Question

A random sample of n measurements was selected from a population with unknown mean μ and standard deviation σ = 15 for each of the situations in parts a through d. Calculate a 99% confidence interval for μ for each of these situations.

a. n = 75, x̄ = 27

b. n = 150, x̄ = 105

c. n = 125, x̄ = 16

d. n = 125, x̄ = 5.37

e. Is the assumption that the underlying population of measurements is normally distributed necessary to ensure the validity of the confidence intervals in parts a through d? Explain.

Answer 1

Since population standard deviation is known and n > 30, the Zcritical (2 tail) for = 0.01 is 2.576

Given = 15

The Confidence Interval is given by ME

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(a) 99%Confidence interval

From the data: = 27, n = 75

The Lower Limit = 27 - 4.46 = 22.54

The Upper Limit = 27 + 4.46 = 31.46

The 99% Confidence Interval is (22.54 , 31.46)

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(b) 99%Confidence interval

From the data: = 105, n = 150

The Lower Limit = 105 - 3.155 = 101.845

The Upper Limit = 105 + 3.155 = 108.155

The 99% Confidence Interval is (101.845 , 108.155)

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(c) 99%Confidence interval

From the data: = 16, n = 125

The Lower Limit = 16 - 3.456 = 12.544

The Upper Limit = 16 + 3.456 = 19.456

The 99% Confidence Interval is (12.544 , 19.456)

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(d) 99%Confidence interval

From the data: = 5.37, n = 125

The Lower Limit = 5.37 - 3.456 = 1.914

The Upper Limit = 5.37 + 3.456 = 8.826

The 99% Confidence Interval is (1.914 , 8.826)

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(e) No, it is not neccessary as sample sizes are large enough, n > 30 and by the central limit theorem, large sample s sized populations approach normality.

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