Question

construct a 95% confidence interval for the population mean, Assume the population has a normal distribution. A sample of 20 part-time workers had a mean annual earnings of $3120 with a standard deviation of $677. Round to the nearest dollar. *Please STEP out the problem clearly for me.

Answer 1

Confidence Interval
$\bar{X} \pm t_{\alpha /2, n-1} S/\sqrt{n}$
$t_{\alpha /2, n-1} = t_{ 0.05 /2, 20- 1 } = 2.093$
$3120 \pm t_{\ 0.05/2, 20 -1} * 677/\sqrt{ 20}$
Lower Limit = $3120 - t_{\ 0.05/2, 20 -1}677/\sqrt{ 20}$
Lower Limit = 2803.1542
Upper Limit = $3120 + t_{\ 0.05/2, 20 -1}677/\sqrt{ 20}$
Upper Limit = 3436.8458

95% Confidence interval is ( 2803 , 3437  )