7 The 2010 mean annual salary of business degree graduates in accounting was $49

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7 The 2010 mean annual salary of business degree graduates in accounting was $49,400. In a follow-up study in June 2012, a sample of n = 120 graduating accounting majors yielded a sample mean of $51,120 and standard deviation of $8,640. Does the 2012 study provide a significant proof that the mean salary in 2012 is higher than in 2010? Perform this test of hypothesis at a 5% level of significance. a p-value = 0.0546 Do not reject H?. Conclude that the mean annual salary in 2012 is no greater than in 2010. b p-value = 0.0546 Reject H?. Conclude that the mean annual salary in 2012 is greater than in 2010. c p-value = 0.0146 Reject H?. Conclude that the mean annual salary in 2012 is greater than in 2010. d p-value = 0.0146 Do not reject H?. Conclude that the mean annual salary in 2012 is no greater than in 2010. 8 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: 32.8 32.1 31.2 31.3 33.4 32.4 32.3 32.3 31.6 32.5 33.5 32.7 31.8 32.9 33.1 31.8 ? = 0.05 TS = ______ a 2.05 Do not reject H?. Do not shut the line down for adjustment. b 2.05 Reject H?. Shut the line down for adjustment. c 2.13 Do not reject H?. Shut the line down for adjustment. d 2.13 Reject H?. Shut the line down for adjustment. 9 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. 198 185 194 184 185 181 181 185 195 197 190 199 192 187 196 191 199 192 200 183 194 181 181 198 189 180 179 186 184 192 187 191 184 199 194 181 195 191 195 180 196 182 191 192 180 186 190 178 193 186 200 180 187 195 194 178 184 193 180 180 180 190 195 199 199 188 191 194 187 197 186 191 188 187 180 190 188 185 184 195 190 178 192 185 193 196 194 198 180 196 198 197 178 195 200 188 193 186 179 180 183 198 186 185 189 185 181 192 184 184 182 193 183 188 186 193 194 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 x? to two decimal points and the standard error to three decimal points. p-value = ______ a 0.026 The evidence is significant at ? = 0.05, but not significant at ? = 0.01. b 0.026 The evidence is significant at ? = 0.10, but not significant at ? = 0.05. c 0.068 The evidence is significant at ? = 0.05, but not significant at ? = 0.01. d 0.068 The evidence is significant at ? = 0.01, but not significant at ? = 0.05. 10 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 110 consecutive, full-term, live-born deliveries from the maternity ward of a hospital in a low-SES area. The mean birth weight is x? = 115.4 oz. with a standard deviation s = 23.6 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 mother with low SES is lower than "normal"? a p-value = 0.0207 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". b p-value = 0.0207 Reject H? at the 5 percent level of significance. Conclude that the mean birth weight of babies born to low-SES mothers is lower than "normal". c p-value = 0.0785 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 no lower than "normal". d p-value = 0.0785 Do not reject H? at the 10 percent level of significance. Conclude that the mean birth weight of babies born to low-SES mothers is no lower than "normal". 11 In a recent study, it was reported that nationwide, undergraduate students have a mean credit card balance of $3,480. A random sample of 150 Indiana undergraduate students revealed a sample mean of $3,675 and standard deviation of $1,285. Does the sample provide significant evidence that the mean credit card balance of Indiana undergraduates is different than the national average? Using the sample information, first build a 95% confidence interval for the mean credit card balance of all Indiana undergraduates. L = U = a The confidence interval captures 7 The 2010 mean annual salary of business degree graduates in accounting was $49,400. In a follow-up study in June 2012, a sample of n = 120 graduating accounting majors yielded a sample mean of $51,120 and standard deviation of $8,640. Does the 2012 study provide a significant proof that the mean salary in 2012 is higher than in 2010? Perform this test of hypothesis at a 5% level of significance. a p-value = 0.0546 Do not reject H?. Conclude that the mean annual salary in 2012 is no greater than in 2010. b p-value = 0.0546 Reject H?. Conclude that the mean annual salary in 2012 is greater than in 2010. c p-value = 0.0146 Reject H?. Conclude that the mean annual salary in 2012 is greater than in 2010. d p-value = 0.0146 Do not reject H?. Conclude that the mean annual salary in 2012 is no greater than in 2010. 