9-2 (p. 292)

diff (A-B): 13, 5, 7, 10, 15
avg diff = 10
sample variance of diff: 17

a.

95% confidence interval:
10 - (2.78)sqrt(17/5), 10 + (2.78)sqrt(17/5)
(4.9, 15.1)

b.

point estimate for average diff (A-B) = 10
sampling allowance = (2.78)sqrt(17/5) = 5.1 (approx)
So, yes, the point estimate exceeds the sampling allowance.

c. Yes. (The confidence interval doesn't include 0.)

d. Yes

e. Null hypothesis: delta = 0 (average difference A-B = 0)

9-9 (p. 304)

a.

null hypothesis: pi = 0.10, proportion of bad glove pairs is the same as before
alternative hypothesis: pi > 0.10, proportion of bad glove pairs is greater than before

b.

Critical value of P:
(P-pi)/sqrt[pi(1-pi)/n] = 1.34
(P-0.10)/sqrt[0.10(0.90)/100] = 1.34
P = 0.1402

c.Any shipment with a proportion of defective gloves larger than 14%, so that would be all shipments listed except for the first (12%) and the third (8%).

9-12
b.

Test statistic: Z = [0.24-0.34]/sqrt[((.24)(.76)/420) + ((.34)(.66)/510)] = -3.38 (approx)
The null hypothesis can be rejected; the difference is statistically discernible (statistically significant) at the 5% level.

9-23 (p. 319)

a.

95% confidence interval for the change in proportion from 1972 to 1975
(.42-.45) - (1.96)sqrt[((.42)(.58)/1500) + ((.45)(.55)/1500)], (.42-.45) + (1.96)sqrt[((.42)(.58)/1500) + ((.45)(.55)/1500)]
(-0.07, 0.01)

c.

(i.) No, the null hypothesis can't be rejected, because 0 falls within the confidence interval. So, the increase is not statistically discernible (statistically significant).

9-25 (p. 319)

a. 95% one-sided confidence interval: mu > 14740 - (1.64)[2000/sqrt(25)]
mu > 14084

c.

(i.) Can reject null hypothesis, because 14000 is not in the confidence interval above. The new process is discernably better than the old.