The National Center for Education surveyed 4400 college graduates about the lengths of time required to earn their bachelor's degrees. THe sample mean is 5.15 years, and assume that the population standard deviation is known to be 1.68 years. Based on gthe sample data, construct the 99% confidence interval for the mean time required by all college graduates.

Also, determine the margin of error, E.

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To construct a confidence interval for the mean time required by all college graduates, we can use the formula:

Confidence Interval = sample mean ± (Z * (population standard deviation / sqrt(sample size)))

First, let's find the Z-score for a 99% confidence level. The Z-score represents the number of standard deviations from the mean. You can find the Z-score using a Z-table or a statistical calculator.

For a 99% confidence level, we want to find the Z-score that includes 99% of the area under the standard normal distribution curve. The Z-score for a 99% confidence level is approximately 2.576.

Now we can substitute the values into the formula:

Confidence Interval = 5.15 ± (2.576 * (1.68 / sqrt(4400)))

Calculating the expression in the parentheses:

Standard Error = (1.68 / sqrt(4400)) ≈ 0.0253

Substituting the value of the standard error into the formula:

Confidence Interval = 5.15 ± (2.576 * 0.0253)

Calculating the expression in parentheses:

2.576 * 0.0253 ≈ 0.0651

Now we can substitute the value of the expression into the formula:

Confidence Interval = 5.15 ± 0.0651

Calculating the lower and upper bounds:

Lower bound = 5.15 - 0.0651 ≈ 5.0849
Upper bound = 5.15 + 0.0651 ≈ 5.2151

Thus, the 99% confidence interval for the mean time required by all college graduates is approximately 5.0849 to 5.2151 years.

To determine the margin of error (E), we can take half of the width of the confidence interval:

Margin of Error (E) = (Upper bound - Lower bound) / 2

Calculating the expression in parentheses:

(5.2151 - 5.0849) / 2 ≈ 0.0651 / 2 ≈ 0.0326

Therefore, the margin of error (E) is approximately 0.0326 years.