8 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: 32.8 32.1 31.2 31.3 33.4 32.4 32.3 32.3 31.6 32.5 33.5 32.7 31.8 32.9 33.1 31.8 ? = 0.05 TS = ______ a 2.05 Do not reject H?. Do not shut the line down for adjustment. b 2.05 Reject H?. Shut the line down for adjustment. c 2.13 Do not reject H?. Shut the line down for adjustment. d 2.13 Reject H?. Shut the line down for adjustment. 9 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. 198 185 194 184 185 181 181 185 195 197 190 199 192 187 196 191 199 192 200 183 194 181 181 198 189 180 179 186 184 192 187 191 184 199 194 181 195 191 195 180 196 182 191 192 180 186 190 178 193 186 200 180 187 195 194 178 184 193 180 180 180 190 195 199 199 188 191 194 187 197 186 191 188 187 180 190 188 185 184 195 190 178 192 185 193 196 194 198 180 196 198 197 178 195 200 188 193 186 179 180 183 198 186 185 189 185 181 192 184 184 182 193 183 188 186 193 194 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 x? to two decimal points and the standard error to three decimal points. p-value = ______ a 0.026 The evidence is significant at ? = 0.05, but not significant at ? = 0.01. b 0.026 The evidence is significant at ? = 0.10, but not significant at ? = 0.05. c 0.068 The evidence is significant at ? = 0.05, but not significant at ? = 0.01. d 0.068 The evidence is significant at ? = 0.01, but not significant at ? = 0.05. 10 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 110 consecutive, full-term, live-born deliveries from the maternity ward of a hospital in a low-SES area. The mean birth weight is x? = 115.4 oz. with a standard deviation s = 23.6 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 mother with low SES is lower than "normal"? a p-value = 0.0207 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". b p-value = 0.0207 Reject H? at the 5 percent level of significance. Conclude that the mean birth weight of babies born to low-SES mothers is lower than "normal". c p-value = 0.0785 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 no lower than "normal". d p-value = 0.0785 Do not reject H? at the 10 percent level of significance. Conclude that the mean birth weight of babies born to low-SES mothers is no lower than "normal". 11 In a recent study, it was reported that nationwide, undergraduate students have a mean credit card balance of $3,480. A random sample of 150 Indiana undergraduate students revealed a sample mean of $3,675 and standard deviation of $1,285. Does the sample provide significant evidence that the mean credit card balance of Indiana undergraduates is different than the national average? Using the sample information, first build a 95% confidence interval for the mean credit card balance of all Indiana undergraduates. L = U = a The confidence interval captures

Explanation / Answer

(7) The test statistic is

Z=(xbar-mu)/(s/vn)

=(51120-49400)/(8640/sqrt(120))

=2.18

It is a one-tailed test.

So P-value= P(Z>2.18) =0.0146 (from standard normal table)

Answer:

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(8)TS =(xbar-mu)/(s/vn)

=(32.356-32)/(0.6976/sqrt(16))

=2.05

Given a=0.05, the critical values are t(0.025, df=n-1=15) = -2.13 or 2.13 (from student t table)

Since t= 2.05 is between -2.13 and 2.13, we do not reject Ho.

Answer:

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(9) The test statistic is

Z=(xbar-mu)/(s/vn)

=(188.855-190)/(6.378/sqrt(117))

=-1.94

It is a one-tailed test.

So the P-value= P(Z<-1.94) =0.0262 (from standard noraml table)

Answer:

c p-value = 0.0146 Reject H?. Conclude that the mean annual salary in 2012 is greater than in 2010.

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