NAME: Enter your LETTER answers HERE 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 1 a b c d 2 a b c d Note: When done, using your last name and firstname, save THIS FILE as, E270Lastname Firstname HW6 (No space between E270 and Last name) E-mail to stowfig@iupui.edu as an attachment. Example: e270Smith Adam HW6 e270 Smith Adam HW6 YES NO PAY ATTENTION Which of the following statements about Type I and Type II errors is correct Type I: Reject a true null hypothesis. Type II: Do not reject a false null hypothesis. Type I: Reject a true alternative hypothesis. Type II: Do not reject a false alternative. Type I: Do not reject a false null hypothesis. Type II: Reject a true null hypothesis. Type I: Reject a false null hypothesis. Type II: Reject a true null hypothesis. You are reading a report that contains a hypothesis test you are interested in. The writer of the report writes that the p-value for the test you are interested in is 0.0851, but does not tell you the value of the test statistic. From this information you can: Not reject the hypothesis at a Probability of Type I error = 0.05, and not reject at a Probability of Type I error = 0.10 Reject the hypothesis at a Probability of Type I error = .05, and reject at a Probability of Type I error = 0.10 Not reject the hypothesis at a Probability of Type I error = .05, but reject the hypothesis at a Probability of Type I error = 0.10 Reject the hypothesis at a Probability of Type I error = .05, but not reject at a Probability of Type I error = 0.10 Next TWO question are based on the random sample below whick is obtained to test the following hypothesis about the population mean. 76 137 87 128 123 58 104 81 89 135 87 105 122 110 130 70 51 59 80 122 117 73 130 89 89 93 60 61 138 137 110 69 140 108 100 85 63 87 139 70 105 116 107 76 78 87 87 136 98 112 66 92 70 91 53 119 50 103 62 93 104 87 139 79 60 56 51 68 120 52 132 88 64 75 139 120 107 103 138 H: 100 H: < 100 3 93 133 124 135 115 60 114 80 137 139 120 71 78 90 55 106 111 91 96 70 64 118 103 83 106 77 72 120 114 54 140 115 83 133 56 97 94 84 50 116 93 84 74 136 61 123 130 108 92 a c d d This hypothesis test is, an two-tail test because the null hypothesis is H: 100. an upper tail test because the alternative hypothesis is H : < 100. an upper tail test because the null hypothesis is H: 100. a lower tail test because the alternative hypothesis is H: < 100. a b c d The value of the test stastic is (if negative use the absolute value), 1.82 1.70 1.64 1.58 a b c d Compute the p-value (probability value). Using a level of significance of = 0.05, the conclusion for the test is: 0.0446 Reject H. Conclude the mean is less than 100. 0.0446 Do not reject H. Do not conclude the mean is less than 100. 0.0571 Reject H. Conclude the mean is less than 100. 0.0571 Do not reject H. Do not conclude the mean is less than 100. 4 5 72 76 94 137 81 95 127 6 a b c d 7 Consider the following hypothesis test. H: = 50 H: 50 A random sample of n = 25 yielded the following data. 25 89 59 88 74 56 81 58 21 69 27 79 44 53 43 29 64 51 71 54 68 50 38 70 71 Use = 0.05 TS = ______ CV = ______ Based on the decision rule, state your conclusion. 2.08 1.64 Reject H. Conclude is different from 50. 2.08 1.711 Reject H. Conclude is different from 50. 1.87 1.96 Do not reject H. Cannot conclude is different from 50. 1.87 2.064 Do not reject H. Cannot conclude is different from 50. In a recent study, a major fast food restaurant had a mean drive through service time of 210 seconds. The company embarks on a quality improvement effort to reduce the service time and has developed improvements to the service process. The new process will be tested in a sample of stores. The new process will be adopted in all of its stores, if it reduced mean service time by more than 30 seconds compared to the current mean service time. To perform the hypothesis test, the sample of 20 stores yields the following data (seconds). 149 143 157 204 173 205 160 146 190 190 153 196 168 152 180 187 160 166 142 185 a b c d Use = 0.05 State the null and alternative hypotheses. |TS| = ______ CV = ______ 2.113 1.729 Reject H. Adopt the new process. 2.113 1.64 Reject H. Do not adopt the new process. 1.542 1.729 Do not reject H. Adopt the new process. 1.542 1.64 Do not reject H. Do not adopt the new process. 8 a b c d According to Automotive News, the mean 3-years-old used car price is $23,400. To compare the average price of similar used cars in central Indiana, a random sample of 125 such cars were selected. The sample mean is $24,920 with a standard deviation of $7,154. Does the sample provide significant evidence that the mean price of 3-years-old used cars in central Indiana is greater than the national mean price? Use = 0.05 p-value = 0.0269. Do not reject H. Do not conclude the average price in central Indiana is greater than 23,400. p-value = 0.0269. Reject H. Conclude the average price in central Indiana is greater than 23,400. p-value = 0.0087. Reject H. Conclude the average price in central Indiana is greater than 23,400. p-value = 0.0537. Reject H. Conclude the average price in central Indiana is greater than 23,400. 