Confidence Intervals

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

A. With 95% probability, the true proportion of very satisfied students falls between 85% and 95%

B. There is a 5% probability that the true proportion of very satisfied students falls between 85% and 95%

C. With 95% probability, random confidence intervals contain the true proportion. Loosely, we say that with 95% confidence, the confidence interval (85%, 95%) contains the

D. The proportion of very satisfied students cannot be off by more than 5%

HINT: is the true proportion a random or the confidence interval?

2.With 80% confidence, for sample proportion 0.40 and sample size 28, what is the upper confidence limit with 2 decimal places?

HINT: What is the critical value for an 80% confidence level?

Margin of Error-Mean

3.What is the margin of error for the population mean at 90% of the confidence level for the information in the data below

Wait Time (minutes)

1.1

6.8

1.9

0.1

0.4

3.1

1.9

2.0

5.1

1.3

3.6

3.6

2.7

1.9

7.5

3.8

2.3

9.2

3.0

3.3

1.8

4.9

6.0

4.1

1.8

5.7

6.7

2.4

3.3

8.6

A. 0.75

B. None of the answers match my calculations

C. 0.72

D. 0.73

E. 0.74

HINT: Accuracy is required because three of the answers are close. It is suggested that you compute using Excel, use cell addresses instead of typing intermediate results, and only round the final results using the Number options in the Home tab.

4.With 90% confidence, for sample mean 345.00, sample standard deviation 13.60, and sample size 35, what is the upper confidence limit with 2 decimal places?

HINT: What is the critical value for an 80% confidence level?

Confidence Interval-Test Mean

5.Bobo'a 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 below

Wait Time (minutes)

2.1

8.0

4.6

2.5

3.5

4.3

3.9

5.0

6.5

3.4

5.6

6.6

4.3

4.9

9.3

5.5

4.3

11.9

5.2

4.8

5.2

7.8

8.0

4.6

4.0

5.9

8.7

3.1

4.9

11.6

Using a 95% confidence level, does the confidence interval support Bobo's claim?

A. 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.

B. The confidence limits (4.77, 6.58) supports the claim of 5 minutes. 5 falls inbetween the confidence limits. Bobo's performance meets expectations.

C.T he confidence limits (1.62, 3.21) reject the claim of 5 minutes. 5 is greater than the upper confidence limit. Bobo's performance is incredible.

D. 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 performance and employees is recommended.

E. None of the answers my calculations.

HINT: compute the confidence limits and decide if the confidence interval rejects 5 minutes

6.Match the following

1.Rejection of H0 when H0 is false, or not rejecting H0 when H0 is true

2.The risk we are willing to take of a type 1 error or the type 1 error rate

3.Rejection of H0 when H0 is true

4.Failure to reject H0 when H0 is false

A.Type 1 error

B. Type 2 Error

C. Not an error

D.

7.Match the following

1.Test of this hypothesis requires 1-tailed test with lower reject region bounded by negative critical value

2.Test of this hypothesis requires 1-tailed test with upper reject region

3.Test of this hypothesis requires 2-tailed test with lower reject region bounded by negative critical value and upper reject region bounded by positive critical value

4.Reject H₀

5.The probability of a type 2 error

6.The power of a test = P(rejecting H₀|H₀ false)

A.

B. 1-

C. pvalue <

D. H₀: 0

E. H₀: 0

F. H₀: =0

8.Match the rules for rejecting H0 at the right to the following tests

1.One-tail test with lower reject region

2.For any test hypothesis ANOVA or Chi Squared, this rule for rejecting H0 always applies

3.Two-tail test with lower and upper reject regions

4.One-tail test with upper reject region

A. Test statistic > positive critical value

B. Test statistic < negative critical value

C. Test statistic outside interval (negative critical value, positive critical value)

D. pvalue <

9.Match problems to procedures

1.An effective vaccine would reduce the proportion of exposed persons contracting a disease. A pharmaceutical company wants to test the effectiveness of a new vaccine in preventing a certain disease. It is expected that 40% of unvaccinated exposed people will contract the disease. A group of 48 exposed persons volunteered to be vaccinated. Later, two of the volunteers contracted the disease. Test H0: 0.40 (vaccine ineffective) at =0.01

2.Test size C battery mean life is at least 25 hours

3.Run times (msec) of a new phone app developed by GotYouGP and established competitor WeTrackYou are logged in a data file. At =0.05, are the run rimes of GotYouGP and WeTrackYou comparable?

