Question

Assume that a sample is used to estimate a population mean . Find the 95% confidence interval for a sample of size 599 with a mean of 27.4 and a standard deviation of 6.9. Enter your answer as a tri-linear inequality accurate to one decimal place (because the sample statistics are reported accurate to one decimal place). Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

Solution :

Given that,

$\bar x$ = 27.4

s = 6.9

n = 599

Degrees of freedom = df = n - 1 = 599 - 1 = 598

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

t$\alpha$ /2,df = t0.025,598 =1.964

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

= 1.964 * (6.9 / $\sqrt$ 599)

= 0.6

The 95% confidence interval estimate of the population mean is,

$\bar x$ - E < $\mu$ < $\bar x$ + E

27.4 - 0.6 < $\mu$ <  27.4 + 0.6

26.8 < $\mu$ < 28.0