1) Match problems to procedures

Do contingency table classification values matter?

Does the data validate the normal probability distribution assumption?

Data were gathered in an experiment comparing the effects of three insecticides in controlling a certain species of parasitic beetle. Each observation represents the number of such insects found dead in a certain fixed area treated with an insecticide.

A multinomial probability distribution describes the distribution of counts across multiple levels of a variable. A special case is the binomial discrete probability distribution. For each level of a variable, which is common to multiple populations, equality of distributions can be tested.

Match with these

1) ANOVA

2) Chi-square test of homogeneity

3) Chi-square test of independence

4) Chi-squa

2). Data for question 2

0:00:00 Time

1.Air quality

0:00:30.Time

1.Air quality

0:01:00.Time

1.Air quality

0:01:30.Time

1.Air quality

0:02:00Time

1.Air quality

0:02:30Time

1.Air quality

0:03:00Time

1.Air quality

0:03:30Time

1.Air quality

0:04:00.Time

1.Air quality

0:04:30.. Time

1.Air quality

0:05:00.Time

1.Air quality

0:05:30.Time

1.Air quality

0:06:00.Time

1.Air quality

0:06:30.Time

1.Air quality

0:07:00.Time

1.Air quality

0:07:30.Time

1.Air quality

0:08:00.Time

1.Air quality

0:08:30.Time

1.Air quality

0:09:00.Time

1.Air quality

0:09:30.Time

1.Air quality

0:10:00.Time

1.Air quality

0:10:30.Time

1.Air quality

0:11:00.Time

1.Air quality

0:11:30.Time

1.Air quality

0:12:00.Time

1.Air quality

0:12:30.Time

1.Air quality

0:13:00.Time

1Air quality

0:13:30.Time

1.Air quality

0:14:00.Time

1.Air quality

0:14:30.Time

1.Air quality

0:15:00.Time

1.Air quality

0:15:30.Time

1.Air quality

0:16:00.

1.Air quality

0:16:30.Time

0.Air quality

0:17:00.Time

1.Air quality

0:17:30.Time

1.Air quality

0:18:00.Time

1.Air quality

0:18:30.Time

1.Air quality

0:19:00.Time

1.Air quality

0:19:30Time

1.Air quality

0:20:00.Time

1.Air quality

0:20:30.Time

1.Air quality

0:21:00

1.Air quality

0:21:30. Time

1.Air quality

0:22:00.Time

1. Air quality

0:22:30.

1.Air quality

0:23:00.Time

1. Air quality

0:23:30.Time

1.Air quality

2). An Air Quality instrument logs 0 when standards are not met and 1 when standards are met. The log is above First, compute the proportion meeting standards as the mean of Air Quality values. Second, at alpha = 0.10 (sensitive, exploratory), test the hypothesis that proportion of times that air quality meets standards is at least 90%.

a)The pvalue of 0.966 indicates that the data provide insignificant evidence against H0: 0.90. H0 is not rejected at = 0.10. The status quo remains unchanged.

b) None of the answers are correct.

c) The pvalue of 0.062 indicates that the data provide weak evidence against H0: 0.90. H0 is rejected at = 0.10.

(d) The pvalue of 0.022 indicates that the data provide strong evidence against H0: 0.90. H0 is rejected at = 0.10. The status quo has changed.

e) The pvalue of 0.006 indicates that the data provide overwhelming evidence against H0: 0.90. H0 is rejected at = 0.10. Send out an air quality alert.

Data for question 3

Yards

298

273

284

292

275

309

292

302

311

287

279

282

297

283

305

299

294

295

286

3) The Lawnpoke Golf Association (LGA) has established rules that manufacturers of golf equipment must meet for their products to be acceptable for LGA events. BatOutaHell Balls uses a proprietary process to produce balls with a mean distances of 295 yards. BatOutaHell is concerned that if the mean distance falls below 295 yards, the word will get out and sales will sag. Further, if the mean distance exceeds 295 yards, their balls may be rejected by LGA. Measurements of the distances are recorded in DATA above . At = 0.05, test the no action hypothesis that the balls have a mean distance of 295 yards.

a) The test statistics of 2.238 is greater than the critical value of 1.734, therefore H0 is rejected.It is reasonable to assume that the distance is not 295 yards.

b) The test statistic is 1.297 and the critical value is 1.734, therefore the test statistics is less than the critical value and the null hypothesis is not rejected.The distance is about 295 yards.

C) The test statistic is 3.003 and the critical value is 1.734, therefore the test statistic is greater than the critical value of 1.734 and the null hypothesis is rejected.The distance is not 295 yards.

d) The test statistic of 1.908 is greater than the critical value of 1.734, therefore H0 is rejected.It is reasonable to assume that the distance is not 295 yards.

