Apr 6, 2017

Hypothesis Testing — practice test 8B

20 cards
Problems with solutions in Elementary Statistics — Basics of Hypothesis Testing; Testing a Claim about a Proportion; Testing a Claim about a Mean: 1) σ known, 2) σ not known; Testing Claim about Variation.

Directions:

• Provide an appropriate response.
• 1. Explain how to determine if a hypothesis test is one-tailed or two-tailed and explain how you know where to shade the critical region. Give an example for each which includes the claim, the hypotheses, and the diagram with critical region shaded.

Examples will vary. Relational operators in the alternative hypotheses indicate whether the test is one-tailed or two-tailed. Strict inequalities determine one-tiled tests, whereas “not equal to” and “different from” determine two-tailed tests. The critical region begins at the CV for one-tailed tests and at both CVs for two-tailed tests. Shading begins at the CV/CVs and extends to the tails of the distribution.

• Solve the problem.
• 2. Write the claim that is suggested by the given statement, then write a conclusion about the claim. Do not use symbolic expressions or formal procedures; use common sense.

Last year an appliance manufacturer received many complaints about the high rate of defects among its washing machines. Approximately 9% of the machines were defective in some way. This year the company tightened up its quality control procedures. The latest shipment of 250 washing machines contained 2 defectives.

The claim is that the defect rate has decreased and is now less than 9%. If the overall defect rate were still 9%, it would be extremely unlikely that a shipment of 250 washing machines would contain as few as 2 defectives. Therefore, the claim that the defect rate has decreased is probably correct.

• Express the null hypothesis H0 and the alternative hypothesis H1 in symbolic form. Use the correct symbol (μ, p, σ) for the indicated parameter.
• 3. A researcher claims that 62% of voters favor gun control.
1. H0: p < 0.62, H1: p ≥ 0.62
2. H0: p = 0.62, H1: p ≠ 0.62
3. H0: p ≠ 0.62, H1: p = 0.62
4. H0: p ≥ 0.62, H1: p < 0.62

• Assume that the data has a normal distribution and the number of observations is greater than fifty. Find the critical z value used to test a null hypothesis.
• 4. α = 0.09 for a right-tailed test.
1. ±1.96
2. 1.96
3. 1.34
4. ±1.34

• Find the value of the test statistic z using z = (p̂ – p) / √(p*q/n).
• 5. The claim is that the proportion of drowning deaths of children attributable to beaches is more than 0.25, and the sample statistics include n = 615 drowning deaths of children with 30% of them attributable to beaches.
1. 2.71
2. –2.86
3. 2.86
4. –2.71

• Use the given information to find the P-value.
• 6. The test statistic in a right-tailed test is z = 0.52.
1. 0.5530
2. 0.3015
3. 0.1915
4. 0.1950

• Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim.
• 7. A psychologist claims that more than 56 percent of the population suffers from professional problems due to extreme shyness. Assume that a hypothesis test of the claim has been conducted and that the conclusion is failure to reject the null hypothesis, state the conclusion in nontechnical terms.
1. There is not sufficient evidence to support the claim that the true proportion is greater than 56 percent.
2. There is sufficient evidence to support the claim that the true proportion is less than 56 percent.
3. There is sufficient evidence to support the claim that the true proportion is greater than 56 percent.
4. There is not sufficient evidence to support the claim that the true proportion is greater than 56 percent.

• Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test.
• 8. The manufacturer of refrigerator system for beer kegs produces refrigerators that are supposed to maintain a true mean temperature, μ, of 42˚F, ideal for a certain type of German pilsner. The owner of the brewery does not agree with the refrigerator manufacturer, and claims he can prove that the true mean temperature is incorrect. Identify the type I error for the test.
1. The error of rejecting the claim that the mean temperature equals 42˚F when it is really different from 42˚F.
2. The error of failing to reject the claim that the mean temperature equals 42˚F when it is really different from 42˚F.
3. The error of rejecting the claim that the mean temperature equals 42˚F when it really does equal 42˚F.

• Solve the problem.
• 9. True or False: In a hypothesis test regarding a population mean, the probability of a type II error, β, depends on the true value of the population mean.
1. True
2. False

• Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.
• 10. A nationwide study of American homeowners revealed that 64% have one or more lawn mowers. A lawn equipment manufacturer, located in Omaha, feels the estimate is too low for households in Omaha. Can the value 0.64 be rejected if a survey of 490 homes in Omaha yields 331 with one or more lawn mowers? Use α = 0.05.

