STAT 200 QUIZ 3 Name___________________________________ Questions 1 - 30 are 2 points each, 31-33 are 5 points each. Show work for partial credit on all problems. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the indicated critical z value. 1) Find the critical value z /2 that corresponds to a 98% confidence level. A) 2.575 B) 2.33 C) 1.75 1) D) 2.05 Express the confidence interval using the indicated format. ^ 2) Express the confidence interval 0.38 < p < 0.54 in the form of p E. A) 0.38 0.16 B) 0.46 0.08 C) 0.38 0.08 2) D) 0.46 0.16 Solve the problem. 3) The following confidence interval is obtained for a population proportion, p: (0.688, 0.724). Use 3) ^ these confidence interval limits to find the point estimate, p. A) 0.724 B) 0.708 C) 0.688 D) 0.706 Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. 4) 95% confidence; the sample size is 9900, of which 30% are successes 4) A) 0.0104 B) 0.0119 C) 0.00677 D) 0.00903 Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. 5) n = 130, x = 69; 90% confidence 5) A) 0.463 < p < 0.599 B) 0.458 < p < 0.604 C) 0.461 < p < 0.601 D) 0.459 < p < 0.603 Solve the problem. Round the point estimate to the nearest thousandth. 6) Find the point estimate of the proportion of people who wear hearing aids if, in a random sample of 898 people, 46 people had hearing aids. A) 0.051 B) 0.050 C) 0.949 D) 0.049 6) Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. 7) Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence 7) interval for the true proportion of all adults in the town who have health insurance. A) 0.582 < p < 0.744 B) 0.536 < p < 0.790 C) 0.566 < p < 0.760 D) 0.548 < p < 0.778 Do one of the following, as appropriate: (a) Find the critical value z /2 , (b) find the critical value t/2 , (c) state that neither the normal nor the t distribution applies. 8) 98%; n = 7; = 27; population appears to be normally distributed. A) t/2 = 2.575 B) z /2 = 2.33 C) t/2 = 1.96 9) 99%; n = 17; is unknown; population appears to be normally distributed. A) t/2 = 2.898 B) z /2 = 2.567 C) z /2 = 2.583 1 Print to PDF without this message by purchasing novaPDF (http://www.novapdf.com/) 8) D) z /2 = 2.05 9) D) t/2 = 2.921 Use the given degree of confidence and sample data to construct a confidence interval for the population mean . Assume that the population has a normal distribution. 10) n = 30, x = 84.6, s = 10.5, 90% confidence A) 80.68 < < 88.52 C) 81.36 < < 87.84 10) B) 81.34 < < 87.86 D) 79.32 < < 89.88 Use the confidence level and sample data to find a confidence interval for estimating the population . Round your answer to the same number of decimal places as the sample mean. 11) Test scores: n = 104, x = 95.3, = 6.5; 99% confidence A) 94.1 < < 96.5 B) 94.2 < < 96.4 C) 93.8 < < 96.8 11) D) 93.7 < < 96.9 Solve the problem. 2 12) Find the critical value R corresponding to a sample size of 6 and a confidence level of 95 percent. A) 12.833 B) 11.07 C) 0.831 12) D) 1.145 Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. 13) Weights of men: 90% confidence; n = 14, x = 161.5 lb, s = 13.7 lb A) 10.8 lb < < 17.7 lb B) 11.1 lb < < 2.7 lb C) 10.4 lb < < 20.3 lb D) 10.2 lb < < 19.3 lb 13) Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data. 14) The football coach randomly selected ten players and timed how long each player took to perform 14) a certain drill. The times (in minutes) were: 9 7 15 6 15 12 8 6 14 5 Find a 95% confidence interval for the population standard deviation . A) 2.7 min < < 6.6 min B) 2.7 min < < 7.2 min C) 2.6 min < < 6.6 min D) 0.8 min < < 2.4 min Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol (, p, ) for the indicated parameter. 15) An entomologist writes an article in a scientific journal which claims that fewer than 16 in ten 15) thousand male fireflies are unable to produce light due to a genetic mutation. Use the parameter p, the true proportion of fireflies unable to produce light. A) H0 : p < 0.0016 B) H 0: p = 0.0016 C) H0 : p = 0.0016 D) H 0: p > 0.0016 H1 : p 0.0016 H 1: p < 0.0016 H1 : p > 0.0016 H 1: p 0.0016 16) Carter Motor Company claims that its new sedan, the Libra, will average better than 23 miles per gallon in the city. Use , the true average mileage of the Libra. A) H0 : < 23 B) H 0: = 23 C) H0 : = 23 D) H 0: > 23 H1 : 23 H 1: < 23 H1 : > 23 2 Print to PDF without this message by purchasing novaPDF (http://www.novapdf.com/) H 1: 23 16) 17) A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a standard deviation different from the = 3.3 mg claimed by the manufacturer. A) H0 : 3.3 mg B) H 0: = 3.3 mg C) H0 : 3.3 mg D) H 0: 3.3 mg H1 : > 3.3 mg H 1: 3.3 mg H1 : < 3.3 mg 17) H 1: = 3.3 mg 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. 