9 The 2014 mean annual salary of graduates with engineering degrees was $63,800. In a follow-up study in June 2015, a sample of n = 110 graduating engineering majors yielded a sample mean of $64,980 and standard deviation of $9,245. Does the 2015 survey provide a significant proof that the mean salary in 2015 is different than the 2014 mean? Perform this test of hypothesis at a 5% level of significance. a b c d 10 a b c d 11 a b c d p-value = 0.0901. 2014 level. p-value = 0.0451. p-value = 0.0901. p-value = 0.1802. 2014 level. Do not reject H. Do not conclude that the mean salary in 2015 is different than the Reject H. Conclude that the mean salary in 2015 is different than the 2014 level. Reject H. Conclude that the mean salary in 2015 is different than the 2014 level. Do not reject H. Do not conclude that the mean salary in 2015 is different than the A production line operates with a mean filling weight of 32 ounces per container. Overfilling or under filling is a serious problem, and the production line should be shut down if either occurs. A quality control inspector samples 16 items every 2 hours and at that time makes the decision of whether to shut the line down for adjustment. One sample provides the following data: Use Excel to find xx and s to avoid rounding problems. 33.3 32.2 32.3 31.6 31.8 33.4 32.1 32.6 31.8 31.0 32.2 33.0 33.3 33.5 31.2 33.0 = 0.05 Decision Rule: Reject H if TS > CV TS = ______ 2.13 Do not reject H. Do not shut the line down for adjustment. 2.13 Reject H. Shut the line down for adjustment. 1.98 Do not reject H. Do not shut the line down for adjustment. 1.98 Reject H. Shut the line down for adjustment. The mean cholesterol level in women ages 21-40 in the United States is 190 mg/dl. A study is conducted to determine the cholesterol levels among recent female Asian immigrants. The following is the cholesterol level of a random sample of 117 recent female Asian immigrants. 116 149 242 145 109 187 202 109 146 201 173 127 172 167 193 135 145 153 176 160 214 176 152 166 126 155 253 203 114 152 117 203 184 249 217 259 257 211 163 194 103 109 114 159 226 181 230 125 115 127 145 120 216 235 258 196 231 152 183 175 185 123 155 157 210 252 156 164 260 222 229 209 218 158 231 257 181 259 231 184 159 212 175 204 143 233 227 183 147 240 233 236 117 170 207 178 220 208 260 186 185 188 152 107 214 223 146 121 242 239 112 205 114 155 140 153 194 Does the sample provide significant evidence that mean cholesterol level of recent female Asian immigrants is lower than the mean cholesterol level among all females in the United States? State the null and alternative hypotheses. Compute the test statistic and the p-value. State the decision rule. Round xx to two decimal points and the standard error to three decimal points. p-value = ______ 0.0622 The evidence is significant at = 0.01, but not significant at = 0.05. 0.0622 The evidence is significant at = 0.10, but not significant at = 0.05. 0.0197 The evidence is significant at = 0.01, but not significant at = 0.05. 0.0197 The evidence is significant at = 0.05, but not significant at = 0.01. 12 We want to test the hypothesis that mothers with low socio-economic status (SES) deliver babies whose birth weights are lower than "normal". To test this hypothesis, a list is obtained of birth weights from 120 consecutive, full-term, live-born deliveries from the maternity ward of a hospital in a low-SES area. The mean birth weight is xx = 114.9 oz. with a standard deviation s = 21.2 oz. Nationwide, the mean birth weight in the United States is 120 oz. At = 0.05, does this sample provide significant evidence that the mean birth weight of babies born to mothers with low SES is lower than "normal" birth weight? a b c d 13 a b c d 14 a b c d e 15 a b c We want to test the hypothesis that mothers with low socio-economic status (SES) deliver babies whose birth weights are lower than "normal". To test this hypothesis, a list is obtained of birth weights from 120 consecutive, full-term, live-born deliveries from the maternity ward of a hospital in a low-SES area. The mean birth weight is xx = 114.9 oz. with a standard deviation s = 21.2 oz. Nationwide, the mean birth weight in the United States is 120 oz. At = 0.05, does this sample provide significant evidence that the mean birth weight of babies born to mothers with low SES is lower than "normal" birth weight? p-value = 0.0041 Reject H at the 1 percent level of significance. Conclude that the mean birth weight of babies born to low-SES mothers is lower than "normal". p-value = 0.0715 Reject H at the 10 percent level of significance. Conclude that the mean birth weight of babies born to low-SES mothers is lower than "normal". p-value = 0.0150 Reject H at the 1 percent level of significance. Conclude that the mean birth weight of babies born to low-SES mothers is lower than "normal". p-value = 0.0424 Do not reject H at the 5 percent level of significance. Conclude that the mean birth weight of babies born to low-SES mothers is not lower than "normal". A random sample of n = 125 Indiana undergraduate students revealed a mean dredit card debt of xx = $4,582 and the standard deviation of s = $1,695. A 95% confidence interval for the mean credit card debt of Indiana undergraduate students is: $4,285 $4,879 $4,137 $5,028 $4,062 $5,102 $3,988 $5,176 Nationwide, the mean credit card debt of undergraduate students is reported to be $4,100. You want to perform a hypothesis test that Indiana mean undergraduate student credit card debt is different from the national mean. Base your conclusion on the interval estimate in the previous question. The 95% confidence interval for Indiana students captures the nationwide mean. Reject the null hypothesis and conclude the Indiana mean is different from the national mean. The 95% confidence interval for Indiana students captures the nationwide mean. Do not reject the null hypothesis and conclude the Indiana mean is not different from the national mean. The 95% confidence interval for Indiana students does not capture the nationwide mean. Reject the null hypothesis and conclude the Indiana mean is different from the national mean. Compared to the MOE in the confidence interval, xx is not within the margin of error. Conclude that the mean credit card balance of Indiana undergraduates is different from the national average. Both c and d are correct Consider the following hypothesis test. H: 0.42 H: > 0.42 A sample of n = 300 provided a sample proportion of px = 0.45. At = 0.05, what is your conclusion? TS = ______ CV = ______ State the decision rule. 1.72 1.64 Do not conclude the population proportion is greater than 0.40. 1.72 1.96 Conclude the population proportion is greater than 0.40. 1.05 1.64 Conclude the population proportion is greater than 0.40. d 1.05 1.64 Do not conclude the population proportion is greater than 0.40. 16 a b c d In the previous question, the prob value for the test is: 0.1762 0.1678 0.1469 0.0427 17 One of the different statistics reported by the Centers for Disease Control regarding incidence of obesity among adults in the United States provides that 28% of men with college degree are obese. The study also reports that 31.5% of men without a college degree are obese. Assume the latter statistic is based on a sample of 1100 men without a college degree. Does the data provide statistically significant evidence that the incidence of obesity among men without a college degree is greater than among those with a college degree? Compute the p-value for this hypothesis test. p-value = ______. 0.0344 The evidence is not statistically significant. Reject H . 0.0344 The evidence is statistically significant. Reject H . 0.0048 The evidence is not statistically significant. Reject H . 0.0048 The evidence is statistically significant. Reject H . a b c d 18 a b c d 19 We want to test the hypothesis that at least 85% of drivers on a freeway violate the speed limit. In a random sample of n = 1,020 vehicles, 83% violated the speed limit. Compute the test statistic. State the null and alternative hypotheses and the decision rule. Use = 0.05. TS = 1.79 Reject H. The proportion of violators is less than 0.85. TS = 1.79 Do not reject H. The proportion of violators is not less than 0.85. TS = 1.45 Do not reject H. The proportion of violators is not less than 0.85. TS = 1.45 Reject H. The proportion of violators is less than 0.85. In a sample of 750 coffee growers from southern Mexico, 435 growers were certified to sell to organic coffee markets. Is there evidence to indicate that fewer than 60% of the coffee growers in southern Mexico are organic certified? State your conclusion so that there is only 5% chance of making Type I error. a p-value = 0.0657 b p-value = 0.1314 c p-value = 0.0657 d p-value = 0.1314 20 a b c d At a 5% level of significance, the evidence that fewer than 60% are organic certified is statistically significant. At a 5% level of significance, the evidence that fewer than 60% are organic certified is not statistically significant. At a 5% level of significance, the evidence that fewer than 60% are organic certified is not statistically significant. At a 5% level of significance, the evidence that fewer than 60% are organic certified is statistically significant. In the previous question, the margin of error for rejecting the null hypothesis is. 0.037 0.033 0.029 0.023