H0: p = 64
H1: p > 064
Test statistic: z = 1.64
P-value: p = 0.0505
Critical value: z = 1.645
Fail to reject null hypothesis. There is not sufficient evidence to warrant refection of the claim that the proportion with lawn mowers in Omaha is 0.64.

• Find the P-value for the indicated hypothesis test.
• 11. Find the P-value for a test of the claim that more than 50% of the people following a particular diet will experience increased energy. Of 100 randomly selected subjects who followed the diet, 47 noticed an increase in their energy level.
1. 0.2257
2. 0.5486
3. 0.2743
4. 0.7257

• Determine whether the given conditions justify testing a claim about a population mean μ.
• 12. The sample size is n = 18, σ is not known, and the original population is normally distributed.
1. Yes
2. No

• Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.
• 13. Various temperature measurements are recorded at different times for a particular city. The mean of 20˚C is obtained for 40 temperatures on 40 different days. Assume that σ = 1.5˚C, test the claim that the population mean is 22˚C. Use a 0.05 significance level.

H0: µ = 22
H1: µ ≠ 22
Test statistic: z = –8.43
P-value: p = 0.0002.
Because the P-value is less than the significance level of α = 0.05, we reject the null hypothesis. There is sufficient evidence to warrant rejection of the claim that the population mean temperature is 22 degrees C.

• Determine whether the hypothesis test involves a sampling distribution of means that is a normal distribution, Student t distribution, or neither.
• 14. Claim: µ = 73. Sample data: n = 24, x̅ = 104, s = 15.1. The sample data appear to come from a population with a distribution that is very far from normal, and σ is unknown.
1. Student t
2. Neither
3. Normal

• Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, P-value, critical value(s), and state the final conclusion.
• 15. Test the claim that for the population of history exams, the mean score is 80. Sample data are summarized as n = 16, x̅ = 84.5, and s = 11.2. Use a significance level of α = 0.01.

α = 0.01
Test statistic: t = 1.607
P-value: 0.10 < P-value < 0.20.
Critical value: t = ±2.947
Because TS t < 2.947, we do not reject the null hypothesis. There is not sufficient evidence to warrant rejection of the claim that the mean score is 80.

• Test the given claim using the traditional method of hypothesis testing. Assume that the sample has been randomly selected from a population with a normal distribution.
• 16. A manufacturer makes ball bearings that are supposed to have a mean weight of 30 g. A retailer suspects that the mean weight is actually less than 30 g. The mean weight for a random sample of 16 ball bearings is 28.8 g with a standard deviation of 3.8 g. At the 0.05 significance level, test the claim that the mean is less than 30 g.

Test statistic: t = –1.263
Critical value: t = –1.753
Fail to reject H0. There is not sufficient evidence to support the claim that the mean is less than 30 g.

1. 17. A cereal company claims that the mean weight of the cereal in its packets is 14 oz. The weights (in ounces) of the cereal in a random sample of 8 of its cereal packets are listed below.
14.6 13.8 14.1 13.7 14.0 14.4 13.6 14.2
Test the claim at the 0.01 significance level.

Test statistic: t = 0.408
Critical value: t = ±3.499
Fail to reject H0: µ = 14 ounces. There is not sufficient evidence to warrant rejection of the claim that the mean weight is 14 ounces.

2. Find the critical value or values of x2 based on the given information.
• 18. H1: σ > 26.1, n = 9, α = 0.01.
1. 20.090
2. 1.646
3. 2.088
4. 21.666

3. Use the traditional method to test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected.
• 19. In one town, monthly incomes for men with college degrees are found to have a standard deviation of \$650. Use a 0.01 significance level to test the claim that for men without college degrees in that town, incomes have a higher standard deviation. A random sample of 22 men without college degrees resulted in incomes with a standard deviation of \$927.

Test statistic: X2 = 42.712
Critical value: X2 = 38.932
Reject H0. There is sufficient evidence to support the claim that incomes of men without college degrees have a standard deviation greater than \$650.

1. 20. For randomly selected adults, IQ scores are normally distributed with a standard deviation of 15. The scores of 14 randomly selected college students are listed below. Use a 0.10 significance level to test the claim that the standard deviation of IQ scores of college students is less than 15.
115 128 107 109 116 124 135
127 115 104 118 126 129 133

Test statistic: X2 = 5.571
Critical value: X2 = 7.042
Reject H0. There is sufficient evidence to support the claim that IQ scores of college students have a standard deviation smaller than 15.