Mann–Introductory Statistics, Fifth Edition, Students Solutions Manual1
14.1Data that are divided into different categories for identification purposes are called categorical data. For example, dividing adults according to whether or not they smoke, or classifying registered voters by their party affiliation: Republican, Democrat, Independent, etc. results in categorical data.
14.3When using the sign test for the median of a single population, Table XI must be used if the sample size n ≤ 25.
14.5a.The rejection region is X ≥ 12.
b.The rejection region is X ≤ 3 andX ≥ 17.
c.The rejection region lies to the left of z = –1.65.
14.7a.Step 1:H0: Median = 28; H1: Median > 28; A one–tailed test.
Step 2:n = 10. Since n < 25, use the binomial distribution.
Step 3:For α = .05, the rejection region is X ≥ 9.
Step 4: The observed value ofX = 8.
Step 5:Do not reject the null hypothesis.
b.Step 1:H0: Median = 100; H1: Median < 100; A one–tailed test.
Step 2:n = 11. Since n < 25, use the binomial distribution.
Step 3:For α = .05, the rejection region is X ≤ 2.
Step 4: The observed value of X = 1.
Step 5:Reject the null hypothesis.
c.Step 1:H0: Median = 180; H1: Median ≠180;A two–tailed test.
Step 2:n = 26. Since n 25, use the normal distribution.
Step 3:For α = .05, the rejection region lies to the left of z = –1.96 and to the right of z = 1.96.
Step 4: For n = 26, p = .50, q = 1 – p = .50.
μ = np = 26 (.50) = 13andσ = = = 2.54950976
X = 3 and = = 13Since X,z = = = –3.73
Step 5:Reject the null hypothesis.
d.Step 1:H0: Median = 55; H1: Median < 55; A one–tailed test.
Step 2:n = 30. Since n 25, use the normal distribution.
Step 3:For α = .05, the rejection region lies to the left of z = –1.65.
Step 4: For p = .50 and n = 30
μ = np = 30 (.50) = 15andσ = = = 2.73861279
X = 6 and = = 15Since X,z = = = –3.10
Step 5:Reject the null hypothesis.
14.9Letpbe the proportion of residents who prefer bottled water, and let B represents bottled water and C represent city water.
Step 1:H0:p = .50; H1:p ≠ .50; A two–tailed test.
Step 2:n = 12. Sincen< 25, use the binomial distribution.
Step 3:For a two–tailed test withn = 12 and α = .05, the rejection region isX≤ 2 andX≥10.
Step 4:Sincepis the proportion of people preferring bottled water, we mark a plus sign for each person who prefers bottled water and a minus sign for each person preferring city water, whichyields the following table.
Person / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12Water Source / B / C / B / C / C / B / C / C / C / C / B / C
Sign / + / – / + / – / – / + / – / – / – / – / + / –
There are 4 plus signs, indicating that 4 of the 12 residents prefer bottled water. Thus, the observed value ofX = 4.
Step 5:The observed value ofXis not in the rejection region, so do not reject H0. Do not conclude that the residents prefer either of these two water sources over the other.
MINITAB: When using MINITAB we can not use plus and minus signs instead we use 0 and 1 to replace them. Here we will use 1 for the residents who drink bottled water and 0 for those who drink city water. So we enter this information in our worksheet and label it Preferences. Now select Stat, Nonparametric and 1 Sample Sign to launch a pop-up menu. In the new menu type the variable name or column number in the box below the word Variables. Then click on the box labeled Test Median and enter the probability of your null hypothesis which in this example is 0.50. Next set the alternative hypothesis to the correct form and in this case that’s not equal to. Finally click on OK and your results appear in the Session window. A copy of the dialog box and the results are shown below.
14.11Letpbe the proportion of all drinkers of JW's beer who can distinguish JW's from the rival brand.
H0:p = .50; H1:p> .50; A right–tailed test.
n = 20. Sincen< 25, use the binomial distribution.
Forn = 20 and α = .025, the rejection region isX≥ 15.
Thirteen drinkers in the sample correctly identified JW's. Thus, the observed value ofX = 13.
Do not reject H0. Do not conclude that drinkers of JW's are more likely to identify it than not.
14.13 Letpbe the proportion of adult North Dakota residents who would prefer to stay in North Dakota.
H0:p = .50; H1:p< .50; A left–tailed test.
Four of the 100 adults have no preference, so the true value of n = 100 – 4 = 96. Sincen> 25, use the normal distribution.
For α = .025, the rejection region lies to the left ofz = –1.96.
n = 96 andp = q = .50
μ = np = 96(.50) = 48 and = = 4.89897949
X = 41 and = = 48SinceX, z = = = 1.33
Do not reject H0. Do not conclude that less than half of all adult residents of North Dakota would prefer to stay.
