Question

In a random sample of 11 people the mean driving distance to work was

18.1 miles and the standard deviation was 7.8 miles. Assume the population is normally distributed and use the t distribution to find the margin of error and construct a 90% confidence interval for the population mean. Interpret the results.

Answer 1

To construct a 90% confidence interval for the population mean driving distance to work, we use the t-distribution with the given sample data. The margin or error is approximately 5.702 miles, and the confidence interval is (12.398, 23.802) miles.

Step-by-step explanation:

To construct a 90% confidence interval for the population mean driving distance to work, we can use the t-distribution since the population standard deviation is unknown. The margin of error can be calculated using the formula:

Margin of Error = t * (standard deviation / sqrt(sample size))

For a 90% confidence level, we can find the t-value with a degrees of freedom of 10 (sample size - 1). From a t-table or using software, the t-value is approximately 1.833. Plugging in the values from the sample, the margin of error is:

Margin of Error = 1.833 * (7.8 / sqrt(11)) ≈ 5.702

To construct the confidence interval, we can subtract and add the margin of error from the sample mean:

Confidence Interval = mean ± margin of error = 18.1 ± 5.702

Therefore, the 90% confidence interval for the population mean driving distance to work is approximately (12.398, 23.802). This means that we can be 90% confident that the true population mean driving distance to work falls within this range.