Section 9-2
1. The samples are simple random samples that are independent. For each of the two groups, the number of successes is at least 5 and the number of failures is at least 5. (Depending on what we call a success, the four numbers are 33, 115, 201,229 and 200,745 and all of those numbers are at least 5.) The requirements are satisfied. 2. n1=201,229, ˆ1 33 0.000163992 201,299 p = = , qˆ1= −1 0.000163992=0.999836, n2=200,745, 2 115 ˆ 0.000572866 200, 745 p = = , qˆ2 = −1 0.000572866=0.999427, 33 115 0.000368183 201, 299 200, 745 p= + = + , and q= −1 0.000368183=0.999632. 3. a. H0: p1=p2. H1: p1<p2.
b. If the P-value is less than 0.001 we should reject the null hypothesis and conclude that there is sufficient evidence to support the claim that the rate of polio is less for children given the Salk vaccine than it is for children given a placebo.
4. a. 0.90, or 90%
b. Because the confidence interval limits do not contain 0, there appears to be a significant difference between the two proportions. Because the confidence interval consists of negative values only, it appears that the first proportion is less than the second proportion. There is sufficient evidence to support the claim that the rate of polio is less for children given the Salk vaccine than it is for children given a placebo.
c. The P-value method and the critical value method are equivalent in the sense that they will always lead to the same conclusion, but the confidence interval method is not equivalent to them.
5. Test statistic: z= −12.39 (rounded). The P-value of 3.137085E–35 is 0.0000 when rounded to four decimal places. There is sufficient evidence to warrant rejection of the claim that the vaccine has no effect. 6. Test statistic: z=2.17. P-value: 0.030. Because the P-value is greater than the significance level of 0.01,
conclude that there is not sufficient evidence to warrant rejection of the claim that for those saying that monitoring e-mail is seriously unethical, the proportion of workers is the same as the proportion of managers.
For Exercises 7 – 18, assume that the data fit the requirements for the statistical methods for two proportions unless otherwise indicated.
7. a. H0: p1=p2. H1: p1>p2. Test statistic: z=6.44. Critical value: z=2.33. P-value: 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that the proportion of people over
55 who dream in black and white is greater than the proportion for those under 25.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs > 0): Z = 6.44 P-Value = 0.000
b. 98% CI: 0.117 <p1−p2< 0.240. Because the confidence interval limits do not include 0, it appears that the two proportions are not equal. Because the confidence interval limits include only positive values, it appears that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25.
MINITAB
Difference = p (1) - p (2)
7. (continued)
c. The results suggest that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25, but the results cannot be used to verify the cause of that difference. 8. a. H0: p1=p2. H1: p1<p2. Test statistic: z= −1.66. Critical value: z= −2.33. P-value: 0.0485
(Tech: 0.0484). Fail to reject H0. There is not sufficient evidence to support the claim that the rate of
dementia among those who use ginkgo is less than the rate of dementia among those who use a placebo. There is not sufficient evidence to support the claim that ginkgo is effective in preventing dementia.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs < 0): Z = -1.66 P-Value = 0.048
b. 98% CI: –0.0542 <p1−p2< 0.00909 (Tech: –0.0541 <p1−p2< 0.00904). Because the confidence interval limits include 0, there does not appear to be a significant difference between dementia rates for those treated with ginkgo and those given a placebo. There is not sufficient evidence to support the claim that the rate of dementia among those who use ginkgo is less than the rate of dementia among those who use a placebo. There is not sufficient evidence to support the claim that ginkgo is effective in preventing dementia.
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (-0.0541115, 0.00904103)
c. The sample results suggest that ginkgo is not effective in preventing dementia.
9. a. H0: p1=p2. H1: p1>p2. Test statistic: z=6.11. Critical value: z=1.645. P-value: 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that the fatality rate is higher for
those not wearing seat belts.
MINITAB
Test for difference = 0 (vs > 0): Z = 6.11 P-Value = 0.000
b. 90% CI: 0.00556 <p1−p2<0.0122. Because the confidence interval limits do not include 0, it appears that the two fatality rates are not equal. Because the confidence interval limits include only positive values, it appears that the fatality rate is higher for those not wearing seat belts.
