From a normal population, a sample of 39 items is taken. The sample mean is 12...

From a normal population, a sample of 39 items is taken. The sample mean is 12 and the sample standard deviation is 2. Construct a 99% confidence interval for the population mean.

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Expert Solution

solution

Given that,

= 12

s =2

n = 39

Degrees of freedom = df = n - 1 =39- 1 = 38

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t /2 df = t0.005,38 = 2.712 ( using student t table)

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

= 2.712 * (2 / $\sqrt$ 39) = 0.87

The 99% confidence interval mean is,

- E < < + E

12 -0.87 < < 12+ 0.87

11.13 < < 12.87

(11.13 , 12.87)

orchestra answered 1 year ago

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