Question

In a study of 24 criminals convicted of antitrust offenses, the average age was 60 years, with a standard deviation of 7.4 years. Construct a 95% confidence interval on the true mean age. (Give your answers correct to one decimal place.)___ to____ years

Answer 1

Answer: 56.9 years to 63.1 years.

Explanation:

Confidence interval for population mean (when population standard deviation is unknown):

`\overline{x}\pm t_(\alpha/2){(s)/(√(n))}`

, where
`\overline{x}`= sample mean, n= sample size, s= sample standard deviation,
`t_(\alpha/2)`= Two tailed t-value for
`\alpha`.

Given: n= 24

degree of freedom = n- 1= 23

`\overline{x}`= 60 years

s= 7.4 years

`\alpha=0.05`

Two tailed t-critical value for significance level of
`\alpha=0.05` and degree of freedom 23:

`t_(\alpha/2)=2.0687`

A 95% confidence interval on the true mean age:

`60\pm (2.0686){(7.4)/(√(24))}\\\\\approx60\pm3.1\\\\=(60-3.1,\ 60+3.1)\\\\=(56.9,\ 63.1)`

Hence, a 95% confidence interval on the true mean age. : 56.9 years to 63.1 years.