MINITAB
Difference = p (1) - p (2)
90% CI for difference: (0.00558525, 0.0122561)
c. The results suggest that the use of seat belts is associated with lower fatality rates than not using seat belts.
10. a. H0: p1=p2. H1: p1≠p2. Test statistic: z=18.26. Critical values: z= ±2.575 (Tech: ±2.576). P- value: 0.0002 (Tech: 0.0000). Reject H0. There is sufficient evidence to warrant rejection of the claim
that the survival rates are the same for day and night.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 18.26 P-Value = 0.000
b. 99% CI: 0.0441 <p1−p2< 0.0579. Because the confidence interval limits do not contain 0, there appears to be a significant difference between the two proportions. There is sufficient evidence to warrant rejection of the claim that the survival rates are the same for day and night.
MINITAB
Difference = p (1) - p (2)
99% CI for difference: (0.0441419, 0.0579311)
c. The data suggest that for in-hospital patients who suffer cardiac arrest, the survival rate is not the same for day and night. It appears that the survival rate is higher for in-hospital patients who suffer cardiac arrest during the day.
11. a. H0: p1=p2. H1: p1≠p2. Test statistic: z=0.57. Critical values: z= ±1.96. P-value: 0.5686 (Tech: 0.5720). Fail to reject H0. There is not sufficient evidence to support the claim that echinacea
treatment has an effect.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 0.57 P-Value = 0.572
b. 95% CI: –0.0798 <p1−p2< 0.149. Because the confidence interval limits do contain 0, there is not a significant difference between the two proportions. There is not sufficient evidence to support the claim that echinacea treatment has an effect.
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (-0.0798112, 0.148851)
c. Echinacea does not appear to have a significant effect on the infection rate. Because it does not appear to have an effect, it should not be recommended.
12. a. H0: p1=p2. H1: p1<p2. Test statistic: z= −2.44. Critical value: z= −2.33. P-value: 0.0074. Reject H0. There is sufficient evidence to support the claim that the incidence of malaria is lower for
infants who use the bednets.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs < 0): Z = -2.44 P-Value = 0.007
b. 98% CI: 0.0950 <p1−p2<–0.00118 (Tech: –0.0950 <p1−p2< –0.00125). Because the confidence interval does not include 0 and it includes only negative values, it appears that the rate of malaria is lower for infants who use the bednets.
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (-0.0949568, -0.00125315)
c. The bednets appear to be effective.
13. a. H0: p1=p2. H1: p1≠p2. Test statistic: z=0.40. Critical values: z= ±1.96. P-value: 0.6892 (Tech: 0.6859). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that
men and women have equal success in challenging calls.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 0.40 P-Value = 0.686
b. 95% CI: –0.0318 <p1−p2<0.0484. Because the confidence interval limits contain 0, there is not a significant difference between the two proportions. There is not sufficient evidence to warrant rejection of the claim that men and women have equal success in challenging calls.
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (-0.0318350, 0.0484421)
c. It appears that men and women have equal success in challenging calls.
14. a. H0: p1=p2. H1: p1≠p2. Test statistic: z=1.91. Critical values: z= ±1.96. P-value: 0.0562 (Tech: 0.0567). Fail to reject H0. There is not sufficient evidence to warrant rejection of the claim that New
York City police and Los Angeles police have the same proportion of hits.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 1.91 P-Value = 0.057
b. 95% CI:–0.000455 <p1−p2<0.130 (Tech: –0.000454 <p1−p2<0.130). Because the confidence interval limits contain 0, there does not appear to be a significant difference between the two
14. (continued)
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (-0.000453716, 0.130358)
c. There does not appear to be a difference between the hit rates of New York City police and Los Angeles police.
