1. With 95% confidence, the program director estimated that the proportion of MGMT650 students that are very satisfied is 90% 5%.

A. The proportion of very satisfied students cannot be off by more than 5 % B. With 95% probability, the true proportion of very satisfied students falls between 85% and 95%. C. There is a 5 % probability that the true proportion of very satisfied students falls between 85% and 95% D. With 95% probability, random confidence intervals contain the true proportion. Loosely we say that the 95% confidence, the confidence interval (855, 95%) contains the true proportion.

hint for Question Is the true proportion a random or the confidence interval?

2. With 80% confidence, for sample proportion 0.42 and sample size 23, what is the upper confidence limit with 2 decimal places?

hint for Question What is the critical value for an 80% confidence level?

3. What is the margin of error for the population mean at 90% confidence level for the information in the ff. Wait times 1.0 5.3 1.1 0.2 0.2 1.8 0.4. 1.5 3.5 0.7 3.0 2.5 0.4 1.3 5.7 2.1 0.8 7.6 1.3 1.2 1.3 3.7 4.2 2.1 0.3 3.3 5.8 0.6 2.0 7.5

4. With 90% confidence, for sample mean 327.50, sample standard deviation 11.40, and sample size 35, what is the upper confidence limit with 2 decimal places? hint for Question What is the critical value for an 80% confidence level?

5. Bobo's Bad Burgers fast food claims you will be in and out in 5 minutes. To test this claim, time in minutes from entering Bobo's to receiving the order was secretly recorded. The results are documented in Wait Time (minutes) 5.1 10.8 6.2 4.3 5.7 7.3 4.7 7.5 8.7 5.5 8.7 7.7 6.5 6.2 11.4 8.1 5.5 12.3 6.4 7.2 7.6 9.2 8.9 6.9 4.8 9.4 11.7 5.6 7.1 13.4 At a 95% confidence level, does the confidence interval support Bobo's claim?

A. None of the answers match my calculations. B. The confidence limits (6.79,8.56) reject the claim of 5 minutes. 5 is less than the lower confidence limit. Audit of Bobo's procedures and employees is recommended. C. The confidence limits (4.77,6.58) support the claim of 5 minutes. 5 falls in between the confidence limits. Bobo's performance meets the expectation. D. The confidence limits (2.77, 4.54) reject the claim of 5 minutes. 5 is greater than the upper confidence limit. Bobo's performance is exemplary. E. The confidence limits (1.62,3.21) reject the claim of 5 minutes. 5 is greater than the upper confidence limit. Performance at Bobo's is incredible.

hint for Question Compute the confidence limits and decide if the confidence interval rejects 5 minutes?

6. Air Quality instrument logs 0 when standards are not met and 1 when standards are met. The log is saved to file Time Air Quality 0:00:00 1 0:00:30 1 0:01:00 0 0:01:30 1 0:02:00 1 0:02:30 1 0:03:00 0 0:03:30 1 0:04:00 1 0:04:30 1 0:05:00 1 0:05:30 0 0:06:00 1 0:06:30 1 0:07:00 1 0:07:30 1 0:08:00 1 0:08:30 1 0:09:00 1 0:09:30 1 0:10:00 1 0:10:30 1 0:11:00 1 0:11:30 1 0:12:00 1 0:12:30 1 0:13:00 1 0:13:30 0 0:14:00 0 0:14:30 1 0:15:00 1

0:15:30 1 0:16:00 1 0:16:30 1 0:17:00 1 0:17:30 0 0:18:00 0 0:18:30 0 0:19:00 1 0:19:30 1 0:20:00 1 0:20:30 1 0:21:00 1 0:21:30 0 0:22:00 1 0:22:30 1 0:23:00 0 0:23:30 1 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 0.062 indicates that the data provide weak evidence against H0: 0.90. H0 is rejected at = 0.10. B. 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. C. he 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. D. None of the answers are correct. E. 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.

hint for Question The sample proportion is the mean of the Air Quality values. Compute the mean to 4 decimal place accuracy. Test H0: 0.90.

7. 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 Yards 293 275 280 290 273 306 287 301 309 285 274 282 294 283 304 297 294 290 283 At = 0.05, test the no action hypothesis that the balls have a mean distance of 295 yards

A. 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. B. 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. C. 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. D. 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.

hint for Question Test H0: = 295.

HA: 295

8. 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 Customer Traffic With Program Without Program 144 140 236 233 108 110 43 42 337 332 134 135 148 151 30 33 181 178 146 147 159 162 248 243 150 149 54 48 349 346 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 rejects H0: Mean difference > 0. B. 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 implemented. C. None of the answers are correct. D. 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. E. 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. F. 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.

hint for Question Is this paired data (dependent on the unit of observation), or data from independent populations?

9. BigDeal Real Estate surveyed prices per square foot in the valley and foothills of Hoke-a-mo, Utah. Based on BD's DATA Valley Foothills 109 82 116 161 106 163 157 112 147 222 105 137 173 226 153 200 137 154 110 176 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 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 B. 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 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 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

hint for Question Is this paired data (2 measurements dependent on the unit of observation), or data from 2 independent populations?

10. 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 Observation Portland Houston Jacksonville 1 85 71 69 2 75 75 74 3 82 73 72 4 76 74 79 5 71 69 85 6 85 82 77 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 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. 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 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