Question

After sampling 16 members of a normal population you find the sample mean equals 50 and the sample standard deviation, s, equals 10. Construct a 95% confidence interval estimate for mu, the population mean.

A. 45.62 – 54.38

B. 44.67 – 55.33

C. 45.10 – 54.90

D. None of the above

Answer 1

Solution :

Given that,

$\bar x$ = 50

s =10

n = Degrees of freedom = df = n - 1 =16 - 1 = 15

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 = t0.025,15 = 2.131 ( using student t table)

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

= 2.131 * ( 10/ $\sqrt$ 16)

= 5.33

The 95% confidence interval mean is,

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

50 - 5.33 < $\mu$ <50 + 5.33

44.67 – 55.33

(44.67 – 55.33 )