15. a. H0: p1=p2. H1: p1>p2. Test statistic: z=9.97. Critical value: z=2.33. P-value: 0.0001 (Tech: 0.0000). Reject H0. There is sufficient evidence to support the claim that the cure rate with oxygen
treatment is higher than the cure rate for those given a placebo. It appears that the oxygen treatment is effective.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs > 0): Z = 9.97 P-Value = 0.000
b. 98% CI: 0.467 <p1−p2<0.687. Because the confidence interval limits do not include 0, it appears that the two cure rates are not equal. Because the confidence interval limits include only positive values, it appears that the cure rate with oxygen treatment is higher than the cure rate for those given a placebo. It appears that the oxygen treatment is effective.
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (0.467454, 0.687321)
c. The results suggest that the oxygen treatment is effective in curing cluster headaches.
16. a. H0: p1=p2. H1: p1<p2. Test statistic: z= −1.85. Critical value: z= −1.645. P-value: 0.0322 (Tech: 0.0324). Reject H0. There is sufficient evidence to support the claim that when given a single
large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs < 0): Z = -1.85 P-Value = 0.032
b. 90% CI: –0.201 <p1−p2<–0.0127. Because the confidence interval does not include 0 and it includes only negative values, it appears that the first proportion is less than the second proportion. There is sufficient evidence to support the claim that when given a single large bill, a smaller proportion of women in China spend some or all of the money when compared to the proportion of women in China given the same amount in smaller bills.
MINITAB
Difference = p (1) - p (2)
90% CI for difference: (-0.200605, -0.0127280)
c. Because the P-value is 0.0322 (Tech: 0.0324), the difference is significant at the 0.05 significance level, but not at the 0.01 significance level. The conclusion does change.
17. a. H0: p1=p2. H1: p1<p2. Test statistic: z= −1.17. Critical value: z= −2.33. P-value: 0.1210 (Tech: 0.1214). Fail to reject H0. There is not sufficient evidence to support the claim that the rate of
left-handedness among males is less than that among females.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs < 0): Z = -1.17 P-Value = 0.121
b. 98% CI: –0.0849 <p1−p2<0.0265 (Tech: –0.0848 <p1−p2< 0.0264). Because the confidence interval limits include 0, there does not appear to be a significant difference between the rate of left- handedness among males and the rate among females. There is not sufficient evidence to support the claim that the rate of left-handedness among males is less than that among females.
17. (continued)
MINITAB
Difference = p (1) - p (2)
98% CI for difference: (-0.0847744, 0.0264411)
c. The rate of left-handedness among males does not appear to be less than the rate of left-handedness among females.
18. a. H0: p1=p2. H1: p1≠p2. Test statistic: z=2.30. Critical values: z= ±2.575 (Tech: ±2.576). P- value: 0.0214 (Tech: 0.0213). Fail to reject H0. There is not sufficient evidence to warrant rejection of
the claim that the rate of those who finish is the same for men and women.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 2.30 P-Value = 0.021
b. 99% CI: –0.000409 <p1−p2<0.00553 (Tech: –0.000409 <p1−p2<0.00554). Because the confidence interval limits contain 0, there does not appear to be a significant difference between the two proportions. There is not sufficient evidence to warrant rejection of the claim that the rate of those who finish is the same for men and women.
MINITAB
Difference = p (1) - p (2)
99% CI for difference: (-0.000386669, 0.00562869)
c. It appears that men and women finish the New York City marathon at the same rate.
19. a. 0.0227 <p1−p2<0.217; because the confidence interval limits do not contain 0, it appears that 1 2
p =p can be rejected.
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (0.0227099, 0.217290)
b. 0.491 <p1<0.629; 0.371 <p2<0.509; because the confidence intervals do overlap, it appears that 1 2 p =p cannot be rejected. MINITAB Sample X N Sample p 95% CI 1 112 200 0.560000 (0.488250, 0.629944) 2 88 200 0.440000 (0.370056, 0.511750
c. H0: p1=p2. H1: p1≠p2. Test statistic: z=2.40. P-value: 0.0164. Critical values: z= ±1.96. Reject H0. There is sufficient evidence to reject p1=p2.
MINITAB
Difference = p (1) - p (2)
Test for difference = 0 (vs not = 0): Z = 2.40 P-Value = 0.016
d. Reject p1=p2. Least effective: Using the overlap between the individual confidence intervals. 20. Hypothesis test: With a test statistic of z= −1.9615, P-value = 0.05 (Tech: 0.0498), reject p1=p2.
