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23 Cards in this Set
- Front
- Back
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A machine is designed to fill toothpaste tubes with 5.8 ounces of toothpaste. The manufacturer does not want any underfilling or overfilling. The correct hypotheses to be tested are
a.H0: μ ≠ 5.8 Ha: μ = 5.8 b.H0: μ = 5.8 Ha: μ ≠ 5.8 Correct c.H0: μ > 5.8 Ha: μ ≤ 5.8 d.H0: μ ≥ 5.8 Ha: μ < 5.8 |
b. H0: μ = 5.8 Ha: μ ≠ 5.8 Correct
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A Type II error is committed when
a.a true alternative hypothesis is mistakenly rejected b.a true null hypothesis is mistakenly rejected c.the sample size has been too small d.not enough information has been available |
a. a true alternative hypothesis is mistakenly rejected Correct
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n = 36 = 24.6 S = 12 H0: μ 20 Ha: μ > 20 Refer to Exhibit 9-1(hint: S is the standard deviation of the sample). If the test is done at 95% confidence, the null hypothesis should a.not be rejected b.be rejected c.Not enough information is given to answer this question. d.None of these alternatives is correct. |
b. be rejected Correct
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n = 36 = 24.6 S = 12 H0: μ 20 Ha: μ > 20
Refer to Exhibit 9-1(hint: S is the standard deviation of the sample). The test statistic is Select one: a.2.3 b.0.38 c.-2.3 d.-0.38 |
a. 2.3 Correct
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n = 64 = 50 s = 16 H0: μ 54 Ha: μ < 54Refer to Exhibit 9-2 (hint: S is the standard deviation of the sample). If the test is done at 95% confidence, the null hypothesis should
a.not be rejected b.be rejected c.Not enough information is given to answer this question. d.None of these alternatives is correct. |
b. be rejected Correct
(50-54)/(16/sqrt(64))=-2. t0.05with df 63=1.669. -2<-1.669, we support alternative and reject null. |
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n = 64 = 50 s = 16 H0: μ 54 Ha: μ < 54Refer to Exhibit 9-2(hint: S is the standard deviation of the sample). The test statistic equals
a.-4 b.-3 c.-2 Correct d.-1 |
c. -2 Correct
(50-54)/(16/sqrt(64))=-2. The correct answer is:-2 |
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The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.
Refer to Exhibit 9-4. At 95% confidence, it can be concluded that the mean of the population is a.significantly greater than 3 Correct b.not significantly greater than 3 c.significantly less than 3 d.significantly greater then 3.18 |
a. significantly greater than 3 Correct
(3.1-3)/(0.5/sqrt(100))=2, which is bigger than 1.660, t0.05 with df=99. So it's significantly bigger than 3 |
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The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took the customers in the sample to check out was 3.1 minutes with a standard deviation of 0.5 minutes. We want to test to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes.Refer to Exhibit 9-4. The test statistic is
a.1.96 b.1.64 c.2.00 d.0.056 |
c. 2.00 Correct
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A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 80%.Refer to Exhibit 9-5. The test statistic is
a.0.80 b.0.05 c.1.25 d.2.00 |
c. 1.25 Correct
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A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in determining whether or not the proportion of the population in favor of Candidate A is significantly more than 80%.
Refer to Exhibit 9-5. At 95% confidence, it can be concluded that the proportion of the population in favor of candidate A a.is significantly greater than 80% b.is not significantly greater than 80% c.is significantly greater than 85% d.is not significantly greater than 85% |
b. is not significantly greater than 80% Correct
(0.85-0.8)/sqrt(0.8*(1-0.8)/100)=1.25, which is smaller than z0.05=1.645, we cannot choose alternative, we cannot reject null. |
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The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200.Refer to Exhibit 9-9. The correct null hypothesis for this problem is
a.μ < 8000 b.μ > 8000 c.μ = 8000 d.μ > 8250 |
a. μ < 8000 Correct
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The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200.Refer to Exhibit 9-9. The value of the test statistic is
a.250 b.8000 c.8250 d.2.0 |
d. 2.0 Correct
(8300-8000)/(1200/sqrt(64))=2 |
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The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200.
At the 5% significance level, the critical value is (Specify to 3 decimal places): At the 10% significance level, the critical value is (Specify to 3 decimal places): |
z0.05=1.645 z 0.10=1.282 *literally look at the z table |
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The sales of a grocery store had an average of $8,000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200.Refer to Exhibit 9-9. At the 5% significant level, what is your conclusion?
a.Reject the null hypothesis. b.Fail to reject the null hypothesis |
a. Reject the null hypothesis. Correct
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For a one-tailed test (lower tail), a sample size of 10 at 90% confidence, the critical value =
a.1.383 b.2.821 c.-1.383 Correct d.-2.821 |
c. -1.383 Correct
check t table, t10% is 1.383 with df 9. Since it's lower tail, ie. alternative is x<#, the critical value is -t.10 |
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For a one-tailed test (lower tail), a sample size of 10 at 90% confidence, the critical value =
a.1.383 b.2.821 c.-1.383 Correct d.-2.821 |
c. -1.383 Correct
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For a one-tailed test (upper tail), a sample size of 18 at 95% confidence, the critical value =
a.2.12 b.-2.12 c.-1.740 d.1.740 |
d. 1.740 Correct
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For a two-tailed test, a sample of 20 at 80% confidence, the critical value =
a.1.328 b.2.539 c.1.325 d.2.528 |
a. 1.328 Correct
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If a hypothesis is not rejected at the 5% level of significance, it
a.will also not be rejected at the 1% level b.will always be rejected at the 1% level c.will sometimes be rejected at the 1% level d.None of these alternatives is correct. |
a. will also not be rejected at the 1% level Correct
If we cannot support alternative hypothesis with 5% chance making mistake, we also cannot support alternative hypothesis with 1% chance making mistake. The critical value for 1% always is bigger than the value for 5%. |
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In hypothesis testing if the null hypothesis is rejected,
a.no conclusions can be drawn from the test b.the alternative hypothesis is true c.the data must have been accumulated incorrectly d.the sample size has been too small |
b. the alternative hypothesis is true Correct
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In order to test the following hypotheses at an α level of significance
H0: μ 800Ha: μ > 800 the null hypothesis will be rejected if the test statistic Z is a. > = Zα b.< Zα c.< -Zα d.= α |
> = Zα Correct
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The average life expectancy of tires produced by the Whitney Tire Company has been 40,000 miles. Management believes that due to a new production process, the life expectancy of their tires has increased. In order to test the validity of their belief, the correct set of hypotheses is
a.H0: μ < 40,000 Ha: μ > = 40,000 b.H0:< = μ 40,000 Ha: μ > 40,000 c.H0: μ > 40,000 Ha: μ 40,000 d.H0: μ 40,000 Ha: μ < 40,000 |
b. H0:< = μ 40,000 Ha: μ > 40,000 Correct
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The level of significance in hypothesis testing is the probability of
a.accepting a true null hypothesis b.accepting a false null hypothesis c.rejecting a true null hypothesis d.None of these alternatives is correct. |
c. rejecting a true null hypothesis Correct
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