TI-83: Select Stat, TESTS, and 1 PropZTest. Enter in the value of your null hypothesis as the value of p, enter the number of successes for x and then enter n. Next highlight the correct version of the alternative hypothesis, press the ENTER key, highlight Calculate, and press the ENTER key. The results appear below for our example where p = .50, x = 55, n = 100.Comparing our p value to α, we fail to reject the null hypothesis.
1−PropZTestprop ≠ .5
z = 1
p = .8413447404
= .55
n = 100
MINITAB: Select Stat, Basic Statistics, and 1 Proportion to generate the first pop-up menu. In the pop-up menu click beside Summarized Data, and beside the words Number of trials insert your n while beside the words Number of events enter your x. Now click on the word Options and in the second pop-up menu enter in the value of p in your null hypothesis as the Test Proportion, select the correct alternative hypothesis, click beside the words Use test and interval based on normal distribution, and then click on OK. Once back in the first pop-up menu click on OK again. A copy of the pop-up menu and the results follow for this example where p = .50, x = 95, n = 100. Comparing our p value to α, we fail to reject the null hypothesis.
14.15Letpbe the proportion of adults that frequently experience stress.
H0:p = .50; H1:p> .50; A right–tailed test.
n = 700. Sincen> 25, use the normal distribution.
For α = .01, the rejection region lies to the right ofz = 2.33.
n = 700 andp = q = .50
μ = np = 700(.50) = 350and = = 13.22875656
X = 370 and = = 350SinceX,z = = = 1.47
Do not reject H0.Do not conclude that over half of adults frequently experience stress in their daily lives.
14.17Step 1:H0: Median = 12 ounces; H1: Median ≠ 12 ounces; A two–tailed test.
Step 2:Since one of the 10 bottles had exactly 12.00 ounces, the true sample size isn = 10 − 1 = 9.
Sincen≤25, use the binomial distribution.
Step 3:Forn = 9 and α = .05, the rejection region isX≤1 andX≥ 8.
Step 4:We mark a plus sign for each bottle that has more than 12 ounces, a minus sign for each bottle that has less than 12 ounces, and a zero for each bottle holding exactly 12 ounces. This yields the following table.
Bottle / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10Amount / 12.10 / 11.95 / 12.00 / 12.01 / 12.02 / 12.05 / 12.02 / 12.03 / 12.04 / 12.06
Sign / + / – / 0 / + / + / + / + / + / + / +
There are eight plus signs, corresponding to the eight bottles which had more than 12 ounces of soda. Thus, the observed value ofX = 8.
Step 5:Since the observed value ofXis in the rejection region, Reject H0. Conclude that the median amount of soda in all such bottles differs from 12 ounces.
Note that we could have based the test on the number of minus signs which would have givenX = 1. Since this value is in the lower part of the rejection region, we would reject H0, as we did in Step 5 above.
Note: When using MINITAB, follow the instructions with problem 14.9 only this time instead of entering your data as zero or one in your worksheet this time enter the Medians. Everything else is the same.
14.19H0: Median = 4 minutes; H1: Median > 4 minutes;A right–tailed test.
Two of the 28 response times are exactly 4 minutes, so the true sample size isn = 28 − 2 = 26.
Sincen> 25, use the normal distribution.
For a right–tailed test with α = .01, the rejection region lies to the right ofz = 2.33.
Letpbe the proportion of response times that exceed 4 minutes. If H0 is true, we would expect about half of the times to exceed 4 minutes.
Thus,n = 26 andp = q = .50, so
μ = np = 26(.50) = 13 and = = 2.54950976
If we assign a plus sign to every response time above 4 minutes, a minus sign to every time below 4 minutes, and a zero to every time of exactly 4 minutes, there will be 21 plus signs, 5 minus signs, and 2 zeroes.
For a right–tailed test, we use the larger of the values (21 and 5) as the observed value of X, so X = 21. = = 13 SinceX,z = = = 2.94
Reject H0. Conclude that the median response time to all 911 calls in the inner city is greater than 4 minutes.
Note: When using MINITAB or Excel, follow the instructions with problem 14.13 only this time successes are values higher than the hypothesized median. Everything else is the same.
14.21H0: Median = 42 months; H1: Median < 42 months;A left–tailed test.
n = 35. Sincen> 25, use the normal distribution.
For α = .01, the rejection region lies to the left ofz = –2.33.
n = 35 andp = q = .50, so
μ = np = 35(.50) = 17.5 and = = 2.95803989
If we mark a plus sign for every time that exceeds 42 months and a minus sign for every time that
is less than 42 months, we obtain 10 plus signs and 25 minus signs. Since the test is left–tailed, we use the smaller value,X = 10.
