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).