4) Customer Traffic

With Program

Without Program

144.With Program

140.Without Program

236. With Program

233.Without Program

108. With Program

110.Without Program

43.With Program

42.Without Program

337. With Program

332.Without Program

134.With Program

135.Without Program

148.With Program

151.Without Program

30.With Program

33.Without Program

181.With Program

178.Without Program

146.With Program

147.Without Program

159.With Program

162.Without Program

248.With Program

243.Without Program

150.With Program

149. Without Program

54.With Program

48. Without Program

349. With Program

346. Without Program

4) The Fast N' Hot food chain wants to test if their "Buy One, Get One Free" program increases customer traffic enough to support the cost of the program. For each of 15 stores, one day is selected at random to record customer traffic with the program in effect, and one day is selected at random to record customer traffic with program not in effect. The results of the experiment are documented in DATA. For each store, compute difference = traffic with program minus traffic without program. At = 0.05, test the hypothesis that the mean difference is at most 0 (at best the program makes no difference, or worse it decreases traffic) against the alternative that the mean difference > 0 (the program increases traffic).

a) The pvalue of 0.033 provides strong evidence against H0. H0 is rejected at = 0.05. You decide to recommend further evaluation of the program.

b) The pvalue of 0.084 provides weak evidence against H0. H0 is not rejected at = 0.05. You decide the evidence is not strong enough to recommend further evaluation of the program.

c) None of the answers are correct.

d) The pvalue of 0.221 indicates that the data provide insignificant evidence against H0. H0 is not rejected at = 0.05. You decide to conclude the study and not to recommend the program.

e)?The pvalue of 0.002 provides overwhelming evidence against H0. H0 is rejected at = 0.05. You decide that the program results in increased customer traffic, overall, and recommend the program be implement

f) The pvalue rejects H0: Mean difference > 0.

5)

Valley

Foothills

109.Valley

103.Foothills

116 Valley

182.Foothills

106.Valley

184. Foothills

157.Valley

133.Foothills

147.Valley

243.Foothills

105.Valley

158.Foothills

173.Valley

247.Foothills

153.Valley

221.Foothills

137.Valley

175.Foothills

110.Valley

197.Foothills

5) BigDeal Real Estate surveyed prices per square foot in the valley and foothills of Hoke-a-mo, Utah. Based on BD's DATA above are prices per square foot equal at = 0.01?

a) The critical value is 2.977 since this is a two-tail scenario. The test statistic is 1.936.Since the test statistic < the critical value, the test statistic does not lie in the area of rejection.Do not reject the null hypothesis.The prices per square foot are equal at alpha = .01

b) The critical value is 2.977 since this is a two-tail scenario. The test statistic is 2.239.Since the test statistic < the critical value, the test statistic does not lie in the area of rejection.Do not reject the null hypothesis.The prices per square foot are equal at alpha = .01

C) The critical value is 2.977 since this is a two-tail scenario. The test statistic is 1.513.Since the test statistic < the critical value, the test statistic does not lie in the area of rejection.Do not reject the null hypothesis.The prices per square foot are equal at alpha = .01

d) The critical value is 2.977 since this is a two-tail scenario. The test statistic is 3.207.Since the test statistic > the critical value, the test statistic does lie in the area of rejection.Reject the null hypothesis.The prices per square foot are not equal at alpha = .01

6)

Observation

Portland

Houston

Jacksonville

1.Observation

85.Portland

71.Houston

64.Jacksonville

2.Observation

75.Portland

75.Houston

69.Jacksonville

3.Observation

82.Portland

73.Houston

67.Jacksonville

4.Observation

76.Portland

74.Houston

74.Jacksonville

5.Observation

71.Portland

69.Houston

80.Jacksonville

6.Observation

85.Portland

82.Houston

72.Jacksonville

National Bearings manufactures bearings at plants located in Portland Oregon, Houston Texas, and Jacksonville Florida. To measure employee knowledge of Total Quality Management (TQM), six employees were randomly selected at each plant and tested. The test scores for these employees are given in DATA above Managers want to know if, on average, knowledge of TQM is equal across the 3 plants. Test equality of mean scores at = 0.05.

a) The F value of 3.419 is < the F critical value of 3.682, therefore do not reject equality of means.Knowledge of TQM is equal across the 3 plants.

b) The F value of 1.326 is < the F critical value of 3.682, therefore do not reject equality of means.Knowledge of TQM is equal across the 3 plants.

c) The F value of 9 is > the F critical value of 3.682, therefore reject the equality of means.Knowledge of TQM is not equal across the 3 plants.

d) The F value of 6.349 is > the F critical value of 3.682, therefore reject the equality of means.Knowledge of TQM is not equal across the 3 plants.

7)

Manufacturer

A

B

C

D

25. A

23.B

25.C

27.D

23.A

21.B

25.C

26.D

21.A

23.B

25.C

27.D

23.A

24.B

21.C

26.D

To test whether the mean time to mix a batch of adhesive is the same for machines produced by four manufacturers, TiteBondMax obtained DATA above on the time (minutes) needed to mix the materials. Test whether the machines have equal mean mixing time at = 0.01.

a) None of the answers are correct.

b) The pvalue 0.0001 is extreme evidence that the machines are not all the same.

C) The data provide insignificant evidence against H0: Equal means at pvalue 0.514. The machine are considered equal.

d) The data provide weak evidence against H0: Equal means at pvalue 0.072. Equality of means remains plausible.

e) The data provide extreme evidence against H0: Equal means at pvalue 0.001. The mixing machines are not all equal.

F) The data provide strong evidence against H0: Equal means at pvalue 0.013. The mixing machines are not all equal.