the left and right tail where the 99% interval did not cover while
`(\alpha )/(2)` is the area under the normal distribution curve for just one tail and we need the value for one tail in order to calculate the confidence interval

Now the margin of error is obtained as

`MOE = Z_{(\alpha )/(2) } * (\sigma )/(√(n) )`

substituting values

`MOE = 2.58* (1095 )/(√(19) )`

`MOE = 648.12`

The 99% confidence interval for the population mean yearly premium is mathematically represented as

`\= x -MOE < \mu < \= x +MOE`

substituting values

`10800 -648.12 < \mu < 10800 + 648.12`

`10800 -648.12 < \mu < 10800 + 648.12`

$10,151
`< \mu <` $11448.12

The largest sample needed is mathematically evaluated as

`n = [\frac{Z_{(\alpha )/(2) } * \sigma }{\mu} ]`

substituting values

`n = [ ( 2.58 * 1095)/(225) ]^2`

`n = 158`