Question

We wish to create a 98% confidence interval for the mean. A sample of 24 gives an average of 72 and a standard deviation of 6.35 . Find the lower value for the confidence interval. Round to 2 decimal places.

Answer 1

Solution :

Given that,

= 72

s = 6.35

n = 24

Degrees of freedom = df = n - 1 = 24 - 1 = 23

At 98% confidence level the t is ,

= 1 - 98% = 1 - 0.98 = 0.02

= 0.02

t _,df = t_0.02,23 =2.177

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

= 2.177 * (6.35 / 24) = 2.82

The 98% confidence interval estimate of the population mean is,

- E

72 - 2.82

69.18

lower value confidence interval is = 69.18