Inferences Based on Two Samples:
Confidence Intervals and
Tests of Hypotheses Chapter 9
1
9.2a. = μ1 = 12 = .5
b. = μ2 = 10 = .375
c. = 12 = 12 10 = 2
= .625
d.Since n1 30 and n2 30, the sampling distribution of is approximately normal by the Central Limit Theorem.
9.4Assumptions about the two populations:
1. Both sampled populations have relative frequency distributions that are approximately normal.
2.The population variances are equal.
Assumptions about the two samples:
The samples are randomly and independently selected from the population.
9.6a. = = 110
b. = = 14.5714
c. = = .1821
d. = = 2741.9355
e. falls near the variance with the larger sample size.
9.8a. = .5
b.The sampling distribution of
is approximately normal
by the Central Limit Theorem
since n1 30 and n2 30.
= 12 = 10
c. = 15.5 26.6 = 11.1
Yes, it appears that = 11.1 contradicts the null hypothesis H0: 12 =10.
d.The rejection region requires /2 = .025 = .05/2 in each tail of the z-distribution. From Table IV, Appendix B, z.025 = 1.96. The rejection region is z1.96 or z1.96.
e.H0: 12 = 10
Ha: 12 10
The test statistic is z = = 42.2
The rejection region is z1.96 or z > 1.96. (Refer to part d.)
Since the observed value of the test statistic falls in the rejection region (z = 42.2 <1.96), H0 is rejected. There is sufficient evidence to indicate the difference in the population means is not equal to 10 at = .05.
f.The form of the confidence interval is:
For confidence coefficient .95, = 1 .95 = .05 and /2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The confidence interval is:
(15.5 26.6) 1.9611.1 .98 (12.08, 10.12)
We are 95% confident that the difference in the two means is between 12.08 and 10.12.
g.The confidence interval gives more information.
9.10Some preliminary calculations:
= 29.3167
= 53.625
= 29.3167
= 29.6328
a.H0: 21 = 10
Ha: 21 > 10
The test statistic is t = = .013
The rejection region requires = .01 in the upper tail of the t-distribution with df = n1+n2 2 = 15 + 16 2 = 29. From Table VI, Appendix B, t.01 = 2.462. The rejection region is t > 2.462.
Since the test statistic does not fall in the rejection region (t=.013 2.462), H0 is not rejected. There is insufficient evidence to conclude 21 > 10 at = .01.
b.For confidence coefficient .98, = .02 and /2 = .01. From Table VI, Appendix B, with df= n1 + n2 2 = 15 + 16 2 = 29, t.01=2.462. The 98% confidence interval for (21) is:
(53.625 43.6) 2.462
10.025 4.817 (5.208, 14.842)
We are 98% confident that the difference between the mean of population 2 and the mean of population 1 is between 5.208 and 14.842.
9.12a.Let 1 = mean carat size of diamonds certified by GIA and 2 = mean carat size of
diamonds certified by HRD. For confidence coefficient .95, = .05 and /2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval is:
- We are 95% confident that the difference in mean carat size between diamonds certified by GIA and those certified by HRD is between -.1969 and -.0843.
- Let 3 = mean carat size of diamonds certified by IGI.
- We are 95% confident that the difference in mean carat size between diamonds certified by GIA and those certified by IGI is between .2438 and .3678.
e.
- We are 95% confident that the difference in mean carat size between diamonds certified by HRD and those certified by IGI is between .3837 and .5091.
9.14a.Let 1 = mean score for males and 2 = mean score for females. For confidence
coefficient .90, = .10 and /2 = .10/2 = .05. From Table IV, Appendix B, z.025 = 1.645. The 90% confidence interval is:
We are 90% confident that the difference in mean service-rating scores between males and females.
- Because 0 falls in the 90% confidence interval, we are 90% confident that there is no difference in the mean service-rating scores between males and females.
