Question

In each of the studies summarized, determine which type of confidence interval (for p or μ) is appropriate, then compute a 99% confidence interval for the underlying population characteristic. (a) In a recent study, a random sample of 260 adults was asked if they eat healthy foods when they dine out at restaurants. Seventy adults indicated they eat healthy foods at restaurants. (b) A random sample of 21 washing machines was obtained, and the length (in minutes) of each wash cycle was recorded. The sample means was 37.8 minutes, and the sample standard deviation was 5.9 minutes. (Assume the underlying distribution of main wash cycle times is normal.)

Answer 1

We have to compute a 99% confidence interval for the underlying population characteristic.

a)

Sample size = n = 260

Sample proportion is

We have to construct 99% confidence interval for the population proportion.

Formula is

Here E is a margin of error.

Zc = 2.58

So confidence interval is ( 0.70 - 0.0733 , 0.70 + 0.0733) => ( 0.6267 , 0.7733)

-----------------------------------------------------------------------------------------------------------------------------------------

b)

Sample size = n = 21

Sample mean = = 37.8

Standard deviation = s = 5.9

We have to construct 99% confidence interval.

Formula is

Here E is a margin of error.

Degrees of freedom = n - 1 = 21 - 1 = 20

Level of significance = 0.01

tc = 2.845 ( Using t table)

So confidence interval is ( 37.8 - 3.6629 , 37.8 + 3.6629) = > ( 34.1371 , 41.4629)