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Question
A random sample of two variables, x and y, produced the following observations:| x | y | |
| 19 | 7 | |
| 13 | 9 | |
| 17 | 8 | |
| 9 | 11 | |
| 12 | 9 | |
| 25 | 6 | |
| 20 | 7 | |
| 17 | 8 |
Test to determine whether the population correlation coefficient is negative. Use a significance level of 0.05 for the hypothesis test.
A) Because t = -4.152 < -1.9432, reject the null hypothesis. Because the null hypothesis is rejected, the sample data does support the hypothesis that there is a negative linear relationship between x and y.
B) Because t = -4.152 < -1.9432, do not reject the null hypothesis. Because the null hypothesis is not rejected, the sample data support the hypothesis that there is a negative linear relationship between x and y.
C) Because t = -9.895 < -1.9432, reject the null hypothesis. Because the null hypothesis is rejected, the sample data does support the hypothesis that there is a negative linear relationship between x and y.
D) Because t = -9.895 < -1.9432, do not reject the null hypothesis. Because the null hypothesis is not rejected, the sample data support the hypothesis that there is a negative linear relationship between x and y.
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.
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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.
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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
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C) 0.0537
D) 0.1412
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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.
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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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For the following hypothesis test: With n= 0.42 and p = 0.42, state the conclusion
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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.
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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
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