= = 17.5SinceX,z = = = –2.37
Reject H0. Conclude that the median time served by all such prisoners is less than 42 months.
14.23For each employee: Paired difference = score before course − score after course
Let M denote the difference in median test scores before and after the course, where a plus sign
indicates an employee's score was lower after the course. Thus, if the course improves test scores, we would expect M < 0.
Step 1: H0:M = 0; H1:M < 0; A left–tailed test.
Step 2: One of the employee's scores were the same before and after the course, so the true sample size isn = 6.
Step 3: Forn = 6 and α = .05, the rejection region isX = 0.
Step 4: We assign a plus sign to each employee whose score decreased, a minus sign to each employee whose score increased, and a zero to each employee whose score was the same before and after the course. The results are shown in the following table.
Before / 8 / 5 / 4 / 9 / 6 / 9 / 5After / 10 / 8 / 5 / 11 / 6 / 7 / 9
Sign / – / – / – / – / 0 / + / –
We find one plus sign, five minus signs, and one zero, indicating that one employee's score decreased, five increased, and one remained the same after the course. Since the test is left–tailed, we use the smaller number of signs for X. Thus, the observed value ofX = 1.
Step 5:X = 1 is not in the rejection region so do not reject H0. Do not conclude that attending this course increases the median self–confidence test score of all employees.
14.25For each employee: Paired difference = (Bikes assembled before)–(Bikes assembled after new system)
H0:M = 0; H1:M≠ 0;A two–tailed test.
Since one worker assembled the same number of bikes under both systems, the true value of
n = 27 – 1 = 26.
n = 26 > 25, so use the normal distribution.
For α = .02, the rejection region lies to the left ofz = –2.33 and to the right ofz = 2.33.
n = 26 andp = q = .50, so,
μ = np = 26(.50) = 13 and = = 2.54950976
Assigning a plus sign to each employee who produced more under the new payment system, and a minus sign to each employee who produced less, results in 19 plus signs and 7 minus signs with one zero, as oneemployee produced the same number of bikes . If we use the larger number of signs,
X = 19. = = 13SinceX,z = = = 2.16
Do not reject H0. Do not conclude that the median number of bikes assembled differs from the median before the new payment system was instituted.
If we had used the smaller number of signs (7 minus signs) we would have z = and our conclusion would be the same (Do not reject H0).
14.27For each matched pair of cows:
Paired difference = Milk production of cow without hormone –Milk production of cow with hormone.
H0:M = 0; H1:M≠ 0; A two–tailed test.
Since milk production was the same for two pairs of cows, the true value ofn = 30 − 2 = 28.
n> 25, so use the normal distribution.
For a two–tailed test with α = .05, the rejection region lies to the left ofz = –1.96 and to the right of z = 1.96.
n = 28 andp = q = .50, so
μ = np = 28(.50) = 14 and = = 2.64575131
Assigning a plus sign to each pair of cows in which the cow taking the hormone produced less milk, a minus sign to each pair in which the cow taking the hormone produced more milk, and a zero to each pair in which both cows produced the same amount of milk results in 9 plus signs, 19 minus signs, and two zeroes.
Using the larger number of signs,X = 19.
= = 14SinceX,z = = = 1.70
Do not reject H0. Do not conclude that the hormone changes the milk production of such cows.
Note that if we had used the smaller number of signs (X = 9), thenz= and the conclusion would be the same (do not reject H0).
14.29 The null hypothesis of the Wilcoxon signed–rank test usually states that the medians of the two population distributions are equal.
14.31a.The rejection region is T≤11.
b.The rejection region is T≤7.
c.The rejection region lies to the left ofz = –1.96.
d.The rejection region lies to the right ofz = 2.33.
14.33a. For each salesperson:
Paired difference = (Number of contacts before) − (Number of contacts after)
Let MA and MB be the median number of contacts by all such salespersons after and before
the installation of governors, respectively.
Step 1: H0:MA = MB; H1:MAMB; A left–tailed test.
Step 2:n = 7. Sincen< 15, use the Wilcoxon signed–rank test for the small–sample case.
Step 3: For a one–tailed test withn = 7 and α = .05, the rejection region is T≤ 4.
Step 4: The signed ranks are calculated in the following table.
