Chapter 8
The comparison of two populations
8-1. n = 25 [pic] = 19.08 [pic] = 30.67 H0: [pic]D = 0 H1: [pic]D [pic] 0 t (24) = [pic] = 3.11 Reject H0 at [pic] = 0.01.
|Paired Difference Test | | |
|Evidence | | | |
| |Size |25 |n |Assumption |
| |Average Difference |19.08 |μD |Populations Normal |
| |Stdev. of Difference |30.67 |sD | |
| | | | |Note: Difference has been defined as |
| |Test Statistic |3.1105 |t | |
| |df |24 | | |
|Hypothesis Testing | | |At an α of |
| |Null Hypothesis |p-value |5% |
| |H0: μ1 − μ2 = |0 |0.0048 |Reject |
8-2. n = 40 [pic] = 5 [pic] = 2.3 H0: [pic]D = 0 H1: [pic]D [pic] 0 t(39) = [pic] = 13.75 Strongly reject H0. 95% C.I. for [pic]D [pic] 2.023(2.3/[pic]) = [4.26, 5.74].
8-3. n = 12 [pic] = 3.67 [pic] = 2.45 (D = Movie – Commercial) H0: [pic]D = 0 H1: [pic]D [pic] 0 (template: Testing Paired Difference.xls, sheet: Sample Data)
|Paired Difference Test |
| |Data | | | | | |
| |Current |Previous |Evidence | | | |
| |M |C | |Size |9 |n |Assumption |
|1|15 |10 | |Average Difference |3.66667 |μD |Populations Normal |
|2|17 |9 | |Stdev. of Difference |2.44949 |sD | |
|3|25 |21 | | | | |Note: Difference has been defined as |
|4|17 |16 | |Test Statistic |4.4907 |t | |
|5|14 |11 | |df |8 | | |
|6|18 |12 |Hypothesis Testing | | |At an α of |
|7|17 |13 | |Null Hypothesis |p-value |5% | |
| | | | |H0: μ1 − μ2 0 t(24) = [pic] = 1.549. At [pic] = 0.05, we cannot reject H0.
|Paired Difference Test | | |
|Evidence | | | |
| |Size |60 |n |Assumption |
| |Average Difference |0.2 |μD |Populations Normal |
| |Stdev. of Difference |1 |sD | |
| | | | |Note: Difference has been defined as |
| |Test Statistic |1.5492 |t | |
| |df |59 | | |
|Hypothesis Testing | | |At an α of |
| |Null Hypothesis |p-value |5% |
| |H0: μ1 − μ2 = |0 |0.1267 | |
| |H0: μ1 − μ2 >= |0 |0.9367 | |
| |H0: μ1 − μ2 0 t (14) = [pic] = 1.469 There is no evidence that the shelf facings are effective.
8-6. n = 12 [pic] = 37.08 [pic] = 43.99 H0: [pic]D = 0 H1: [pic]D [pic] 0 (template: Testing Paired Difference.xls, sheet: Sample Data)
|Paired Difference Test |
| |Data | | | | | |
| |Current |Previous |Evidence | | | |
| |France |Spain | |Size |12 |n |Assumption |
|1 |258 |214 | |Average Difference |37.0833 |μD |Populations Normal |
|2 |289 |250 | |Stdev. of Difference |43.9927 |sD | |
|3 |228 |190 | | | | |Note: Difference has been defined as |
|4 |200 |185 | |Test Statistic |2.9200 |t | |
|5 |190 |114 | |df |11 | | |
|6 |350 |285 |Hypothesis Testing | | |At an α of |
|7 |310 |378 | |Null Hypothesis |p-value |5% | |
|10 |175 |120 | |H0: μ1 − μ2 0 C = [pic] + 2.326([pic]) = 0.30029 We need: P([pic] > C | [pic]D = 0.1) = P([pic] > 0.30029 | [pic]D = 0.1) = P[pic] = P(Z > 1.551) = 0.0604
8-8. n = 20 [pic] = 1.25 [pic] = 42.896 H0: [pic]D = 0 H1: [pic]D [pic] 0 t (19) = [pic] = 0.13 Do not reject H0; no evidence of a difference.