18) = 0.05 for a left-tailed test. 18) A) -1.645 B) 1.96 C) -1.96 D) 1.645 ^ Find the value of the test statistic z using z = p-p . pq n 19) A claim is made that the proportion of children who play sports is less than 0.5, and the sample statistics include n = 1320 subjects with 30% saying that they play a sport. A) 14.53 B) -29.66 C) 29.66 D) -14.53 19) Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). 20) The test statistic in a left-tailed test is z = -1.83. 20) A) 0.0672; fail to reject the null hypothesis B) 0.0672; reject the null hypothesis C) 0.0336; reject the null hypothesis D) 0.9664; fail to reject the null hypothesis 21) The test statistic in a two-tailed test is z = 1.95. A) 0.0256; reject the null hypothesis C) 0.0512; fail to reject the null hypothesis 21) B) 0.9744; fail to reject the null hypothesis D) 0.0512; reject the null hypothesis 22) With H1 : p 0.377, the test statistic is z = 3.06. 22) A) 0.0011; reject the null hypothesis C) 0.0022; fail to reject the null hypothesis B) 0.0022; reject the null hypothesis D) 0.0011; fail to reject the null hypothesis 23) With H1 : p < 2/3, the test statistic is z = -1.89. 23) A) 0.0294; fail to reject the null hypothesis C) 0.0588; fail to reject the null hypothesis B) 0.9706; fail to reject the null hypothesis D) 0.0294; reject the null hypothesis Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. 24) An entomologist writes an article in a scientific journal which claims that fewer than 3 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is to reject the null hypothesis, state the conclusion in nontechnical terms. A) There is not sufficient evidence to support the claim that the true proportion is less than 3 in ten thousand. B) There is not sufficient evidence to support the claim that the true proportion is greater than 3 in ten thousand. C) There is sufficient evidence to support the claim that the true proportion is less than 3 in ten thousand. D) There is sufficient evidence to support the claim that the true proportion is greater than 3 in ten thousand. 3 Print to PDF without this message by purchasing novaPDF (http://www.novapdf.com/) 24) 25) A skeptical paranormal researcher claims that the proportion of Americans that have seen a UFO, p, is less than 2 in every ten thousand. Assuming 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. A) There is sufficient evidence to support the claim that the true proportion is less than 2 in ten thousand. B) There is not sufficient evidence to support the claim that the true proportion is less than 2 in ten thousand. C) There is sufficient evidence to support the claim that the true proportion is greater than 2 in ten thousand. D) There is not sufficient evidence to support the claim that the true proportion is greater than 2 in ten thousand. 25) Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. 26) A medical researcher claims that 10% of children suffer from a certain disorder. Identify the type I 26) error for the test. A) Reject the claim that the percentage of children who suffer from the disorder is different from 10% when that percentage really is different from 10%. B) Fail to reject the claim that the percentage of children who suffer from the disorder is equal to 10% when that percentage is actually different from 10%. C) Fail to reject the claim that the percentage of children who suffer from the disorder is equal to 10% when that percentage is actually 10%. D) Reject the claim that the percentage of children who suffer from the disorder is equal to 10% when that percentage is actually 10%. 27) The principal of a school claims that the percentage of students at his school that come from single-parent homes is 11%. Identify the type II error for the test. A) Fail to reject the claim that the percentage of students that come from single-parent homes is equal to 11% when that percentage is actually different from 11%. B) Reject the claim that the percentage of students that come from single-parent homes is equal to 11% when that percentage is actually 11%. C) Fail to reject the claim that the percentage of students that come from single-parent homes is equal to 11% when that percentage is actually 11%. D) Reject the claim that the percentage of students that come from single-parent homes is equal to 11% when that percentage is actually less than 11%. Find the P-value for the indicated