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 H0, since H0 must have an equal sign in it. That leaves "µ ≠ 60" as the H1. 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):
H0: µ = 60 , H1: µ ≠ 60 (two tail test)
Choose alpha = 0.05 (that was actually given)
DR: If zX-bar < -1.96 or zX-bar > 1.96 then reject H0 (1.96 was the z-score that chopped off the upper and lower 2.5% of area)
We find zX-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 H0.
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 H1, since it does not have an equal sign in it. That leaves "µ = 0" as the H0. 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):
H0: µ = 0 , H1: µ < 0 (one tail test)
Choose alpha = 0.05 (that was actually given)
DR: If zX-bar < -1.645 then reject H0 (-1.645 was the z-score that chopped off the lowest 5% of area)
We find zX-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 H0.
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 H0, since H0 must have an equal sign in it. That leaves "µ = 98.6 " as the H0. 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):
H0: µ = 98.6 , H1: µ < 98.6 (one tail test)
Choose alpha = 0.05 (that was actually given)
DR: If tX-bar < -1.660 then reject H0 (-1.660 was the t-score with 100 degrees of freedom that chopped off the lower 5% of area)
We find tX-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 H0.
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 H0, since H0 must have an equal sign in it. That leaves "µ = 3103 " as the H0. 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):
H0: µ = 3103 , H1: µ < 3103 (one tail test)
Choose alpha = 0.01 (that was actually given)
DR: If tX-bar < -2.345 then reject H0 (-2.345 was the t-score closest to 189 degrees of freedom that chopped off the lower 1% of area)
We find tX-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 H0.
So the mean weight of these babies is indeed below 3103.

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

 


Return to the stats page