9.16a.The descriptive statistics are:
Descriptive Statistics: US, Japan
Variable N Mean Median TrMean StDev SE Mean
US 5 6.562 6.870 6.562 1.217 0.544
Japan 5 3.118 3.220 3.118 1.227 0.549
Variable Minimum Maximum Q1 Q3
US 4.770 8.000 5.415 7.555
Japan 1.920 4.910 1.970 4.215
To determine if the mean annual percentage turnover for U.S. plants exceeds that for Japanese plants, we test:
H0: 12 = 0
Ha: 12 > 0
The test statistic is t= 4.456
The rejection region requires = .05 in the upper tail of the t-distribution with df = n1+n2 2 = 5 + 5 2 = 8. From Table VI, Appendix B, t.05 = 1.860. The rejection region is t> 1.860.
Since the observed value of the test statistic falls in the rejection region (t = 4.46 >1.860), H0 is rejected. There is sufficient evidence to indicate the mean annual percentage turnover for U.S. plants exceeds that for Japanese plants at = .05.
b.The p-value = P(t 4.456). Using Table VI, Appendix B, with df = + n22 = 5 + 5 – 2
= 8, .005 < P(t 4.456) < .001. Since the p-value is so small, there is evidence to reject
H0 for > .005.
c.The necessary assumptions are:
1.Both sampled populations are approximately normal.
2.The population variances are equal.
3.The samples are randomly and independently sampled.
There is no indication that the populations are not normal. Both sample variances are similar, so there is no evidence the population variances are unequal. There is no indication the assumptions are not valid.
9.18a.Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Females, Males
Variable N Mean Median TrMean StDev SE Mean
Females 90 3.678 4.000 3.763 1.356 0.143
Males 44 3.909 4.000 3.950 0.960 0.145
Variable Minimum Maximum Q1 Q3
Females 1.000 5.000 2.000 5.000
Males 2.000 5.000 3.000 5.000
Let 1 = mean relationship score for females and μ2 = mean relationship score for males. To determine if male students tend to have better relationships, on average, with their fathers than female students, we test:
H0: 1 = 2
Ha: 12
The test statistic is
The rejection region requires = .01 in the lower tail of the zdistribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z2.33.
Since the observed value of the test statistic does not fall in the rejection region (z = 1.14 2.33), H0 is not rejected. There is insufficient evidence to indicate male students tend to have better relationships, on average, with their fathers than female students at = .01.
b.The p-value is P(z1.14) = .5 .3729 = .1271. (Using Table IV, Appendix B.).
c.We have to assume that the samples are independent.
d.Since the sample sizes were both greater than 30 (n1 = 90 and n2 = 44), the Central Limit Theorem applies. Thus, we know the sampling distribution of is approximately normal regardless of the distributions being sampled from.
9.20a.The first population is the set of responses for all business students who have access to lecture notes and the second population is the set of responses for all business students not having access to lecture notes.
b.To determine if there is a difference in the mean response of the two groups, we test:
H0: 12 = 0
Ha: 12 0
The test statistic is z = = 2.19
The rejection region requires /2 = .01/2 = .005 in each tail of the z-distribution. From Table IV, Appendix B, z.005=2.58. The rejection region is z2.58 or z2.58.
Since the observed value of the test statistic does not fall in the rejection region (z = 2.19 2.58), H0 is not rejected. There is insufficient evidence to indicate a difference in the mean response of the two groups at = .01.
c.For confidence coefficient .99, = .01 and /2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The confidence interval is:
(8.48 7.80)
.68 ± .801 (.121, 1.481)
We are 99% confident that the difference in the mean response between the two groups is between .121 and 1.481.
d.A 95% confidence interval would be smaller than the 99% confidence interval. The z value used in the 95% confidence interval is z.025 = 1.96 compared with the z value used in the 99% confidence interval of z.005 = 2.58.
9.22a.The bacteria counts are probably normally distributed because each count is the median of fivemeasurements from the same specimen.
b.Let 1 = mean of the bacteria count for the discharge and 2=mean of the bacteria count upstream. Since we want to test if the mean of the bacteria count for the discharge exceeds the mean of the count upstream, we test:
H0: 12 = 0
Ha: 12 > 0
c.Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Plant, Upstream
Variable N Mean Median TrMean StDev SE Mean
Plant 6 32.10 31.75 32.10 3.19 1.30
Upstream 6 29.617 30.000 29.617 2.355 0.961
Variable Minimum Maximum Q1 Q3
Plant 28.20 36.20 29.40 35.23
Upstream 26.400 32.300 27.075 31.850
The test statistic is
No level was given, so we will use = .05. The rejection region requires = .05 in the upper tail of the t-distribution with df = n1 + n22 = 6 + 6 – 2 = 10. From Table VI, Appendix B, t.05 = 1.812. The rejection region is t > 1.812.