4.Given weights are 31, 29, 26, 33, 40, 28, 30, and 25 at =0.05 test that the population mean is 35.

5.Testing that a gasoline additive increases mileage, we compute mpg with additive miune mpg without additive for each car. If the mean of the difference is positive enough, action will be required to implement the additive.

A. 1-tailed test of proportion with negative critical mean

B. 2-tailed test of mean with both negative and positive critical values

C. 1-tailed test of mean with negative critical values

D. 1-tailed test of paired-data with positive critical value

E. 2-tailed test of means of 2 independent populations

10.Match problems to procedures

1.Does the data validate the normal probability distribution assumption?

2.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.

3.Do contingency table classification values matter?

4.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. variable, which is common to multiple is common to multiple populations, equality of distribution can be tested.

A. ANOVA

B. Chi-square test of homogeneity

C. Chi-square test of independence

D. Chi-square test of goodness of fit

Test H0-Proportion

11.An Air Quality instrument logs 0 when standards are not met and 1 when standards are met. The log file is below

Time Air Quality

0:00:00 1

0:00:30 0

0:01:00 1

0:01:30 1

0:02:00 0

0:02:30 1

0:03:00 1

0:03:30 1

0:04:00 1

0:04:30 1

0:05:00 1

0:05:30 1

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 0

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 1

0:14:00 1

0:14:30 1

0:15:00 1

0:15:30 0

0:16:00 0

0:16:30 0

0:17:00 1

0:17:30 1

0:18:00 1

0:18:30 1

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 0

0:23:00 1

0:23:30 1

First, compute the proportion meeting the standards as the mean of Air Quality values. Second, at =0.10 (sensitive, exploratory), test the hypothesis that proportion of times that air quality meets standard is at least 90%.

A. The pvalue of 0.966 indicates that the data provide insignifance evidence against H0: 0.90. H0 is not rejects at =0.10. The status quo remains unchanged.

B. None of the answers are correct

C. 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.

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

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.

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

Test H0-Mean

12.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. BatOuaHell is concerned that is 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. Measures of the distance are listed below

Yards

290

272

277

287

274

305

288

297

307

284

272

280

293

279

303

295

292

289

279

At =0.05, test the no action hypothesis that the balls have a mean distance of 295 yards.

A. 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.

B. The test statistic is 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 statistic is 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 13.003 and the critical value is 1.734, therefore the test statistics is greater than the critical value of 1.734 and the null hypothesis is rejected. The distance is not 295 yards.

HINT:

Test H0: =295

HA: = 295

Test H0-Paired Data

13.The Fast N' Hot food chain wants to test if their "BOGO" 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 are below

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

C. 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

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

E. None of the answers are correct

F. 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.

HINT: Is this paired data (dependent on the unit observation), or data from independent populations?

Test H0-Mean

14.BigDeal Real Estate surveyed prices per square foot in the valley and foothills of Hoke-a-mo, Utah. BD's results are below

Valley Foothills

109 75

116 154

106 156

157 105

147 215

105 130

173 219

153 193

137 147

110 169

Based on the results, are prices per square foot equal at =0.01?

A. The critical value is 2.977 since the two-tailed 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 =0.01

B. The critical value is 2.977 since the two-tailed 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 =0.01

C. The critical value is 2.977 since the two-tailed 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 =0.01

D. The critical value is 2.977 since the two-tailed 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 =0.01

HINT: Is this paired data (2 measurement dependent on the unit of observation), or data from 2 independent populations?