Question

Given a sample of 65 college graduates who have taken a statistics course, with a mean starting salary of $42,429 and a sample standard deviation of $8,838, you want to construct confidence intervals using 99% confidence.

(a) Calculate the margin of error (E) for this 99% confidence interval. Provide your answer to at least 2 decimal places.

(b) Determine the 99% confidence interval for the mean starting salary (μ). Express the interval as <μ<, and use 3 decimal places in your critical value.

Answer 1

The margin of error (E) for a 99% confidence interval for the mean starting salary is calculated as $2,825.85. Using this margin of error, the 99% confidence interval for the mean starting salary (μ) is determined to be between $39,603.150 and $45,254.850.

Step-by-step explanation:

To calculate the margin of error (E) for a 99% confidence interval, you first need to use the Z-score that corresponds to the desired confidence level. For 99% confidence, the Z-score is typically found using a standard normal distribution table or statistical software and is approximately 2.576.

The formula for the margin of error (E) is:

E = Z * (s / √n)

Where Z is the Z-score, s is the sample standard deviation, and n is the sample size. Plugging in the values given:

E = 2.576 * (8838 / √65) = 2.576 * (1097.47) ≈ 2825.85 (rounded to two decimal places)

For part (b), to determine the 99% confidence interval for the mean starting salary (μ), we use the sample mean and the margin of error:

Confidence Interval = (x - E, x + E)

Plugging in the sample mean and margin of error:

Confidence Interval = (42,429 - 2825.85, 42,429 + 2825.85)

= (39,603.15, 45,254.85)

Thus, with 99% confidence, we estimate that the true mean starting salary for college graduates who have taken a statistics course is between \$39,603.150 and \$45,254.850.