hypothesis test. 28) In a sample of 47 adults selected randomly from one town, it is found that 9 of them have been exposed to a particular strain of the flu. Find the P-value for a test of the claim that the proportion of all adults in the town that have been exposed to this strain of the flu is 8%. A) 0.0524 B) 0.0262 C) 0.0024 D) 0.0048 29) A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 32% of them plan to go into general practice. Find the P-value for a test of the school's claim. A) 0.3461 B) 0.3078 C) 0.1539 D) 0.1635 4 Print to PDF without this message by purchasing novaPDF (http://www.novapdf.com/) 27) 28) 29) Find the critical value or values of 2 based on the given information. 30) H0 : = 8.0 n = 10 = 0.01 A) 23.209 B) 21.666 C) 1.735, 23.589 30) D) 2.088, 21.666 SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. 31) A manufacturer considers his production process to be out of control when defects exceed 31) 3%. In a random sample of 85 items, the defect rate is 5.9% but the manager claims that this is only a sample fluctuation and production is not really out of control. At the 0.01 level of significance, test the manager's claim. Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. 32) A large software company gives job applicants a test of programming ability and the 32) mean for that test has been 160 in the past. Twenty-five job applicants are randomly selected from one large university and they produce a mean score and standard deviation of 183 and 12, respectively. Use a 0.05 level of significance to test the claim that this sample comes from a population with a mean score greater than 160. Use the P-value method of testing hypotheses. Use the traditional method to test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected. 33) For randomly selected adults, IQ scores are normally distributed with a standard 33) 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. Round the sample standard deviation to three decimal places. 115 128 107 109 116 124 135 127 115 104 118 126 129 133 5 Print to PDF without this message by purchasing novaPDF (http://www.novapdf.com/) httptf/Enm - StatistlalTablapdf x ' (- 'l' SWItch totab: httpw'fhnmaubaltledux'ntsbarsh/Builnessrstan'StatlstlalTables.pdf e I ' Q Searrh ' ' a ' '5? v o IQ + Autm'r'uatl:Eat-m 3 STATISTICAL TABLES Cumulative normal distribution Critical values of the tdistribution Critical values of the F distribution Critical values of the chi-squared distribution . E El '4 I" hum/econ - StatistiaiTablapdf x SWItrh to tab: httpEffhnmaubaltedu/ntsharsh/BuslnessVstatJStatlstiaiTahles.pdf (3' I i 0' Seorrh + Automati : TABLE A.1 Cumulative Standardized Normal Distribution A(2) is the integral of the standardized normal distribution from co to z (in other words, the area under the curve to the le of 2). It gives the probability of a normal random variable not being more than 2 standard deviations above its mean. Values of z of particular importance: 2 AEZ 1.645 0.95 00 Lower hmit ofrigh't 5% tall 1.960 0.9750 Lower limit ofright 2.5%tail 2.326 0.9900 Lower limit efright1% tail 2.576 0.9950 Lower limit efright 0.5%tail 3.090 0.9990 Lower limit ofright 0.1' tail 3.291 0.9995 Lower Limit ofright 0.05% tail . men \f1. Find the critical value z/2 that corresponds to a 98% confidence level. = 1 - 0.98 = 0.02 /2 = 0.02/2 = 0.01 z/2 = 2.33. 2. Express the confidence interval 0.38 < p < 0.54 in the form of p ^ E. p = (0.38 + 0.54)/2 = 0.46 E = 0.46 - 0.38 = 0.08 In the form p ^ E, is 0.46 0.08 3. The following confidence interval is obtained for a population proportion, p: (0.688, 0.724). Use these confidence interval limits to find the point estimate, p pp = (0.688 + 0.724)/2 = 0.706 4. Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places: 95% confidence; the sample size is 9900, of which 30% are successes Margin of error = z/2 * ^ ^) P (1P n pp = 0.3, n = 9900 To obtain z/2: = 1 - 0.95 = 0.05 /2 = 0.05/2 = 0.025 z/2 = 1.96. Margin of error = 1.96 * 0.3( 10.3) 9900 = 0.00903 5. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 130, x = 69; 90% confidence pp = 69/130 = 0.5308 Confidence interval = pp margin of error Margin of error = z/2 * ^ ^) P (1P n To obtain z/2: = 1 - 0.90 = 0.10 /2 = 0.1/2 = 0.05 z/2 = 1.65. Margin of error = 1.65 * 0.5308( 10.5308) 130 = 0.0722 Confidence interval = 0.5308 0.0722 = (0.459 < p < 0.603) 6. Find the point estimate of the proportion of people who wear hearing aids if, in a random sample of 898 people, 46 people had hearing aids. pp = 46/898 = 0.051 7. