1. A random sample of 40 college student students shows that the score of a College...

1. A random sample of 40 college student students shows that the score of a College Statistics is normally distributed with its mean, 81 and standard deviation, 8.4. Find 99 % confidence interval estimate for the true mean.

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Expert Solution

Solution :

Given that,

= 81

s = 8.4

n = 40

Degrees of freedom = df = n - 1 = 40 - 1 = 39

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005.39 = 2.708

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

= 2.708 * (8.4 / $\sqrt$ 40)

= 3.6

The 99% confidence interval estimate of the population mean is,

- E < < + E

81 - 3.6 < < 81 + 3.6

77.4 < < 84.6

(77.4,84.6)

The 99% confidence interval 77.4 to 84.6

orchestra answered 11 months ago

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