A study conducted by an airline showed that a random sample of nine of its passengers disembarking at the Cleveland airport, took an average of 24.1 minutes to claim their luggage. From a previous survey it was willing to assume that time to claim luggage is normally distributed with a variance of 18 (min)2. A 95% confidence interval for the mean time to claim one's luggage has endpoints.

Question

A study conducted by an airline showed that a random sample of nine of its passengers disembarking at the Cleveland airport, took an average of 24.1 minutes to claim their luggage. From a previous survey it was willing to assume that time to claim luggage is normally distributed with a variance of 18 (min)2. A 95% confidence interval for the mean time to claim one's luggage has endpoints.

24.1 ± 8.32

24.1 ± 3.92

24.1 ± 2.77

24.1 ± 3.26

24.1 ± 9.78

Answer 1

To calculate the confidence interval for the mean time to claim luggage, we can use the formula:

Confidence interval = sample mean ± (critical value x standard error)

First, let's find the critical value. Since the sample size is small (n = 9) and we don't know the population standard deviation, we need to use the t-distribution.

For a 95% confidence level and 8 degrees of freedom (n - 1), the t-critical value is approximately 2.306.

Next, let's find the standard error. The formula for the standard error is:

Standard error = sqrt(variance / sample size)

Given that the variance is 18 and the sample size is 9:

Standard error = sqrt(18 / 9) = sqrt(2) ≈ 1.41

Now we can calculate the confidence interval:

Confidence interval = 24.1 ± (2.306 x 1.41)
Confidence interval = 24.1 ± 3.25

Therefore, the 95% confidence interval for the mean time to claim one's luggage is 24.1 ± 3.26 (option D).

Answer 2

To find the confidence interval for the mean time to claim one's luggage, we can use the formula:

Confidence Interval = sample mean ± (critical value) x (standard deviation / square root of sample size)

Given:
Sample mean (x̄) = 24.1
Standard deviation (σ) = sqrt(18) = 4.24 (since variance = 18)

The critical value depends on the desired confidence level. For a 95% confidence level, we use a critical value of 1.96.

Plugging in the values into the formula:

Confidence Interval = 24.1 ± (1.96) x (4.24 / sqrt(9))

Simplifying the equation:

Confidence Interval = 24.1 ± (1.96) x (4.24 / 3)

Confidence Interval = 24.1 ± 2.77

Therefore, the 95% confidence interval for the mean time to claim one's luggage has endpoints of 24.1 ± 2.77.

The correct answer is 24.1 ± 2.77.

Answer 3

CI95 = mean ± 1.96(sd/√n)

Substitute the values given into the formula to choose your answer.

Note: standard deviation is the square root of the variance.