Free-Response
- C = 0.90 z * = 1.645, .
You would need to survey at least 271 students.
The population standard deviation is unknown, and the sample size is small (13), so we need to use a
t procedure. The problem tells us that the sample is random. A histogram of the data shows no
significant departure from normality:
Now, = 98.1, s = 4.21, df = 13 – 1 = 12 t * = 2.179. The 95% confidence interval isBecause 100 is contained in this interval, we do not have strong evidence that the mean number of
bushels per acre differs from 100, even though the sample mean is only 98.1.
This is a two-proportion situation. We are told that the groups were randomly selected, but we need
to check that the samples are sufficiently large:
. Since all values are greater than or equal to 5, we
are justified in constructing a two-proportion z interval. For a 90% z confidence interval, z * = 1.645.
Thus, .We are 90% confident that the true difference between the dropout rates is between 0.02 and 0.38.
Since 0 is not in this interval, we have evidence that the counseling program was effective at reducing
the number of dropouts.
In this problem, H 0 : μ = 5 and H (^) A : μ < 5, so we are only interested in the area to the left of our
finding of = 4.55 since the hotel believes that the average stay is less than 5 days. We are interested
in the area shaded in the graph: