High School

Assume simple random sampling. Calculate the 95% confidence interval estimate for the population mean for each of the following:

a. \( N = 1700; \, n = 70; \, s = 9; \, \bar{x} = 143 \)

b. \( N = 1725; \, n = 80; \, \bar{x} = 232.2 \)

c. \( N = 4200; \, n = 200; \, S^2 = 126; \, \bar{X} = 59.1; \, \sigma^2 = 81 \)

Answer :

a) The 95% confidence interval is (137.6, 148.4). b) The 95% confidence interval is (227.3, 237.1). c) The 95% confidence interval is (58.1, 60.1).

a. To calculate the 95% confidence interval estimate for the population mean in part (a), we can use the following formula:

(X - 1.96σ/√n, X + 1.96σ/√n)

where:

X is the sample mean

σ is the population standard deviation (we are assuming that we know the population standard deviation in this case)

n is the sample size

In this case, we have:

X = 143

σ = 9

n = 70

Plugging these values into the formula, we get the following confidence interval:

(143 - 1.96 × 9/√70, 143 + 1.96 × 9/√70)

(137.6, 148.4)

Therefore, we can be 95% confident that the true population mean is between 137.6 and 148.4.

b. To calculate the 95% confidence interval estimate for the population mean in part (b), we can use the following formula:

(X - 1.96σ/√n, X + 1.96σ/√n)

where:

X is the sample mean

σ is the sample standard deviation

n is the sample size

In this case, we have:

X = 232.2

σ = 9

n = 80

Plugging these values into the formula, we get the following confidence interval:

(232.2 - 1.96 × 9/√80, 232.2 + 1.96 × 9/√80)

(227.3, 237.1)

Therefore, we can be 95% confident that the true population mean is between 227.3 and 237.1.

c. To calculate the 95% confidence interval estimate for the population mean in part (c), we can use the following formula:

(X - 1.96√[tex]\sigma^2[/tex]/√n, X + 1.96√[tex]\sigma^2[/tex]/√n)

where:

X is the sample mean

σ^2 is the population variance (we are not given the population standard deviation in this case, so we have to estimate it using the sample standard deviation s)

n is the sample size

In this case, we have:

X = 59.1

[tex]s^2[/tex] = 126

n = 200

We can estimate the population variance [tex]\sigma^2[/tex] as follows:

[tex]\sigma^2[/tex] = [tex]s^2[/tex]/n = 126/200 = 0.63

Plugging these values into the formula, we get the following confidence interval:

(59.1 - 1.96√0.63/√200, 59.1 + 1.96√0.63/√200)

(58.1, 60.1)

Therefore, we can be 95% confident that the true population mean is between 58.1 and 60.1.

To learn more about confidence interval here:

https://brainly.com/question/34861571

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