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

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