Question

[t-interval] If sample mean = 33.6, s = 1.5, sample size is 18. Determine 95% confidence interval.

Select the closest answer from below

	A.	(32.85 , 34.35)
	B.	(32.91 , 34.29)
	C.	(33.02 , 34.18)
	D.	(32.98 , 34.21)

Answer 1

Solution :

Given that,

= 33.6

s =1.5

n =18

Degrees of freedom = df = n - 1 =18 - 1 = 17

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,17 = 2.110 ( using student t table)

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

= 2.110 * (1.5 / 18)

= 0.75

The 95% confidence interval is,

- E < < + E

33.6 - 0.75 < < 33.6+ 0.75

(32.85 , 34.35)