Question

Thirty-four items are randomly selected from a population of 260 items. The sample mean is 33, and the sample standard deviation 6. Develop a 95% confidence interval for the population mean. (Round the t-value to 3 decimal places. Round the final answers to 2 decimal places.)  

The confidence interval is between and.

Answer 1

Solution :

Given that,

= 33

s =6

n = Degrees of freedom = df = n - 1 =34 - 1 = 35

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

  / 2= 0.05 / 2 = 0.025

t /2,df = t0.025,35 = 2.030 ( using student t table)

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

= 2.030* (6 / 35)

= 2.06

The 95% confidence interval is,

- E < < + E

33 - 2.06 < <33 +2.06

30.94 < < 35.06