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Question
A decision maker wishes to test the following null and alternative hypotheses using an alpha level equal to 0.05:
H0 : 
1 - 2 = 0
HA : 1 - 2 
0
The population standard deviations are assumed to be known. After collecting the sample data, the test statistic is computed to be z = 1.78
Using the p-value approach, what decision should be reached about the null hypothesis?
A) Since p-value = 0.0018 < 
/2 = 0.025, reject the null hypothesis
B) Since p-value = 0.0415 > /2 = 0.025, do not reject the null hypothesis.
C) Since p-value = 0.0033 < /2 = 0.025, reject the null hypothesis.
D) Since p-value = 0.0375 > /2 = 0.025, do not reject the null hypothesis.
Answer
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According to CNN business partner Careerbuilder.com, the average starting salary for accounting graduates in 2008 was at least $47,413. Suppose that the American Society for Certified Public Accountants planned to test this claim by randomly sampling 200 accountants who graduated in 2008.
Compute the power of the hypothesis test to reject the null hypothesis if the true average starting salary is only $47,000. Assume that the population standard deviation is known to be $4,600 and the test is to be conducted using an alpha level equal to 0.01.
A) 0.0872
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D) 0.9812
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Swift is the holding company for Swift Transportation Co., Inc., a truckload carrier headquartered in Phoenix, Arizona. Swift operates the largest truckload fleet in the United States. Before Swift switched to its current computer-based billing system, the average payment time from customers was approximately 40 days. Suppose before purchasing the present billing system, it performed a test by examining a random sample of 24 invoices to see if the system would reduce the average billing time. The sample indicates that the average payment time is 38.7 days.
The company that created the billing system indicates that the system would reduce the average billing time to less than 40 days. Conduct a hypothesis test to determine if the new computer-based billing system would reduce the average billing time to less than 40 days. Assume the standard deviation is known to be 6 days. Use a significance level of 0.025.
A) Since z = -0.423 > -1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days.
B) Since z = -1.0614 > -1.96, we will not reject H0, there is not sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days.
C) z = -1.0231 > -1.96, we will reject H0, there is sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days.
D) z = 0.341 > -1.96, we will reject H0, there is sufficient evidence to conclude that the new computer-based billing system would reduce the average billing time to less than 40 days.
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According to data from the Environmental Protection Agency, the average daily water consumption for a household of four people in the United States is approximately at least 243 gallons. (Source: http://www.catskillcenter.org/programs/csp/H20/Lesson3/house3.htm) Suppose a state agency plans to test this claim using an alpha level equal to 0.05 and a random sample of 100 households with four people.
Calculate the probability of committing a Type II error if the true population mean is 230 gallons. Assume that the population standard deviation is known to be 40 gallons.
A) 0.0331
B) 0.0712
C) 0.0537
D) 0.1412
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You are given the following null and alternative hypotheses: If the true population mean is 4,345, determine the value of beta. Assume the population standard deviation is known to be 200 and the sample size is 100.
A) 0.9192
B) 0.8233
C) 0.6124
D) 0.0314
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Q:
A major issue facing many states is whether to legalize casino gambling. Suppose the governor of one state believes that more than 55% of the state's registered voters would favor some form of legal casino gambling. However, before backing a proposal to allow such gambling, the governor has instructed his aides to conduct a statistical test on the issue. To do this, the aides have hired a consulting firm to survey a simple random sample of 300 voters in the state. Of these 300 voters, 175 actually favored legalized gambling.
Assuming that a significance level of 0.05 is used, what conclusion should the governor reach based on these sample data?
A) Since z = 1.1594 < 1.645, do not reject the null hypothesis.
The sample data do not provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling.
B) Since z = 2.1316 > 1.645, reject the null hypothesis.
The sample data provide sufficient evidence to conclude that more than 55 percent of the population favor legalized gambling.
C) Since z = 1.1594 < 1.645, do not reject the null hypothesis.
The sample data do not provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling.
D) Since z = 2.1316 > 1.645, reject the null hypothesis.
The sample data provide sufficient evidence to conclude that more than 58 percent of the population favor legalized gambling.
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A) 2.7299
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C) 1.4919
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B) Because the calculated value of the test statistic, t=1.7291, is neither greater than 2.013 nor less than -2.013, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.
C) Because the calculated value of the test statistic, z = 1.2412, is neither greater than 2.575 nor less than -2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.
D) Because the calculated value of the test statistic, z = 0.3266, is neither greater than 2.575 nor less than -2.575, do not reject the null hypothesis and conclude that the population proportion is not different from 0.40.
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A) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type II error.
B) The null hypothesis, the specification not being met, was not rejected when in fact it was not being met, this is a Type I error.
C) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type II error.
D) The null hypothesis, the specification is being met, was rejected when in fact it was being met, this is a Type I error.
Q:
The Center on Budget and Policy Priorities (www.cbpp.org) reported that average out-of-pocket medical expenses for prescription drugs for privately insured adults with incomes over 200% of the poverty level was $173 in 2002. Suppose an investigation was conducted in 2012 to determine whether the increased availability of generic drugs, Internet prescription drug purchases, and cost controls have reduced out-of-pocket drug expenses. The investigation randomly sampled 196 privately insured adults with incomes over 200% of the poverty level, and the respondents' 2012 out-of-pocket medical expenses for prescription drugs were recorded. These data are in the file Drug Expenses. Based on the sample data, can it be concluded that 2012 out-of-pocket prescription drug expenses are lower than the 2002 average reported by the Center on Budget and Policy Priorities? Use a level of significance of 0.01 to conduct the hypothesis test.
A) Because t = -2.69 is less than -2.3456, do not reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average.
B) Because t = -2.69 is less than -2.3456, reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average.
C) Because t = -1.69 is less than -0.8712, reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are lower than the 2002 average.
D) Because t = -1.69 is less than -0.8712, do not reject H0
Conclude that 2012 average out-of-pocket prescription drug expenses are not lower than the 2002 average.
Q:
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A) p-value = 0.0872 > 0.025; therefore do not reject H0
B) p-value = 0.0422 > 0.005; therefore do not reject H0
C) p-value = 0.0314 < 0.105; therefore reject H0
D) p-value = 0.0121< 0. 0805; therefore reject H0
Q:
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Assume that a standard deck of 52 playing cards is randomly shuffled and the first 2 cards are dealt to you. What is the probability that you have a blackjack? A blackjack is where one card is an ace and the other card is worth 10 points. The 10-point cards are kings, queens, jacks and 10's.