Question

Construct a 99% confidthence interval for the population mean .Assume the population has a normal distribution. A group of 19 randomly selected employees has a mean age of 22.4 years with a standard deviation of 3.8 years. Round to the nearest tenth.
A) Determine the critical value ta/2 with n-the 1 degrees of freedom
B) Determine the lower and upper bound of the confidence interval
C) Interpret the confidence interval.

Answer 1

Answer with explanation:

Confidence interval for 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

$t_{\alpha/2}$ = Critical t-value for n-1 degrees of freedom

We assume the population has a normal distribution.

Given, n= 19 , s= 3.8 , $\overline{x}=22.4$

$\alpha=1-0.99=0.01$

A) Critical t value for $\alpha/2=0.005$ and degree of 18 freedom

$t_{\alpha/2}$ = 2.8784

B) Required confidence interval:

$22.4\pm ( 2.8744)\dfrac{3.8}{\sqrt{19}}\\\\=22.4\pm2.5058\\\\=(22.4-2.5058,\ 22.4+2.5058)=(19.8942,\ 24.9058)\approx(19.9,\ 24.9)$

Lower bound = 19.9 years

Uppen bound = 24.9 years

C) Interpretation: We are 99% confident that the true population mean of lies in (19.9, 24.9) .