Since the observed value of the test statistic does not fall in the rejection region (t = 1.53 1.812), H0 is not rejected. There is insufficient evidence to indicate the mean bacteria count for the discharge exceeds the mean of the count upstream at = .05.
d.We must assume:
1.The mean counts per specimen for each location is normally distributed.
2.The variances of the 2 distributions are equal.
3.Independent and random samples were selected from each population.
9.24a.A 95% confidence interval cannot be determined because no standard deviations are reported.
b.Let μ1 = mean salary for male entry level managers with a CPA and μ2 = mean salary for female entry level managers with a CPA.
For confidence coefficient .95, α = .05 and α/2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The 95% confidence interval is:
(40,084 35,377) 1.96
4,707 ± 1,472.09 (3,234.91, 6,179.09)
We are 95% confident that the difference between the mean salaries of male and female entry level managers who earned a CPA degree is between $3,234.91 and $6,179.09.
c.The 95% confidence interval is:
(40,084 35,377) 1.96
4,707 5,888.37 (1,181.37, 10,595.37)
We are 95% confident that the difference between the mean salaries of male and female entry level managers who earned a CPA degree is between $1,181.37 and $10,595.37.
d.Since the interval in part b did not contain 0, there is evidence that the mean salary for male entry level managers who earned a CPA is greater than that for females. Since the interval in part c did contain 0, there is no evidence that the mean salary for male entry level managers who earned a CPA is different than that for females.
e.Let us assume that = = 2. In order to find a significant difference between the two means, the test statistic must fall in the rejection region. The rejection region requires /2 = .05/2 = .025 in each tail of the z-distribution. From Table IV, Appendix B, z.025 = 1.96. The rejection region is z1.96 or z > 1.96.
The test statistic is z =
To find , z > 1.96 > 1.96
> 1.96 > 1.96
816 > .3142 < 2,597.17
f.This value of the standard deviation seems rather small, considering the values of the mean salaries.
9.26a.
Pair / Difference1 / 3
2 / 2
3 / 2
4 / 4
5 / 0
6 / 1
= = 2
= = 2
b.d = 12
c.For confidence coefficient .95, = .05 and /2 = .025. From Table VI, Appendix B, with df = nD 1 = 6 1 = 5, t.025 = 2.571. The confidence interval is:
2 1.484 (.516, 3.484)
d.H0: d = 0
Ha: d 0
The test statistic is t = = 3.46
The rejection region requires /2 = .05/2 = .025 in each tail of the t-distribution with df = nD 1 = 6 1 = 5. From Table VI, Appendix B, t.025 = 2.571. The rejection region is t2.571 or t > 2.571.
Since the observed value of the test statistic falls in the rejection region (3.46 > 2.571), H0 is rejected. There is sufficient evidence to indicate that the mean difference is different from 0 at = .05.
9.28a.H0: 12 = 0
Ha: 12 < 0
The rejection region requires = .10 in the lower tail of the z-distribution. From Table IV, Appendix B, z.10 = 1.28. The rejection region is z1.28.
b.H0: 12 = 0
Ha: 12 < 0
The test statistic is .
The rejection region is z < 1.28 (Refer to part a.)
Since the observed value of the test statistic falls in the rejection region (z = 4.71 < 1.28), H0 is rejected. There is sufficient evidence to indicate 12 < 0 at = .10.
c.Since the sample size of the number of pairs is greater than 30, we do not need to assume that the population of differences is normal. The sampling distribution of is approximately normal by the Central Limit Theorem. We must assume that the differences are randomly selected.
d.For confidence coefficient .90, = .10 and /2 = .10/2 = .05. From Table IV, Appendix B, z.05 = 1.645. The 90% confidence interval is:
e.The confidence interval provides more information since it gives an interval of possible values for the difference between the population means.
