To determine the views of a students on a particular campus about fraternities, an 11-point attitude scale was administered to a random sample of 40 students. This survey yielded a sample mean of 6 (the higher the score, the more favorable the view of the fraternities) and a standard deviation of 1.5

A) Estimate the standard error of the mean
B) Find the 95% confidence interval for the population mean
C)Find the 99% confidence interval for the population mean

A) SEm = SD/√n

B) 95% = mean ± 1.96 SEm

C) Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.005) and its Z score. Substitute that for the 1.96 in B equation.

A) To estimate the standard error of the mean, we use the formula:

Standard Error of the Mean (SE) = Standard Deviation (σ) / Square Root of Sample Size (n)

Given:
Standard Deviation (σ) = 1.5
Sample Size (n) = 40

Using the formula, we can calculate the standard error of the mean:

SE = 1.5 / √40 ≈ 0.237

Therefore, the estimate of the standard error of the mean is approximately 0.237.

B) To find the 95% confidence interval for the population mean, we use the formula:

Confidence Interval = Sample Mean ± (Critical Value * Standard Error)

Given:
Sample Mean = 6
Standard Error (SE) ≈ 0.237

First, we need to find the critical value for the 95% confidence level. Since the sample size is 40, we can use a t-distribution. Looking up the critical value in the t-table (with degrees of freedom = n-1 = 39) for a 95% confidence level, the two-tailed critical value is approximately 2.022.

Confidence Interval = 6 ± (2.022 * 0.237)

Confidence Interval ≈ 6 ± 0.479

The 95% confidence interval for the population mean is approximately (5.521, 6.479).

C) To find the 99% confidence interval for the population mean, we follow the same process as in part B.

First, we need to find the critical value for the 99% confidence level using the t-table. The critical value for a 99% confidence level with 39 degrees of freedom is approximately 2.704.

Confidence Interval = 6 ± (2.704 * 0.237)

Confidence Interval ≈ 6 ± 0.641

The 99% confidence interval for the population mean is approximately (5.359, 6.641).

To answer these questions, we can follow these steps:

A) Estimate the standard error of the mean:

The standard error of the mean (SE) can be calculated using the formula:

SE = standard deviation / square root of the sample size

In this case, the standard deviation is 1.5 and the sample size is 40.

SE = 1.5 / √40 ≈ 0.238

So, the estimated standard error of the mean is approximately 0.238.

B) Find the 95% confidence interval for the population mean:

To find the 95% confidence interval, we can use the formula:

CI = sample mean ± (critical value * standard error)

The critical value for a 95% confidence level is found in the t-distribution table for a sample size of 40, which is approximately 2.021.

CI = 6 ± (2.021 * 0.238)

CI = 6 ± 0.482

The lower bound of the confidence interval is 6 - 0.482 = 5.518
The upper bound of the confidence interval is 6 + 0.482 = 6.482

So, the 95% confidence interval for the population mean is approximately (5.518, 6.482).

C) Find the 99% confidence interval for the population mean:

Similarly, to find the 99% confidence interval, we can use the formula:

CI = sample mean ± (critical value * standard error)

The critical value for a 99% confidence level is found in the t-distribution table for a sample size of 40, which is approximately 2.704.

CI = 6 ± (2.704 * 0.238)

CI = 6 ± 0.643

The lower bound of the confidence interval is 6 - 0.643 = 5.357
The upper bound of the confidence interval is 6 + 0.643 = 6.643

So, the 99% confidence interval for the population mean is approximately (5.357, 6.643).