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. pp = 61/92 = 0.663 Confidence interval = pp margin of error Margin of error = z/2 * ^ ^) P (1P n To obtain z/2: = 1 - 0.90 = 0.10 /2 = 0.1/2 = 0.05 z/2 = 1.65. Margin of error = 1.65 * 0.663(10.663) 92 = 0.0813 Confidence interval = 0.663 0.0813 = (0.582 < p < 0.744) 8. Do one of the following, as appropriate: (a) Find the critical value z/2, (b) find the critical value t/2, (c) state that neither the normal nor the t distribution applies. 98%; n = 7; = 27; population appears to be normally distributed We find critical value z/2 since we know the population standard deviation. = 1 - 0.98 = 0.02 /2 = 0.02/2 = 0.01 z/2 = 2.33. 9. 99%; n = 17; is unknown; population appears to be normally distributed We find the critical value t/2 since is unknown Degrees of freedom = n - 1 = 17 - 1 = 16 Using df = 16 and /2 = 0.01/2 = 0.005, t/2 = 2.921 10. Use the given degree of confidence and sample data to construct a confidence interval for the population mean . Assume that the population has a normal distribution. n = 30, x = 84.6, s = 10.5, 90% confidence Confidence interval = margin of error Margin of error = t/2 * s n To obtain t/2: = 1 - 0.90 = 0.10 /2 = 0.10/2 = 0.05 Get the critical t/2 corresponding to = 0.05, with degrees of freedom = 30 - 1 = 29, t/2 = 1.699. Margin of error = 1.699 * 10.5 30 = 3.26 Confidence interval = 84.6 3.26= (81.34 < < 87.86) 11. Use the confidence level and sample data to find a confidence interval for estimating the population . Round your answer to the same number of decimal places as the sample mean. Test scores: n = 104, x = 95.3, = 6.5; 99% confidence. Confidence interval = margin of error Margin of error = z/2 * n To obtain z/2: = 1 - 0.99 = 0.01 /2 = 0.01/2 = 0.005 Get the critical z/2 corresponding to = 0.005, z/2 = 2.58. Margin of error = 2.58 * 6.5 104 = 1.644 Confidence interval = 95.3 1.644= (93.7 < < 96.9) 12. Find the critical value 2 R corresponding to a sample size of 6 and a confidence level of 95 percent Degrees of freedom = 6 - 1 = 5 The critical at = (1 - 0.95)/2 = 0.025 is 12.833 13. Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. Weights of men: 90% confidence; n = 14, x = 161.5 lb, s = 13.7 lb Confidence interval for the population standard deviation 2 2 (n1)s (n1)s = Square root ( X 2 ( /2) < ^2 < X 2 (1 /2) ) /2 = (1 - 0.9)/2 = 0.05 Degrees of freedom = 14 - 1 = 13 2 X (0.05) = 22.362 X 2 (10.05) = 5.892 (141) 13.72 Confidence interval = square root ( 22.362 < ^2< (141) 13.72 5.892 ) = 10.4 < < 20.3 14. Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data. The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 9 , 7, 15, 6, 15, 12, 8, 6, 14, 5 x Sample variance, S2 = 1 n1 ( ) 2 x 2 x2 = x = 9 + 7 + . . . + 14 + 5 = 97 9 2 + 7 2 2 + . . . + 14 n = 10 2 S = 1 101 972 1081 ( 10 ) = 15.57 2 + 5 = 1081 = 1 - 0.95 = 0.05 (n1) s2 20.05/ 2 ,( n1) Lower limit = 2 0.05 /2 ,(101) 20.025 ,(9) = 19.023 Lower limit = = 7.366 2 10.05 /2 ,(n1) (n1) s 2 Upper limit = 20.975 ,(9) Upper limit ( 101 ) 15.57 19.023 = 2.7 = ( 101 ) 15.57 2.7 = 51.9 95% confidence interval for the population variance = (7.366, 51.9) To obtain 95% confidence interval for the population standard deviation, take the square roots to give, 2.7 < < 7.2 15. Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol (, p, ) for the indicated parameter. An entomologist writes an article in a scientific journal which claims that fewer than 16 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Use the parameter p, the true proportion of fireflies unable to produce light p = 16/1000 = 0.016 H 0 : p = 0.0016 H 0 : p < 0.0016 16. Carter Motor Company claims that its new sedan, the Libra, will average better than 23 miles per gallon in the city. Use , the true average mileage of the Libra C) H0: = 23 H1: > 23 17. A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a standard deviation different from the = 3.3 mg claimed by the manufacturer. H0: = 3.3 mg H1: 3.3 mg 18. 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. = 0.05 for a left-tailed test. At = 0.05 , Z = 1.645 But since it is left tailed, Z = - 1.65 19. A claim is made that the proportion of children who play sports is less than 0.5, and the sample statistics include n = 1320 subjects with 30% saying that they play a sport p = 0.5, pp = 0.3, q = 1 - p = 1- 0.5 = 0.5 test statistic z using z = 0.30.5 0.50.5 1320 = - 14.53 Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). 