9.30a.The data should be analyzed as a paired difference experiment because each actor who
won an Academy Award was paired with another actor with similar characteristics who did not win the award.
b.Let 1 = mean life expectancy of Academy Award winners and 2 = mean life expectancy of non-Academy Award winners. To compare the mean life expectancies of Academy Award winners and non-winners, we test:
H0: 12 = d = 0
Ha: d 0
c.Since the p-value was so small, there is sufficient evidence to indicate the mean life expectancies of the Academy Award winners and non-winners are different for any value of > .003. Since the sample mean life expectancy of Academy Award winners is greater than that for non-winners, we can conclude that Academy Award winners have a longer mean life expectancy than non-winners.
9.32a.Let 1 = mean driver chest injury rating and 2 = mean passenger chest injury rating.
Because the data are paired, we are interested in 12 = d, the difference in mean chest injury ratings between drivers and passengers.
b.The data were collected as matched pairs and thus, must be analyzed as matched pairs. Two ratings are obtained for each car – the driver’s chest injury rating and the passenger’s chest injury rating.
c.Using MINITAB, the descriptive statistics are:
Descriptive Statistics: DrivChst, PassChst, diff
Variable N Mean Median TrMean StDev SE Mean
DrivChst 98 49.663 50.000 49.682 6.670 0.674
PassChst 98 50.224 50.500 50.148 7.107 0.718
diff 98 -0.561 0.000 -0.420 5.517 0.557
Variable Minimum Maximum Q1 Q3
DrivChst 34.000 68.000 45.000 54.000
PassChst 35.000 69.000 45.000 55.000
diff -15.000 13.000 -4.000 3.000
For confidence coefficient .99, = .01 and /2 = .01/2 = .005. From Table IV, Appendix B, z.005 = 2.58. The 99% confidence interval is:
d.We are 99% confidence that the difference between the mean chest injury ratings of drivers and front-seat passengers is between 1.999 and 0.877. Since 0 is in the confidence interval, there is no evidence that the true mean driver chest injury rating exceeds the true mean passenger chest injury rating.
e.Since the sample size is large, the sampling distribution of is approximately normal by the Central Limit Theorem. We must assume that the differences are randomly selected.
9.34a.
b.We do not need to estimate anything – we know the parameter’s value.
c.Using MINITAB, the descriptive statistics are:
Descriptive Statistics: 1990 SAT, 2000 SAT, Diff
Variable N Mean Median TrMean StDev SE Mean
1990 SAT 51 1043.7 1034.0 1042.4 59.5 8.3
2000 SAT 51 1066.1 1054.0 1064.1 65.9 9.2
Diff 51 22.45 19.00 21.29 17.66 2.47
Variable Minimum Maximum Q1 Q3
1990 SAT 942.0 1172.0 993.0 1089.0
2000 SAT 966.0 1197.0 1007.0 1120.0
Diff -8.00 70.00 11.00 30.00
Let 1 = mean SAT score in 2000 and 2 = mean SAT score in 1999. Then d =12. To determine if the true mean SAT score in 2000 differs from that in 1999, we test:
H0: d = 0
Ha: d 0
The test statistic is
The rejection region requires /2 = .10/2 = .05 in each tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z1.645 or z > 1.645.
Since the observed value of the test statistic falls in the rejection region (z = 9.08 > 1.645), H0 is rejected. There is sufficient evidence to indicate the true mean SAT score in 2000 is different than that in 1999 at = .10.
9.36a.To determine whether male students' attitudes toward their fathers differ from their attitudes toward their mothers, on average, we test:
H0: d = 0
Ha: d 0
b.Using MINITAB, the descriptive statistics are:
Descriptive Statistics: Father, Mother, Diff
Variable N Mean Median TrMean StDev SE Mean
Father 13 3.846 4.000 3.909 1.068 0.296
Mother 13 4.154 4.000 4.182 0.801 0.222
Diff 13 -0.308 -1.000 -0.273 1.032 0.286
Variable Minimum Maximum Q1 Q3
Father 2.000 5.000 3.000 5.000
Mother 3.000 5.000 3.500 5.000
Diff -2.000 1.000 -1.000 1.000
The test statistic is
The rejection region requires /2 = .05/2 = .025 in each tail of the t-distribution with df
= nd – 1 = 13 – 1 = 12. From Table VI, Appendix B, t.025 = 2.179. The rejection region is t2.179 or t > 2.179.
