Question

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 ?.

(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 = = 44500

sample standard deviation = s = 6800

sample size = n = 64

Degrees of freedom = df = n - 1 = 63

At 99.7% confidence level the t is ,

= 1 - 99.7% = 1 - 0.997 = 0.003

/ 2 = 0.003 / 2 = 0.0015

t /2,df = t0.0015,63 = 3.087

Margin of error = E = t/2,df * (s /n)

= 3.087* (6800 / 64)

= 2623.950

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

- E < < + E

44500 - 2623.950 < < 44500 + 2623.950

41876.050 < < 47123.950

Lower-bound: 41876.050

Upper-bound: 47123.950