20. The test statistic in a left-tailed test is z = -1.83. p-value = 0.0336, reject the null hypothesis. (since 0.0336 < 0.05) 21. The test statistic in a two-tailed test is z = 1.95. p-value = 0.0512, fail to reject the null hypothesis. (since 0.0512 > 0.05) 22. With H1: p 0.377, the test statistic is z = 3.06. p-value = 0.0022, reject the null hypothesis. (since 0.0022 < 0.05) 23. With H1: p < 2/3, the test statistic is z = -1.89. p-value = 0.0294., reject the null hypothesis. (since 0.0294.< 0.05) Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. 24. An entomologist writes an article in a scientific journal which claims that fewer than 3 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is to reject the null hypothesis, state the conclusion in nontechnical terms. C. There is sufficient evidence to support the claim that the true proportion is less than 3 in ten thousand. 25. A skeptical paranormal researcher claims that the proportion of Americans that have seen a UFO, p, is less than 2 in every ten thousand. Assuming 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 B. There is not sufficient evidence to support the claim that the true proportion is less than 2 in ten thousand. Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. 26. A medical researcher claims that 10% of children suffer from a certain disorder. Identify the type I error for the test D. Reject the claim that the percentage of children who suffer from the disorder is equal to 10% when that percentage is actually 10%. 27. The principal of a school claims that the percentage of students at his school that come from single-parent homes is 11%. Identify the type II error for the test. A. Fail to reject the claim that the percentage of students that come from singleparent homes is equal to 11% when that percentage is actually different from 11%. Find the P-value for the indicated hypothesis test. 28. In a sample of 47 adults selected randomly from one town, it is found that 9 of them have been exposed to a particular strain of the flu. Find the P-value for a test of the claim that the proportion of all adults in the town that have been exposed to this strain of the flu is 8%. p = 0.08 , pp = 9/47 = 0.191, q = 1 - p = 1- 0.08 = 0.92 test statistic z = 0.910.08 0.080.92 47 = 2.81 Test statistic, 2.81 p-value = 0.0048 29. A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 32% of them plan to go into general practice. Find the P-value for a test of the school's claim. p = 0.28 , pp = 0.32, q = 1 - p = 1- 0.28 = 0.72 test statistic z = 0.320.28 0.280.72 130 = 0.988 Test statistic = 0.988 p-value = 0.1635 30. Find the critical value or values of 2 based on the given information. H0: = 8.0 n = 10 = 0.01 = 0.01/2 = 0.005, and at 1 - 0.005 = 0.995, degrees of freedom, 10 - 1 = 9, Critical value = 1.735, 23.589 Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. 31. A manufacturer considers his production process to be out of control when defects exceed 3%. In a random sample of 85 items, the defect rate is 5.9% but the manager claims that this is only a sample fluctuation and production is not really out of control. At the 0.01 level of significance, test the manager's claim. Null and alternative hypothesis H 0 : p = 0.03 H 0 : p > 0.03 , pp = 0.059, q = 1 - p = 1- 0.03 = 0.97 test statistic z using z = 0.0590.03 0.030.97 85 = 1.57 Test statistic, 1.57 p-value = 0.0582, fail to reject the null hypothesis since 0.0582 > 0.01 There is not sufficient evidence to conclude that the production process is out of control. Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. 32. A large software company gives job applicants a test of programming ability and the mean for that test has been 160 in the past. Twenty-five job applicants are randomly selected from one large university and they produce a mean score and standard deviation of 183 and 12, respectively. Use a 0.05 level of significance to test the claim that this sample comes from a population with a mean score greater than 160. Use the P-value method of testing hypotheses. Null and alternative hypothesis H 0 : = 160 H 0 : p > 160 Test statistic 183160 t = 12/ 25 = 9.58 Test statistic, 9.58 and degrees of freedom = 25 - 1 = 24 p-value = .00001. , Reject the null hypothesis since .00001 < 0.05 There is sufficient evidence to conclude that the sample comes from a population with a mean score greater than 160. Use the traditional method to test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected. 33. 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. Round the sample standard deviation to three decimal places. 115, 128, 107, 109, 116, 124, 135, 127, 115, 104, 118, 126, 129, 133, x Sample variance, S2 = x2 = 2 115 1 n1 + 128 ( 2 ) 2 x 2 2 + . . . + 129 2 + 133 = 204296 x = 115 + 128 + . . . + 129 + 133 = 1686 n = 14 2 S = 1 141 16862 204296 ( ) = 96.418 14 Standard deviation = 96.418 = 9.819 Null and alternative hypotheses H 0 : = 15 H 1 : < 15 Test statistic , Chi square = (n1) s 2 2 = ( 141 ) 96.418 225 = 5.571 Critical value Df = 14 - 1 = 13 Left tailed critical value at 0.10 significance level = 7.042 We fail to reject the null hypothesis since 5.571 < 7.042. There is not sufficient evidence to conclude hat the standard deviation of IQ scores of college students is less than 15. httptf/Enm - StatistlalTablapdf x ' (- 'l' SWItch totab: httpw'fhnmaubaltledux'ntsbarsh/Builnessrstan'StatlstlalTables.pdf e I ' Q Searrh ' ' a ' '5? v o IQ + Autm'r'uatl:Eat-m 3 STATISTICAL TABLES Cumulative normal distribution Critical values of the tdistribution Critical values of the F distribution Critical values of the chi-squared distribution . E El '4 I" hum/econ - StatistiaiTablapdf x SWItrh to tab: httpEffhnmaubaltedu/ntsharsh/BuslnessVstatJStatlstiaiTahles.pdf (3' I i 0' Seorrh + Automati : TABLE A.1 Cumulative Standardized Normal Distribution A(2) is the integral of the standardized normal distribution from co to z (in other words, the area under the curve to the le of 2). It gives the probability of a normal random variable not being more than 2 standard deviations above its mean. Values of z of particular importance: 2 AEZ 1.645 0.95 00 Lower hmit ofrigh't 5% tall 1.960 0.9750 Lower limit ofright 2.5%tail 2.326 0.9900 Lower limit efright1% tail 2.576 0.9950 Lower limit efright 0.5%tail 3.090 0.9990 Lower limit ofright 0.1' tail 3.291 0.9995 Lower Limit ofright 0.05% tail . men \f1. Find the critical value z/2 that corresponds to a 98% confidence level. = 1 - 0.98 = 0.02 /2 = 0.02/2 = 0.01 z/2 = 2.33. 2. Express the confidence interval 0.38 < p < 0.54 in the form of p ^ E. p = (0.38 + 0.54)/2 = 0.46 E = 0.46 - 0.38 = 0.08 In the form p ^ E, is 0.46 0.08 3. The following confidence interval is obtained for a population proportion, p: (0.688, 0.724). Use these confidence interval limits to find the point estimate, p pp = (0.688 + 0.724)/2 = 0.706 4. Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places: 95% confidence; the sample size is 9900, of which 30% are successes Margin of error = z/2 * ^ ^) P (1P n pp = 0.3, n = 9900 To obtain z/2: = 1 - 0.95 = 0.05 /2 = 0.05/2 = 0.025 z/2 = 1.96. Margin of error = 1.96 * 0.3( 10.3) 9900 = 0.00903 5. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. n = 130, x = 69; 90% confidence pp = 69/130 = 0.5308 Confidence interval = pp margin of error Margin of error = z/2 * ^ ^) P (1P n To obtain z/2: = 1 - 0.90 = 0.10 /2 = 0.1/2 = 0.05 z/2 = 1.65. Margin of error = 1.65 * 0.5308( 10.5308) 130 = 0.0722 Confidence interval = 0.5308 0.0722 = (0.459 < p < 0.603) 6. Find the point estimate of the proportion of people who wear hearing aids if, in a random sample of 898 people, 46 people had hearing aids. pp = 46/898 = 0.051 7. Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Of 92 adults selected randomly from one town, 61 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. pp = 61/92 = 0.663 Confidence interval = pp margin of error Margin of error = z/2 * ^ ^) P (1P n To obtain z/2: = 1 - 0.90 = 0.10 /2 = 0.1/2 = 0.05 z/2 = 1.65. Margin of error = 1.65 * 0.663(10.663) 92 = 0.0813 Confidence interval = 0.663 0.0813 = (0.582 < p < 0.744) 8. Do one of the following, as appropriate: (a) Find the critical value z/2, (b) find the critical value t/2, (c) state that neither the normal nor the t distribution applies. 98%; n = 7; = 27; population appears to be normally distributed We find critical value z/2 since we know the population standard deviation. = 1 - 0.98 = 0.02 /2 = 0.02/2 = 0.01 z/2 = 2.33. 9. 99%; n = 17; is unknown; population appears to be normally distributed We find the critical value t/2 since is unknown Degrees of freedom = n - 1 = 17 - 1 = 16 Using df = 16 and /2 = 0.01/2 = 0.005, t/2 = 2.921 10. Use the given degree of confidence and sample data to construct a confidence interval for the population mean . Assume that the population has a normal distribution. n = 30, x = 84.6, s = 10.5, 90% confidence Confidence interval = margin of error Margin of error = t/2 * s n To obtain t/2: = 1 - 0.90 = 0.10 /2 = 0.10/2 = 0.05 Get the critical t/2 corresponding to = 0.05, with degrees of freedom = 30 - 1 = 29, t/2 = 1.699. Margin of error = 1.699 * 10.5 30 = 3.26 Confidence interval = 84.6 3.26= (81.34 < < 87.86) 11. Use the confidence level and sample data to find a confidence interval for estimating the population . Round your answer to the same number of decimal places as the sample mean. Test scores: n = 104, x = 95.3, = 6.5; 99% confidence. Confidence interval = margin of error Margin of error = z/2 * n To obtain z/2: = 1 - 0.99 = 0.01 /2 = 0.01/2 = 0.005 Get the critical z/2 corresponding to = 0.005, z/2 = 2.58. Margin of error = 2.58 * 6.5 104 = 1.644 Confidence interval = 95.3 1.644= (93.7 < < 96.9) 12. Find the critical value 2 R corresponding to a sample size of 6 and a confidence level of 95 percent Degrees of freedom = 6 - 1 = 5 The critical at = (1 - 0.95)/2 = 0.025 is 12.833 13. Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation. Weights of men: 90% confidence; n = 14, x = 161.5 lb, s = 13.7 lb Confidence interval for the population standard deviation 2 2 (n1)s (n1)s = Square root ( X 2 ( /2) < ^2 < X 2 (1 /2) ) /2 = (1 - 0.9)/2 = 0.05 Degrees of freedom = 14 - 1 = 13 2 X (0.05) = 22.362 X 2 (10.05) = 5.892 (141) 13.72 Confidence interval = square root ( 22.362 < ^2< (141) 13.72 5.892 ) = 10.4 < < 20.3 14. Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation . Assume that the population has a normal distribution. Round the confidence interval limits to one more decimal place than is used for the original set of data. The football coach randomly selected ten players and timed how long each player took to perform a certain drill. The times (in minutes) were: 9 , 7, 15, 6, 15, 12, 8, 6, 14, 5 x Sample variance, S2 = 1 n1 ( ) 2 x 2 x2 = x = 9 + 7 + . . . + 14 + 5 = 97 9 2 + 7 2 2 + . . . + 14 n = 10 2 S = 1 101 972 1081 ( 10 ) = 15.57 2 + 5 = 1081 = 1 - 0.95 = 0.05 (n1) s2 20.05/ 2 ,( n1) Lower limit = 2 0.05 /2 ,(101) 20.025 ,(9) = 19.023 Lower limit = = 7.366 2 10.05 /2 ,(n1) (n1) s 2 Upper limit = 20.975 ,(9) Upper limit ( 101 ) 15.57 19.023 = 2.7 = ( 101 ) 15.57 2.7 = 51.9 95% confidence interval for the population variance = (7.366, 51.9) To obtain 95% confidence interval for the population standard deviation, take the square roots to give, 2.7 < < 7.2 15. Express the null hypothesis and the alternative hypothesis in symbolic form. Use the correct symbol (, p, ) for the indicated parameter. An entomologist writes an article in a scientific journal which claims that fewer than 16 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Use the parameter p, the true proportion of fireflies unable to produce light p = 16/1000 = 0.016 H 0 : p = 0.0016 H 0 : p < 0.0016 16. Carter Motor Company claims that its new sedan, the Libra, will average better than 23 miles per gallon in the city. Use , the true average mileage of the Libra C) H0: = 23 H1: > 23 17. A researcher claims that the amounts of acetaminophen in a certain brand of cold tablets have a standard deviation different from the = 3.3 mg claimed by the manufacturer. H0: = 3.3 mg H1: 3.3 mg 18. 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. = 0.05 for a left-tailed test. At = 0.05 , Z = 1.645 But since it is left tailed, Z = - 1.65 19. A claim is made that the proportion of children who play sports is less than 0.5, and the sample statistics include n = 1320 subjects with 30% saying that they play a sport p = 0.5, pp = 0.3, q = 1 - p = 1- 0.5 = 0.5 test statistic z using z = 0.30.5 0.50.5 1320 = - 14.53 Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). 20. The test statistic in a left-tailed test is z = -1.83. p-value = 0.0336, reject the null hypothesis. (since 0.0336 < 0.05) 21. The test statistic in a two-tailed test is z = 1.95. p-value = 0.0512, fail to reject the null hypothesis. (since 0.0512 > 0.05) 22. With H1: p 0.377, the test statistic is z = 3.06. p-value = 0.0022, reject the null hypothesis. (since 0.0022 < 0.05) 23. With H1: p < 2/3, the test statistic is z = -1.89. p-value = 0.0294., reject the null hypothesis. (since 0.0294.< 0.05) Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. 24. An entomologist writes an article in a scientific journal which claims that fewer than 3 in ten thousand male fireflies are unable to produce light due to a genetic mutation. Assuming that a hypothesis test of the claim has been conducted and that the conclusion is to reject the null hypothesis, state the conclusion in nontechnical terms. C. There is sufficient evidence to support the claim that the true proportion is less than 3 in ten thousand. 25. A skeptical paranormal researcher claims that the proportion of Americans that have seen a UFO, p, is less than 2 in every ten thousand. Assuming 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 B. There is not sufficient evidence to support the claim that the true proportion is less than 2 in ten thousand. Assume that a hypothesis test of the given claim will be conducted. Identify the type I or type II error for the test. 26. A medical researcher claims that 10% of children suffer from a certain disorder. Identify the type I error for the test D. Reject the claim that the percentage of children who suffer from the disorder is equal to 10% when that percentage is actually 10%. 27. The principal of a school claims that the percentage of students at his school that come from single-parent homes is 11%. Identify the type II error for the test. A. Fail to reject the claim that the percentage of students that come from singleparent homes is equal to 11% when that percentage is actually different from 11%. Find the P-value for the indicated hypothesis test. 28. In a sample of 47 adults selected randomly from one town, it is found that 9 of them have been exposed to a particular strain of the flu. Find the P-value for a test of the claim that the proportion of all adults in the town that have been exposed to this strain of the flu is 8%. p = 0.08 , pp = 9/47 = 0.191, q = 1 - p = 1- 0.08 = 0.92 test statistic z = 0.910.08 0.080.92 47 = 2.81 Test statistic, 2.81 p-value = 0.0048 29. A medical school claims that more than 28% of its students plan to go into general practice. It is found that among a random sample of 130 of the school's students, 32% of them plan to go into general practice. Find the P-value for a test of the school's claim. p = 0.28 , pp = 0.32, q = 1 - p = 1- 0.28 = 0.72 test statistic z = 0.320.28 0.280.72 130 = 0.988 Test statistic = 0.988 p-value = 0.1635 30. Find the critical value or values of 2 based on the given information. H0: = 8.0 n = 10 = 0.01 = 0.01/2 = 0.005, and at 1 - 0.005 = 0.995, degrees of freedom, 10 - 1 = 9, Critical value = 1.735, 23.589 Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. 31. A manufacturer considers his production process to be out of control when defects exceed 3%. In a random sample of 85 items, the defect rate is 5.9% but the manager claims that this is only a sample fluctuation and production is not really out of control. At the 0.01 level of significance, test the manager's claim. Null and alternative hypothesis H 0 : p = 0.03 H 0 : p > 0.03 , pp = 0.059, q = 1 - p = 1- 0.03 = 0.97 test statistic z using z = 0.0590.03 0.030.97 85 = 1.57 Test statistic, 1.57 p-value = 0.0582, fail to reject the null hypothesis since 0.0582 > 0.01 There is not sufficient evidence to conclude that the production process is out of control. Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim. 32. A large software company gives job applicants a test of programming ability and the mean for that test has been 160 in the past. Twenty-five job applicants are randomly selected from one large university and they produce a mean score and standard deviation of 183 and 12, respectively. Use a 0.05 level of significance to test the claim that this sample comes from a population with a mean score greater than 160. Use the P-value method of testing hypotheses. Null and alternative hypothesis H 0 : = 160 H 0 : p > 160 Test statistic 183160 t = 12/ 25 = 9.58 Test statistic, 9.58 and degrees of freedom = 25 - 1 = 24 p-value = .00001. , Reject the null hypothesis since .00001 < 0.05 There is sufficient evidence to conclude that the sample comes from a population with a mean score greater than 160. Use the traditional method to test the given hypothesis. Assume that the population is normally distributed and that the sample has been randomly selected. 33. 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. Round the sample standard deviation to three decimal places. 115, 128, 107, 109, 116, 124, 135, 127, 115, 104, 118, 126, 129, 133, x Sample variance, S2 = x2 = 2 115 1 n1 + 128 ( 2 ) 2 x 2 2 + . . . + 129 2 + 133 = 204296 x = 115 + 128 + . . . + 129 + 133 = 1686 n = 14 2 S = 1 141 16862 204296 ( ) = 96.418 14 Standard deviation = 96.418 = 9.819 Null and alternative hypotheses H 0 : = 15 H 1 : < 15 Test statistic , Chi square = (n1) s 2 2 = ( 141 ) 96.418 225 = 5.571 Critical value Df = 14 - 1 = 13 Left tailed critical value at 0.10 significance level = 7.042 We fail to reject the null hypothesis since 5.571 < 7.042. There is not sufficient evidence to conclude hat the standard deviation of IQ scores of college students is less than 15