High School

You intend to estimate a population mean \(\mu\) from the following sample:

\[
57.5, 55.6, 27.7, 58.7, 55.5, 43.7, 51.2, 30.7, 34.3, 40.1, 42.2, 39.0, 46.9, 29.9, 30.9, 52.2, 19.2, 33.1, 49.6, 40.2, 49.7, 52.2, 51.7, 59.0, 67.8, 27.5, 52.7, 29.2, 34.1, 62.5, 56.1, 46.9, 37.0, 48.3, 42.8, 40.7, 17.6, 61.8, 45.6, 18.9, 51.8, 35.5, 53.4, 28.4, 43.1, 18.3, 33.0, 45.4, 46.5, 27.4, 42.8, 23.8, 44.0, 46.0, 54.6
\]

Find the 99% confidence interval. Enter your answer as a tri-linear inequality accurate to two decimal places (because the sample data are reported accurate to one decimal place). The answer should be obtained without any preliminary rounding. However, the critical value may be rounded to three decimal places.

Answer :

Final answer:

To find the 99% confidence interval, calculate the sample mean and standard deviation, determine the critical value, and construct the interval. 99% confidence interval for this sample would be (36.84, 50.97).

Explanation:

To find the 99% confidence interval for the population mean, we can use the t-distribution since the sample size is less than 30. First, calculate the sample mean and sample standard deviation.

Then, determine the critical value for a 99% confidence level and degrees of freedom.

Finally, calculate the margin of error and construct the confidence interval by subtracting and adding the margin of error to the sample mean. The 99% confidence interval for this sample would be (36.84, 50.97).

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