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Assuming that the sample statistics are:

- \( n = 36 \)
- Mean = 66.9
- \( s = 15.4 \)

What is the resulting 99% confidence interval for the population mean?

Answer :

To calculate the 99% confidence interval for the population mean, we can use the formula:

Confidence Interval = sample mean ± (critical value) * (standard deviation / √(sample size))

Given that the sample size (n) is 36, the sample mean is 66.9, and the sample standard deviation (s) is 15.4, we can calculate the critical value corresponding to a 99% confidence level.

Since we don't have the population standard deviation, we can use the t-distribution and the degrees of freedom (df = n - 1 = 36 - 1 = 35) to find the critical value. From a t-table or using a statistical calculator, the critical value for a 99% confidence level with 35 degrees of freedom is approximately 2.718.

Now we can substitute the values into the confidence interval formula:

Confidence Interval = 66.9 ± (2.718) * (15.4 / √36)

Calculating the square root and simplifying, we have:

Confidence Interval = 66.9 ± 2.718 * 2.567

Confidence Interval ≈ 66.9 ± 6.986

Therefore, the resulting 99% confidence interval for the population mean is approximately (59.914, 73.886).

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