|Paired Difference Test | | |
|Evidence | | | |
| |Size |20 |n |Assumption |
| |Average Difference |1.25 |μD |Populations Normal |
| |Stdev. of Difference |42.89 |sD | |
| | | | |Note: Difference has been defined as |
| |Test Statistic |0.1303 |t | |
| |df |19 | | |
|Hypothesis Testing | | |At an α of |
| |Null Hypothesis |p-value |5% |
| |H0: μ1 − μ2 = |0 |0.8977 | |
8-9. [pic] = 100 [pic] = 100 [pic] = 76.5 [pic] = 88.1 [pic] = 38 [pic] = 40 H0: [pic]0 H1: [pic] 0
(Template: Testing Population Means.xls, sheet: Z-test from Stats) (need to use the t-test since the population std. dev. is unknown)
|Evidence | | | |Assumptions | | | | |
| | | |Sample1 |Sample2 | |Populations Normal | | | | |
| | |Siz|100 |10|n |H0: | | | |
| | |e | |0 | |Populati| | | |
| | | | | | |on | | | |
| | | | | | |Variance| | | |
| | | | | | |s Equal | | | |
| |Pooled Variance |1522 |s2p | | | |
| |Nu|p-value |5% | |1 − α |Co| |
| |ll| | | | |nf| |
| |Hy| | | | |id| |
| |po| | | | |en| |
| |th| | | | |ce| |
| |es| | | | |In| |
| |is| | | | |te| |
| | | | | | |rv| |
| | | | | | |al| |
| | | | | | | | |
| | | |Test Statistic |-10.7845 |t | | | |
| | | |Null Hypothesis |p-value |5% | |1 |Confidence Interval |
| | | | | | | |− | |
| | | | | | | |α | |
| | | |Null |p-value |5% | |
| | | |Hypothesis | | | |
| | | |Sample1 |Sample2 | |Populations Normal | | |
| | |Size |40 |40 |n |H0: Population Variances Equal |
| | |Mea|15 |6.2|x-b| |F |1.36111 | |
| | |n | | |ar | |ra| | |
| | | | | | | |ti| | |
| | | | | | | |o | | |
| |Pooled Variance |10.625 |s2p | | | |
| |Null |p-value |5% |
| |Hypothesis | | |
| | |Sample1 |Sample2 | |
| |Size |128 |212 |n |
| |Mean |23.5 |18 |x-bar |
| | |Popn. 1 |Popn. 2 | |
| |Popn. Std. Devn. |12.2 |10.5 |σ |
| | | | |€ |
|Hypothesis Testing | | |
| | | | | |
| |Test Statistic |4.2397 |z | |
| | | | |At an α of |
| |Null Hypothesis |p-value |5% |
| |H0: μ1 − μ2 = |0 |0.0000 |Reject |
8-14. [pic] = 13 [pic] = 13 [pic] = 20.385 [pic] = 10.385 [pic] = 7.622 [pic] = 4.292 [pic] = .05 H0: u1 = u2 H1: u1 [pic] u2
[pic]
Use a critical value of 2.064 for a two-tailed test. Reject H0. The two methods do differ.
8-15. Liz (1): [pic] = 32 [pic] = 4,238 [pic] = 1,002.5 Calvin (2): [pic] = 37 [pic] = 3,888.72 [pic] = 876.05 a. one-tailed: H0: [pic] [pic] 0 H1: [pic] > 0 b. z = [pic] = 1.53 c. At [pic] = 0.5, the critical point is 1.645. Do not reject H0 that Liz Claiborne models do not get more money, on the average. d. p-value = .5 ( .437 = .063 (It is the probability of committing a Type I error if we choose to reject and H0 happens to be true.) e. [pic]
8-16. (Template: Testing Population Means.xls, sheet: t-test from Stats) (need to use the t-test since the population std. dev. is unknown) H0: [pic] = 0 H1: [pic] [pic] 0
|t-Test for Difference in Population Means | | | | | |
| | | | | | | | |
| | | | |Sample1 |Sample2 | |Populations Normal | | |
| | | |Size |28 |28 |n |H0:| | |
| | | | | | | |Pop| | |
| | | | | | | |ula| | |
| | | | | | | |tio| | |
| | | | | | | |n | | |
| | | | | | | |Var| | |
| | | | | | | |ian| | |
| | | | | | | |ces| | |
| | | | | | | |Equ| | |
| | | | | | | |al | | |
| |Pooled Variance |29.3642 |s2p | | | |
| |Null |p-value |1% |
| |Hypothesis | | |
| | | |Sample1 |Sample2 | |
| | |Size |25 |20 |n |
| | |Mean |87 |64 |x-bar |
| | |Std. Deviation |12 |23 |s |
| | | | | | |
| |Assuming Population Variances are Equal |
| | |Pooled Variance |314.116 |s2p | |
| | |Test Statistic |4.3257 |t | |
| | |df |43 | | |
| | | | | |At an α of |
| | |Null Hypothesis |p-value |5% |
| | |H0: μ1 − μ2 = |0 |0.0001 |Reject |
8-19. With training (1): [pic] = 13 [pic] = 55 [pic] = 8 Without training (2): [pic] = 15 [pic] = 48 [pic] = 6 H0: [pic] [pic] 4,000 H1: [pic] > 4,000 t (26) = [pic] = 1.132 The critical value at [pic] = .05 for t (26) in a right-hand tailed test is 1.706. Since 1.132 < 1.706, there is no evidence at [pic] = .05 that the program executives get an average of $4,000 per year more than other executives of comparable levels.