Confidence interval: –0.422 <p1−p2<0.0180, which suggests that we should not reject p1=p2 (because 0 is included). The hypothesis test and confidence interval lead to different conclusions about the equality of p1=p2.
MINITAB
Difference = p (1) - p (2)
95% CI for difference: (-0.422046, 0.0180456)
Test for difference = 0 (vs not = 0): Z = -1.96 P-Value = 0.050
21. 2 2 / 2 2 2 1.645 3383 2 2 0.02 α z n E = = = ⋅ (Tech: 3382)
Section 9-3
1. Independent: b, d, e
2. –17.32 cm <μ1−μ2< –11.61 cm
3. Because the confidence interval does not contain 0, it appears that there is a significant difference between the mean height of women and the mean height of men. Based on the confidence interval, it appears that the mean height of men is greater than the mean height of women.
4. a. Yes.
b. 90%
5. a. H0: μ1=μ2. H1: μ1≠μ2. Test statistic: t= −2.979. Critical values: t= ±2.032 (Tech: ±2.002). P- value < 0.01 (Tech: 0.0042). Reject H0. There is sufficient evidence to warrant rejection of the claim
that the samples are from populations with the same mean. Color does appear to have an effect on creativity scores. Blue appears to be associated with higher creativity scores.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = -2.98 P-Value = 0.004 DF = 58
b. 95% CI: –0.98 <μ1−μ2< –0.18 (Tech: –0.97 <μ1−μ2< –0.19)
MINITAB
Difference = mu (1) - mu (2)
95% CI for difference: (-0.970, -0.190)
6. a. H0: μ1=μ2. H1: μ1≠μ2. Test statistic: t=2.647. Critical values: t= ±2.032 (Tech: ±1.995). P- value < 0.02 (Tech: 0.0101). Reject H0. There is sufficient evidence to warrant rejection of the claim
that the samples are from populations with the same mean. Color does appear to have an effect on word recall scores. Red appears to be associated with higher word memory recall scores.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = 2.65 P-Value = 0.010 DF = 68
b. 95% CI: 0.83 <μ1−μ2< 6.33 (Tech: 0.88 <μ1−μ2< 6.28)
MINITAB
Difference = mu (1) - mu (2) 95% CI for difference: (0.88, 6.28)
7. a. H0: μ1=μ2. H1: μ1>μ2. Test statistic: t=0.132. Critical value: t=1.729. P-value > 0.10 (Tech: 0.4480). Fail to reject H0. There is not sufficient evidence to support the claim that the magnets are
effective in reducing pain. It is valid to argue that the magnets might appear to be effective if the sample sizes are larger.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs >): T-Value = 0.13 P-Value = 0.448 DF = 33
b. 90% CI: –0.61 <μ1−μ2< 0.71 (Tech: –0.59 <μ1−μ2< 0.69)
MINITAB
Difference = mu (1) - mu (2)
90% CI for difference: (-0.592, 0.692)
8. a. H0: μ1=μ2. H1: μ1<μ2. Test statistic: t= −0.676. Critical value: t= −2.345 (Tech: –2.337). P- value > 0.10 (Tech: 0.2499). Fail to reject H0. There is not sufficient evidence to support the claim that
the mean number of words spoken in a day by men is less than that for women.
MINITAB
Difference = mu (1) - mu (2)
8. (continued)
b. 98% CI: –2443.6 words <μ1−μ2< 1350.6 words (Tech: –2436.8 words <μ1−μ2< 1343.8 words)
MINITAB
Difference = mu (1) - mu (2)
98% CI for difference: (-2437, 1344)
9. a. The sample data meet the loose requirement of having a normal distribution. H0: μ1=μ2. H1: μ1>μ2. Test statistic: t=0.852. Critical value: t=2.426 (Tech: 2.676). P-value > 0.10 (Tech: 0.2054). Fail to reject H0. There is not sufficient evidence to support the claim that men have a higher mean body
temperature than women.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs >): T-Value = 0.85 P-Value = 0.206 DF = 12
b. 98% CI: –0.54DF <μ1−μ2< 1.02 D F (Tech: –0.51DF <μ1−μ2< 0.99 D F) MINITAB Difference = mu (1) - mu (2) 98% CI for difference: (-0.515, 0.995)
10. a. H0: μ1=μ2. H1: μ1≠μ2. Test statistic: t=1.559. Critical values: t= ±2.977 (Tech: ±2.789). P- value > 0.10 (Tech: 0.1316). Fail to reject H0. There is not sufficient evidence to support the claim that
men and women have different mean body temperatures.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = 1.56 P-Value = 0.132 DF = 24
b. 99% CI: –0.19°F <μ1−μ2< 0.61°F (Tech: –0.17°F <μ1−μ2< 0.59°F)
MINITAB
Difference = mu (1) - mu (2)
99% CI for difference: (-0.167, 0.587)
11. a. H0: μ1=μ2. H1: μ1<μ2. Test statistic: t= −3.547. Critical value: t= −2.462 (Tech: –2.392). P- value < 0.005 (Tech: 0.0004). Reject H0. There is sufficient evidence to support the claim that the
mean maximal skull breadth in 4000 b.c. is less than the mean in a.d. 150.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs <): T-Value = -3.55 P-Value = 0.000 DF = 57
b. 98% CI: –8.13 mm <μ1−μ2< –1.47 mm (Tech: –8.04 mm <μ1−μ2< –1.56 mm)
MINITAB
Difference = mu (1) - mu (2) 98% CI for difference: (-8.04, -1.56)
12. a. H0: μ1=μ2. H1: μ1≠μ2. Test statistic: t= −0.941. Critical value: t= ±2.201 (Tech: 2.080). P-value > 0.20 (Tech: 0.3573). Fail to reject H0. There is not sufficient evidence to warrant rejection of
the claim that Flight 1 and Flight 3 have the same mean arrival delay time.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = -0.94 P-Value = 0.358 DF = 20
b. 95% CI: –18.1 min <μ1−μ2< 7.3 min (Tech: –17.4 min<μ1−μ2< 6.6 min)
MINITAB
Difference = mu (1) - mu (2)
13. a. H0: μ1=μ2. H1: μ1<μ2. Test statistic: t= −3.142. Critical value: t= −2.462 (Tech: –2.403). P- value < 0.005 (Tech: 0.0014). Reject H0. There is sufficient evidence to support the claim that students
taking the nonproctored test get a higher mean than those taking the proctored test.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs <): T-Value = -3.17 P-Value = 0.001 DF = 49
b. 98% CI: –25.54 <μ1−μ2< –3.10 (Tech: –25.27 <μ1−μ2< –3.37)
MINITAB
Difference = mu (1) - mu (2)
98% CI for difference: (-25.27, -3.37)
14. a. H0: μ1=μ2. H1: μ1≠μ2. Test statistic: t= −0.770. Critical values: t= ±2.756 (Tech: ±2.666). P- value > 0.20 (Tech: 0.4443). Fail to reject H0. There is not sufficient evidence to warrant rejection of
the claim that the samples are from populations with the same mean.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = -0.77 P-Value = 0.444 DF = 56
b. 99% CI: –18.17 <μ1−μ2< 10.23 (Tech: –17.71 <μ1−μ2< 9.77)
MINITAB
Difference = mu (1) - mu (2)
99% CI for difference: (-17.71, 9.77)
15. a. H0: μ1=μ2. H1: μ1≠μ2. Test statistic: t=1.274. Critical values: t= ±2.023 (Tech: ±1.994). P- value > 0.20 (Tech: 0.2066). Fail to reject H0. There is not sufficient evidence to warrant rejection of
the claim that males and females have the same mean BMI.
MINITAB
Difference = mu (1) - mu (2)
T-Test of difference = 0 (vs not =): T-Value = 1.27 P-Value = 0.207 DF = 71
b. 95% CI: –1.08 <μ1−μ2< 4.76 (Tech: –1.04 <μ1−μ2< 4.72)
MINITAB