Question

Assume that a sample is used to estimate a population mean μ . Find the 99.9% confidence interval for a sample of size 68 with a mean of 78.3 and a standard deviation of 15.6. Enter your answer as an open-interval (i.e., parentheses) accurate to one decimal place (because the sample statistics are reported accurate to one decimal place).

99.9% C.I. =

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

To find the 99.9% confidence interval for the population mean, we use the formula (sample mean - (critical value)(standard deviation)/sqrt(sample size), sample mean + (critical value)(standard deviation)/sqrt(sample size)). Given the sample mean of 78.3, standard deviation of 15.6, and a sample size of 68, we find a confidence interval of approximately (73.9, 82.7).

Step-by-step explanation:

To find the 99.9% confidence interval, we use the formula:

(sample mean - (critical value)(standard deviation)/sqrt(sample size), sample mean + (critical value)(standard deviation)/sqrt(sample size))

Given that the sample mean is 78.3, the standard deviation is 15.6, and the sample size is 68, we need to find the critical value for a 99.9% confidence level.

The critical value can be found using a z-table or calculator, and it corresponds to the area under the standard normal distribution curve. The critical value for a 99.9% confidence level is approximately 3.290. Plugging in these values into the formula, we get:

(78.3 - (3.290)(15.6)/sqrt(68), 78.3 + (3.290)(15.6)/sqrt(68))

Simplifying the expression gives us an open-interval of approximately (73.9, 82.7).