Question

(10 points) Starting salaries of 64 college graduates who have taken a statistics course have a mean of $44,500 with a standard deviation of $6,800. Find a 99.7% confidence interval for u. (NOTE: Do not use commas or dollar signs in your answers. Round each bound to three decimal places.) Lower-bound: Upper-bound:

Answer 1

Solution :

Given that,

Point estimate = sample mean = $\bar x$ = 44500

sample standard deviation = s = 6800

sample size = n = 64

Degrees of freedom = df = n - 1 = 64 - 1 = 63

At 99.7% confidence level

$\alpha$ = 1 - 99.7%

$\alpha$ =1 - 0.997 =0.003

$\alpha$/2 = 0.0015

t$\alpha$/2,df = t_0.0015,63 = 3.087

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 3.087 * ( 6800 / $\sqrt$ 64 )

Margin of error = E = 2623.950

The 99.7% confidence interval estimate of the population mean is,

$\bar x$  ± E

= 44500  ± 2623.950

= ( 41876.050, 47123.950 )

lower bound = 41876.050

upper bound = 47123.950