Since the observed value of the test statistic does not fall in the rejection region (t = 1.08 2.179), H0 is not rejected. There is insufficient evidence to indicate that malestudents’ attitudes toward their fathers differ from their attitudes toward their mothers at
= .05.
c.In order for the above test to be valid, we must assume that
1. The population of differences is normal
2. The differences are randomly selected
d.A stemandleaf display of the differences is:
Character StemandLeaf Display
Stemandleaf of Diff N = 13
Leaf Unit = 0.10
1 2 0
(6) 1 000000
6 0 00
4 1 0000
We know that the data are not normal, because there are only 4 different values for the differences. Although the stemandleaf display does not look particularly normal, it is rather moundshaped. The ttest procedure works pretty well even if the data are not exactly normal.
9.38a.Let 1 = mean trait for senior year in high school and 2 = mean trait during sophomore year of college. Also, let d = 12. To determine if there is a decrease in the mean self-concept of females between the senior year in high school and the sophomore year in college, we test:
H0: d = 0
Ha: d > 0
b.Since the same female students were used in each study, the data should be analyzed as paired differences. The samples are not independent.
c.Since the number of pairs is nd = 133, no assumptions about the population of differences is necessary. We still assume that a random sample of differences is selected.
d.Leadership: The p-value > .05. There is no evidence to reject H0 for = .05. There is no evidence to indicate that there is a decrease in mean self-concept of females on leadership between the senior year in high school and the sophomore year in college for = .05.
Popularity: The p-value < .05. There is evidence to reject H0 for = .05. There is evidence to indicate that there is a decrease in mean self-concept of females on popularity between the senior year in high school and the sophomore year in college for = .05.
Intellectual self-confidence: The p-value < .05. There is evidence to reject H0 for =.05. There is evidence to indicate that there is a decrease in mean self-concept of females on intellectual self-concept between the senior year in high school and the sophomore year in college for = .05.
9.40a.From the exercise, we know that x1 is a binomial random variable with the number of trials equal to n1. From Chapter 7, we know that for large n, the distribution of is approximately normal. Since x1 is simply multiplied by a constant, x1 will also have a normal distribution. Similarly, the distribution of is approximately normal, and thus, the distribution of x2 is approximately normal.
- The Central Limit Theorem is necessary to find the sampling distributions of and when n1 and n2 are large. Once we have established that both and have normal distributions, then the distribution of their difference will also be normal.
9.42a.The rejection region requires = .01 in the lower tail of the z-distribution. From Table IV, Appendix B, z.01 = 2.33. The rejection region is z2.33.
b.The rejection region requires = .025 in the lower tail of the z-distribution. From Table IV, Appendix B, z.025 = 1.96. The rejection region is z1.96.
c.The rejection region requires = .05 in the lower tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z1.645.
d.The rejection region requires = .10 in the lower tail of the z-distribution. From Table IV, Appendix B, z.10 = 1.28. The rejection region is z1.28.
9.44For confidence coefficient .95, = 1 .95 = .05 and /2 = .05/2 = .025. From Table IV, Appendix B, z.025 = 1.96. The95%confidence interval for p1p2 is approximately:
a. (.65 .58) 1.96
.07 .067 (.003, .137)
b. (.31 .25) 1.96
.06 .086 (.026, .146)
c. (.46 .61) 1.96
.15 .131 (.281, .019)
9.46 = = .65 = 1 = 1 .65 = .35
H0: p1p2 = 0
Ha: p1p2 > 0
The test statistic is z = = 1.14
The rejection region requires = .05 in the upper tail of the z-distribution. From Table IV, Appendix B, z.05 = 1.645. The rejection region is z > 1.645.
Since the observed value of the test statistic does not fall in the rejection region (z = 1.14 1.645), H0 is not rejected. There is insufficient evidence to indicate the proportion from population 1 is greater than that for population 2 at = .05.
9.48a.Let p1 = proportion of men who prefer to keep track of appointments in their head and
p2 = proportion of women who prefer to keep track of appointments in their head. To determine if the proportion of men who prefer to keep track of appointments in their head is greater than that of women, we test:
H0: p1p2 = 0
Ha: p1p2 > 0
b. and
The test statistic is