8-20. (Use template: “testing difference in means.xls”) (need to use the t-test since the population std. dev. is unknown)
H0: μP - μL= 0 H1: μP - μL[pic] 0
t-Test for Difference in Population Means
|Evidence | | | |Assumptions | | |
| | | |Sample1 |Sample2 | |Populations Normal | | |
| | |Size |20 |20 |n |H0: Population Variances Equal |
| | |Mea|1 |6 |x-| |F |5.16529 | |
| | |n | | |ba| |ra| | |
| | | | | |r | |ti| | |
| | | | | | | |o | | |
| |Test Statistic |-8.1868 |t | | | |
| |Null |p-value |5% | |1 −|Confidence |
| |Hypothesis | | | |α |Interval |
| | | |US |China | |Populations Normal | | |
| | |Size |15 |18 |n |H0: Population Variances Equal |
| | |Mea|3.8 |6.1|x-b| |F |5.80372 | |
| | |n | | |ar | |ra| | |
| | | | | | | |ti| | |
| | | | | | | |o | | |
| |Test Statistic |-1.676 |t | | | |
| |Null |p-value |1% | |1 −|Confidence |
| |Hypothesis | | | |α |Interval |
| | | |Sample1 |Sample2 | |Populations Normal | | |
| | |Size |20 |20 |n |H0: Population Variances Equal |
| |
| |Pooled Variance |9.04 |s2p | | |
| |Test Statistic |-1.3463 |t | | |
| |df |38 | | | |
| | | | |At an α of | |
| |Null Hypothesis |p-value |5% | |
| |H0: μ1 − μ2 = |0 |0.1862 | | |
| |H0: μ1 − μ2 >= |0 |0.0931 | | |
| |H0: μ1 − μ2 0 t(48) = [pic] = 2.785 At [pic] = 0.05, reject H0. Also reject at [pic] = 0.01. p-value = 0.0038.
|Evidence | | | |
| | | |Sample1 |Sample2 | |
| | |Size |25 |25 |n |
| | |Mean |12 |13.5 |x-bar |
| | |Std. Deviation |2.5 |1 |s |
| | | | | | |
| |Assuming Population Variances are Equal |
| | |Pooled Variance |3.625 |s2p | |
| | |Test Statistic |-2.7854 |t | |
| | |df |48 | | |
| | | | | |At an α of |
| | |Null Hypothesis |p-value |5% |
| | |H0: μ1 − μ2 = |0 |0.0076 |Reject |
| | |H0: μ1 − μ2 >= |0 |0.0038 |Reject |
8-26. H0: [pic] = 0 H1: [pic] [pic] 0 z = [pic] = 0.8887 Do not reject H0. There is no evidence of a difference in average stock returns for the two periods.
8-27. (Use template: “testing difference in means.xls”) (need to use the t-test since the population std. dev. is unknown) H0: μN - μO ≤ 0 H1: μN - μO > 0
|Evidence | | | |Assumptions | | |
| | | |Sample1 |Sample2 | |Populations Normal | | |
| | |Size |8 |10 |n |H0: Population Variances Equal |
| |
| |Pooled Variance |4.23063 |s2p | | |
| |Test Statistic |0.7175 |t | | |
| |df |16 | | | |
| | | | |At an α of | |
| |Null Hypothesis |p-value |5% | |
| |H0: μ1 − μ2 = |0 |0.4834 | | |
| |H0: μ1 − μ2 >= |0 |0.7583 | | |
| |H0: μ1 − μ2 0 z = [pic] = [pic] = 2.835 Reject H0. On-time departure percentage has probably declined after NW’s merger with Republic. p-value = 0.0023.
|Evidence | |Sample 1 |Sample 2 | |
| |Size |100 |100 |n |
| |#Successes |85 |68 |x |
| |Proportion |0.8500 |0.6800 |p-hat |
| | | | | |
|Hypothesis Testing | | | |
|Hypothesized Difference Zero | | |
| | | | | |
| |Pooled p-hat |0.7650 | | |
| |Test Statistic |2.8351 |z | |
| | | | |At an α of |
| |Null Hypothesis |p-value |5% |
| |H0: p1 - p2 = 0 |0.0046 |Reject |
| |H0: p1 - p2 >= 0 |0.9977 | |
| |H0: p1 - p2 0 z = [pic] = 4.677 Reject H0. There is strong evidence that the percentage of word-of-mouth recommendations in small towns is greater than it is in large metropolitan areas.
8.31. [pic] = 31 x 1 = 11 [pic] = 50 x 2= 19 H0: p1 – p2 = 0 H1: p1 – p2 [pic] 0 z = [pic] = (0.228 Do not reject H0. There is no evidence that one corporate raider is more successful than the other.
8-32. Before campaign (1): [pic] = 2,060 [pic] = 0.13 After campaign (2): [pic] = 5,000 [pic] = 0.19 H0: p2 ( p1 [pic] .05 H1: p2 – p1 > .05 z = [pic] = [pic] = 1.08 No evidence to reject H0; cannot conclude that the campaign has increased the proportion of people who prefer California wines by over 0.05.
8-33. 95% C.I. for p2 ( p1: ([pic] ( [pic]) [pic] 1.96[pic] = .06 [pic] 1.96[pic] = [0.0419, 0.0781] We are 95% confident that the increase in the proportion of the population preferring California wines is anywhere from 4.19% to 7.81%.
|Confidence Interval | | | | | | |
|1 − α |Confidence Interval | | | |
| |Size |120 |200 |n |
| |#Successes |34 |41 |x |
| |Proportion |0.2833 |0.2050 |p-hat |
| | | | | |
|Hypothesis Testing | | | |
|Hypothesized Difference Zero | | |
| | | | | |
| |Pooled p-hat |0.2344 | | |
| |Test Statistic |1.6015 |z | |
| | | | |At an α of |
| |Null Hypothesis |p-value |5% |
| |H0: p1 - p2 = 0 |0.1093 | |
| |H0: p1 - p2 >= 0 |0.9454 | |
| |H0: p1 - p2 p 2 [pic] = (101 +110)/320 = .659 z = [pic] = 5.33 Strongly reject H0; Motorola’s system is superior (p-value is very small).
8-40. Old method (1): [pic] = 40 [pic] = 1,288 New method (2): [pic] = 15 [pic] = 1,112 H0: [pic]12 [pic] [pic]22 H1: [pic]12 > [pic]22 use [pic] = .05 F (39,14) = s 12/s 22 = 1,288/1,112 = 1.158 The critical point at [pic] = .05 is F (39,14) = 2.27 (using approximate df in the table). Do not reject H0. There is no evidence that the variance of the new production method is smaller.
|F-Test for Equality of Variances |
| | | | |
| | |Sample 1 |Sample 2 |
| |Size |40 |15 |
| |Variance |1288 |1112 |
| | | | |
| |Test Statistic |1.158273 |F |
| |df1 |39 | |
| |df2 |14 | |
| | | | |
| | | |At an α of |
| |Null Hypothesis |p-value |5% |
| |H0: σ21 - σ22 = 0 |0.7977 | |
| |H0: σ21 - σ22 >= 0 |0.6012 | |
| |H0: σ21 - σ22 = 0 |1.0000 | |
| |H0: σ21 - σ22 = 0 |0.9945 | |
| |H0: σ21 - σ22 = 0 |0.9258 | |
| |H0: σ21 - σ22 = |0 |0.1389 | | |
| |H0: μ1 − μ2 = 0 |1.0000 | | |
| |H0: p1 - p2 = |0 |0.5432 | | |
| |H0: μ1 − μ2 = |0 |0.9914 | |
| |H0: μ1 − μ2 0.20.) The solution of Problem 8-61 is valid from the equal-variance requirement.
8-63. A = After: nA = 16 [pic] = 91.75 sA = 5.0265959 B = Before: nB = 15 [pic] = 84.7333 sB = 5.3514573 H0: [pic] [pic] 5 H1: [pic] > 5 t (29) = [pic] = 1.08 Do not reject H0. There is no evidence that advertising is effective.
8-64. H0: [pic]12 = [pic]22 H1: [pic]12 [pic] [pic]22 F (14,15) = (5.3514573)2/(5.0265959)2 = 1.133 Do not reject H0 at [pic] = 0.10. There is no evidence that the population variances are not equal.
|F-Test for Equality of Variances |
| | | | |
| | |Sample 1 |Sample 2 |
| |Size |15 |16 |
| |Variance |28.6381 |25.26667 |
| | | | |
| |Test Statistic |1.133434 |F |
| |df1 |14 | |
| |df2 |15 | |
| | | | |
| | | |At an α of |
| |Null Hypothesis |p-value |10% |
| |H0: σ21 - σ22 = 0 |0.8100 | |
| |H0: σ21 - σ22 >= 0 |0.5950 | |
| |H0: σ21 - σ22 = |0 |0.1725 | |
| |H0: μ1 − μ2 = |0 |0.0702 | | |
| |H0: μ1 − μ2 = |0 |1.0000 | | |
| |H0: μ1 − μ2 = 0 |0.9962 | | |
| | |H0: p1 - p2 = |0 |0.9766 | |
| | |H0: μ1 − μ2 = |0 |0.9626 | |
| |H0: μ1 − μ2 