Section 8.4

#13. Given info: n = 40, X-bar = 58.3, sigma = 9.5

We are to "test the claim that the population mean is equal to 60 ...," so one of our hypotheses must be "µ = 60 ." This must be H_{0}, since H_{0} must have an equal sign in it. That leaves "µ ≠ 60" as the H_{1}. Notice that we are given a value for sigma, so we can use a z-test. Also, the sample size is large, so we need not make any assumptions of normality.

Our test (using the classical approach):

H_{0}: µ = 60 , H_{1}: µ ≠ 60 (two tail test)

Choose alpha = 0.05 (that was actually given)

DR: If z_{X-bar} < -1.96 or z_{X-bar} > 1.96 then reject H_{0} (1.96 was the z-score that chopped off the upper and lower 2.5% of area)

We find z_{X-bar} = (X-bar - µ)/[sigma/sqrt(n)] = (58.3 - 60)/[9.5/sqrt(40)] = -1.7/1.502 = -1.13.

Since -1.13 is NOT less than -1.96, we fail to reject H_{0}.

So the population mean is indeed 60.

Follow-up question: Yes, the overall perception of 1 minute seems to be accurate.

#15. Given info: n = 40, X-bar = -2.1, sigma = 4.8

We are to "test the claim that the mean weight change is less than 0," so one of our hypotheses must be "µ < 0 ." This must be H_{1}, since it does not have an equal sign in it. That leaves "µ = 0" as the H_{0}. Notice that we are given a value for sigma, so we can use a z-test. Also, the sample size is large, so we need not make any assumptions of normality.

Our test (using the classical approach):

H_{0}: µ = 0 , H_{1}: µ < 0 (one tail test)

Choose alpha = 0.05 (that was actually given)

DR: If z_{X-bar} < -1.645 then reject H_{0} (-1.645 was the z-score that chopped off the lowest 5% of area)

We find z_{X-bar} = (X-bar - µ)/[sigma/sqrt(n)] = (-2.1 - 0)/[4.8/sqrt(40)] = -2.1/0.7589 = -2.77.

Since -2.77 is less than -1.645, we reject H_{0}.

So the mean weight change is indeed less than 0.

Follow-up question: It is true that the weight loss plan seems to be effective, but the average weight loss is not so great that it is worth much extra cost or effort.

Section 8.5

#17. Given info: n = 106, X-bar = 98.20, s = 0.62

We are to "test the claim that the mean body temperature is less than 98.6," so one of our hypotheses must be "µ < 98.6." This must not be H_{0}, since H_{0} must have an equal sign in it. That leaves "µ = 98.6 " as the H_{0}. Notice that we are not given a value for sigma, so we must use a t-test. Also, the sample size is large, so we need not make any assumptions of normality.

Our test (using the classical approach):

H_{0}: µ = 98.6 , H_{1}: µ < 98.6 (one tail test)

Choose alpha = 0.05 (that was actually given)

DR: If t_{X-bar} < -1.660 then reject H_{0} (-1.660 was the t-score with 100 degrees of freedom that chopped off the lower 5% of area)

We find t_{X-bar} = (X-bar - µ)/[s/sqrt(n)] = (98.20 - 98.6)/[0.62/sqrt(106)] = -0.4/0.0602 = -6.64.

Since -6.64 is less than -1.660, we reject H_{0}.

So the population mean is below 98.6.

Follow-up question: Yes, it appears the common mean of 98.6 is wrong.

#19. Given info: n = 190, X-bar = 2700, s = 645

We are to "test the claim that weights of these babies have a mean that is less than 3103," so one of our hypotheses must be "µ < 3103." This must not be H_{0}, since H_{0} must have an equal sign in it. That leaves "µ = 3103 " as the H_{0}. Notice that we are not given a value for sigma, so we must use a t-test. Also, the sample size is large, so we need not make any assumptions of normality.

Our test (using the classical approach):

H_{0}: µ = 3103 , H_{1}: µ < 3103 (one tail test)

Choose alpha = 0.01 (that was actually given)

DR: If t_{X-bar} < -2.345 then reject H_{0} (-2.345 was the t-score closest to 189 degrees of freedom that chopped off the lower 1% of area)

We find t_{X-bar} = (X-bar - µ)/[s/sqrt(n)] = (2700 - 3103)/[645/sqrt(190)] = -403/46.793 = -8.612.

Since -8.612 is less than -2.345, we reject H_{0}.

So the mean weight of these babies is indeed below 3103.

Follow-up question: Yes, it appears that weights are affected by cocaine use.