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.

Step-by-step explanation:

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

$\overline{x}\pm t_{\alpha/2}{\dfrac{s}{\sqrt{n}}}$

, 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:

$60\pm (2.0686){\dfrac{7.4}{\sqrt{24}}}\\\\\approx60\pm3.1\\\\=(60-3.1,\ 60+3.1)\\\\=(56.9,\ 63.1)$

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