Differences / Absolute / Ranks of / SignedBefore / After / (Before–After) / Differences / Differences / Ranks
50 / 49 / +1 / 1 / 1 / +1
63 / 60 / +3 / 3 / 2 / +2
42 / 47 / −5 / 5 / 4.5 / −4.5
55 / 51 / +4 / 4 / 3 / +3
44 / 50 / −6 / 6 / 6 / −6
65 / 60 / +5 / 5 / 4.5 / +4.5
66 / 58 / +8 / 8 / 7 / +7
Sum of positive ranks = 1 + 2 + 3 + 4.5 + 7 = 17.5
Sum of absolute values of negative ranks = 4.5 + 6 = 10.5
For a left–tailed test, T is the sum of the absolute values of the negative ranks. Thus:
The observed value of T = 10.5
Step 5: Since the observed value of T is not in the rejection region, do not reject H0. Do not conclude that the use of governors tends to reduce the number of contacts made per week by the Gamma Corporation's salespersons.
b.The conclusion in part a of this exercise is the same as that of the corresponding test of Exercise 10.101 (do not reject H0).
MINITAB: To calculate the p value we enter the values of Before and After in columns C1 and C2. Then select Calc and Calculator to get a pop-up menu. In the menu type C3 into the box marked Store result in variable, in the box below the word Expression type C1 −C2, and click on OK. This gives you a new column of data which you should label Increase. Now select Stat, Nonparametric and 1 Sample Wilcox to launch a pop-up menu.In the new menu type the variable name or column number of the new variable Increase which we created in the last step in the box below the word Variables. Then click on the box labeled Test Median and enter the probability of your null hypothesis which in this example is 0. Next set the alternative hypothesis to the correct form and in this case that’s less than zero. Finally click on OK and your results appear in the Session window. A copy of each of the dialog boxes and the results are shown below.Comparing our p value to α, we fail to reject the null hypothesis.
14.35For each employee: Paired difference = score before course – score after course
Let MA and MBdenote the median self–confidence test scores of all such employees after and before attending the course.
a.H0:MA = MB; H1:MAMB; A right–tailed test.
One of the employees had the same test score before and after attending the course, so the true sample size isn = 7 − 1 = 6. Sincen< 15, use the Wilcoxon signed–rank test procedure for the small–sample case.
Forn = 6, α = .05, and a one–tailed test, the rejection region is T≤ 2.
Differences / Absolute / Ranks of / SignedBefore / After / (Before–After) / Differences / Differences / Ranks
8 / 10 / –2 / 2 / 3 / –3
5 / 8 / –3 / 3 / 5 / –5
4 / 5 / –1 / 1 / 1 / –1
9 / 11 / –2 / 2 / 3 / –3
6 / 6 / – / – / – / –
9 / 7 / +2 / 2 / 3 / 3
5 / 9 / –4 / 4 / 6 / –6
Sum of positive ranks = 3
Sum of absolute values of negative ranks = 3 + 5 + 1 + 3 + 6 = 18
For a right–tailed test, T is the sum of the positive ranks. Thus, the observed value of T = 3
Do not reject H0. Do not conclude that attending this course increases the median self–confidence test scores of employees.
b.In both exercises H0 is not rejected.
14.37For each adult: Paired difference = hiking time before course – hiking time after course
Let MA and MB denote the median time to complete the hike after and before the fitness course
for all such adults.
H0:MA = MB; H1:MAMB; A left–tailed test.
n = 20 > 15, so use the Wilcoxon signed–rank test with the normal distribution approximation.
For α = .025 and a left–tailed test, the rejection region lies to the left ofz = –1.96.
Forn = 20:
μT = =
σT =
Calculating the differences, absolute differences, ranks, and signed ranks yields:
Sum of positive ranks = 182.5
Sum of absolute values of negative ranks = 27.5
Since the test is left–tailed, Tis the sum of the absolute values of the negative ranks. Thus, the observed value of T = 27.5
z =
Reject H0. Conclude that the fitness course tends to reduce the median time required to complete the two–mile hike.
Note: Using technology only simplifies this process a little. We must still calculate μTand σT but we can use technology tofind the p- vales of the z we calculate. The instructions for calculating p values with technology are in the textbook on page 691 or with problem 6.11.
14.39The Wilcoxon signed–rank test is used for paired samples, while the Wilcoxon rank–sum test is used for independent samples.
14.41a.Step 1:H0: The two population distributions are identical.
H1: The two population distributions are different.
Step 2:Because n1 < 10 and n2 < 10, use the Wilcoxon rank–sum test for small samples.
Step 3:The rejection region is T≤ 28 and T≥ 56.
Step 4:The observed value of T = 22.
Step 5:Reject H0.
b.Step 1:H0: The two population distributions are identical.
H1: The distribution of population 1 lies to the right of the distribution of population 2.
Step 2:Because n2 = 12 > 10, use the normal distribution.
Step 3:The rejection region lies to the right ofz = 1.96.
Step 4: