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##### stats problems with solution fall 2014

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Question;1In a test of hypothesis, the symbol represents:a The level of significance, or the probability of not rejecting a false null hypothesis.b The level of significance, or the probability of committing a Type I error.c The power of the test, or the probability of rejecting a false null hypothesis.d The level of significance, or the probability of committing a Type II error.2Which of the following statements about Type I and Type II errors is correcta Type I: Reject a true alternative hypothesis. Type II: Do not reject a false alternative.b Type I: Do not reject a false null hypothesis. Type II: Reject a true null hypothesis.c Type I: Reject a false null hypothesis. Type II: Reject a true null hypothesis.d Type I: Reject a true null hypothesis. Type II: Do not reject a false null hypothesis.3You are reading a report that contains a hypothesis test you are interested in. The writer of the report writes thatthe p-value for the test you are interested in is 0.061, but does not tell you the value of the test statistic. From thisinformation you can:a Not reject the hypothesis at a Probability of Type I error = 0.05, and not reject at a Probability of Type I error =0.10b Reject the hypothesis at a Probability of Type I error =.05, and reject at a Probability of Type I error = 0.10c Not reject the hypothesis at a Probability of Type I error =.05, but reject the hypothesis at a Probability of TypeI error = 0.10dReject the hypothesis at a Probability of Type I error =.05, but not reject at a Probability of Type I error = 0.104abcd5A sample of n = 150 is taken and the value of the sample mean income per year is calculated. You want to testthe hypothesis that H: $50,000. The value of the test statistic is 2.03. The p-value is:0.10.050.0150.0212Consider the following hypothesis test.H: 120H: > 120A sample of n = 150 p rovided a sample mean of 128 and sample standard deviation of 65. Compute the teststatistic. At 5% level of significance, what is the conclusion for the test?a Since the test statistic |z| is less than the critical value 1.64. Do not reject the null hypothesis that the mean isless than or equal to 120.b Since the test statistic |z| is less than the critical value 1.64, reject the null hypothesis that the mean is less than orequal to 120.c Since the test statistic |z| is greater than the critical value 1.64, reject the null hypothesis that the mean is lessthan or equal to 120.d Since the test statistic |z| is greater than the critical value 1.64, do not reject the null hypothesis that the mean isless than or equal to 120.6abcdWhat is the probability value of the test statistic?0.07210.08690.06550.0281abcdIn a recent study, a major fast food restaurant had a mean service time of 164 seconds, ranked fourth in theindustry. The company embarks on a quality improvement effort to reduce the service time and has developedimprovements to the service process. The new process will be tested in a sample of 50 stores. The new processwill be adopted in all of its stores, if it resulted in decreased service time. The null and alternative hypothesesfor this test are:H: 164H: 164H: > 164H: 164H: 4,000b H: > 4,000H: 4,000cH: = 4,000H: 4,000d H: 4,000H: 2H: 2d H: 0.50A sample of 200 provided a sample proportion of p = 0.57. At = 0.05, what is your conclusion?a Test statistic = 1.98, critical value = 1.64. Conclude that the population proportion is greater than 0.50.bTest statistic = 1.98, critical value = 1.64. Do not conclude that the population proportion is greater than 0.50.c Test statistic = 2.98, critical value = 1.96. Conclude that the population proportion is greater than 0.50.dTest statistic = 2.98, critical value = 1.96. Do not conclude that the population proportion is greater than 0.50.25abcdIn the previous question, the prob value for the test is:0.04770.02390.01460.0057Next THREE questions are based on the followingConsider the following hypothesis test.H: 0.75H: 0.40H1: 0.40b H0: 0.40d H0: 0.40H1: < 0.40303132If a sample of 420 Internet users found 189 receiving more than 10 e-mail messages per day, using = 0.05,what is your conclusion?a Prob value = 0.0183. Do not conclude that the proportion of Internet users receiving more than 10 e-mails perday has increased.b Prob value = 0.0183. Conclude that the proportion of Internet users receiving more than 10 e-mails per day hasincreased.c Prob value = 0.0885. Do not conclude that the proportion of Internet users receiving more than 10 e-mails perday has increased.d Prob value = 0.0885. Conclude that the proportion of Internet users receiving more than 10 e-mails per day hasincreased.Drugstore.com was the first e-Commerce company to offer Internet drugstore retailing. Drugstore.comcustomers were provided the opportunity to buy health, beauty, personal care, wellness and pharmaceuticalreplenishment products over the Internet. At the end of 10 months of operation, the company reported 44% oforders were from repeat customers. Assume that Drugstore.com will use a sample of customer orders eachquarter to determine if the proportion of orders from repeat customers has changed from the initial = 0.44.During the first quarter a sample of 500 orders showed 205 repeat customers. What is your conclusion using =0.05?a Prob value = 0.1770. Do not conclude that the proportion of orders from repeat customers has changed.b Prob value = 0.1770. Conclude that the proportion of orders from repeat customers has changed.c Prob value = 0.0404. Do not conclude that the proportion of orders from repeat customers has changed.d Prob value = 0.0404. Conclude that the proportion of orders from repeat customers has changed.In the previous question, during the second quarter a sample of 500 orders showed 245 repeat customers. Whatis your conclusion using = 0.05?a Prob value = 0.0244. Do not conclude that the proportion of orders from repeat customers has changed.b Prob value = 0.0244. Conclude that the proportion of orders from repeat customers has changed.c Prob value = 0.0643. Do not conclude that the proportion of orders from repeat customers has changed.d Prob value = 0.0643. Conclude that the proportion of orders from repeat customers has changed.33At least 20% of all workers are believed to be willing to work fewer hours for less pay to obtain more time forp ersonal and leisure activities. A USA Today/CNN/Gallup poll with a sample of 596 respondents found 83willing to work fewer hours for less pay to obtain more personal and leisure time. At 5% level of significance,does the sample result support the claim that at least 20% of all workers are willing to work fewer hours for lessp ay to obtain more time for personal and leisure activities? Round the proportion to two decimal point.a Prob value = 0.0002. Reject the claim. Conclude that less than 20% of all workers are willing to work fewerhours for less pay to obtain more time for personal and leisure activities.b Prob value = 0.0002. Do not reject the claim. Do not conclude that less than 20% of all workers are willing towork fewer hours for less pay to obtain more time for personal and leisure activities.c Prob value = 0.0436. Reject the claim. Conclude that less than 20% of all workers are willing to work fewerhours for less pay to obtain more time for personal and leisure activities.d Prob value = 0.0436. Do not reject the claim. Do not conclude that less than 20% of all workers are willing towork fewer hours for less pay to obtain more time for personal and leisure activities.

Paper#61315 | Written in 18-Jul-2015

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