Question

A sample of size n = 92 is drawn from a normal population whose standard deviation is 9.5. Construct a 90% confidence interval for μ. Round the answer to at least two decimal places.

a) A 90% confidence interval for the mean is [35.22, 40.06].
b) A 90% confidence interval for the mean is [37.14, 39.14].
c) A 90% confidence interval for the mean is [36.78, 39.50].
d) A 90% confidence interval for the mean is [33.50, 41.78].

Answer 1

To construct a 90% confidence interval for the mean with a sample size of 92 and a known standard deviation of 9.5, we use the formula CI = X ± z * (s/√n). Rounding to two decimal places, the correct confidence interval is [35.22, 40.06]. Therefore, the correct option is a).

Step-by-step explanation:

To construct a confidence interval, we can use the formula:

CI = X ± z * (σ/√n)

Where CI is the confidence interval, X is the sample mean, σ is the population standard deviation, n is the sample size, and z is the z-score corresponding to the desired confidence level.

In this case, the sample mean X is unknown, so we use the formula:

CI = X ± z * (s/√n)

Where s is the sample standard deviation.

Given that the sample size n is 92, the standard deviation s is 9.5, and we want a 90% confidence interval, we can find the z-score by using a z-table or calculator. With a 90% confidence level, the z-score is approximately 1.645.

Now we can calculate the confidence interval:

CI = X ± 1.645 * (9.5/√92)

Rounding to two decimal places:

CI = X ± 1.645 * (0.992)

CI = X ± 1.63

Therefore, the correct option is a) A 90% confidence interval for the mean is [35.22, 40.06].