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Chapter Nine 9.1 a. The null hypothesis is a claim about a population parameter that is assumed to be true until it is declared false. b. An alternative hypothesis is a claim about a population parameter that will be true if the null hypothesis is false. c. The critical point(s) divides the whole area under a distribution curve into rejection and nonrejection regions. d. The significance level, denoted by α, is the probability of making a Type I error, that is, the probability of rejecting the null hypothesis when it is actually true. e. The nonrejection region is the area where the null hypothesis is not rejected. f. The rejection region is the area where the null hypothesis is rejected. g. A hypothesis test is a two-tailed test if the rejection regions are in both tails of the distribution curve; it is a left-tailed test if the rejection region is in the left tail; and it is a right-tailed test if the rejection region is in the right tail. h. Type I error: A type I error occurs when a true null hypothesis is rejected. The probability of committing a Type I error, denoted by α is: α = P(H0 is rejected | H0 is true) Type II error: A Type II error occurs when a false null hypothesis is not rejected. The probability of committing a Type II error, denoted by β, is: β = P(H0 is not rejected |H0 is false) 9.3 A hypothesis test is a two-tailed test if the sign in the alternative hypothesis is "≠ "; it is a left-tailed test if the sign in the alternative hypothesis is " < " (less than); and it is a right-tailed test if the sign in the alternative hypothesis is " > " (greater than). Table 9.3 on page 405 of the text describes these relationships. 9.5 a. Left-tailed test b. Right-tailed test 9.7 a. Type II error b. Type I error 9.9 a. H0: μ = 20 hours; H1: μ ≠ 20 hours; a two-tailed test b. H0: μ = 10 hours; H1: μ > 10 hours; a right-tailed test c. H0: μ = 3 years; H1: μ ≠ 3 years; a two-tailed test 159 c. Two-tailed test 160 9.11 Chapter Nine d. H0: μ = $1000; H1: μ < $1000; a left-tailed test e. H0: μ = 12 minutes; H1: μ > 12 minutes; a right-tailed test For a two-tailed test, the p–value is twice the area in the tail of the sampling distribution curve beyond the observed value of the sample test statistic. For a one-tailed test, the p–value is the area in the tail of the sampling distribution curve beyond the observed value of the sample test statistic. 9.13 a. Step 1: H0: µ = 46; H1: µ≠ 46; A two-tailed test. Step 2: Since n > 30, use the normal distribution. Step 3: s x = s / n = 9.7 / 40 = 1.53370467 z = ( x – μ) / s x = (49.60–46) / 1.53370467 = 2.35 From the normal distribution table, area to the right of z = 2.35 is .5 – .4906 = .0094 approximately. Hence, p–value = 2(.0094) = .0188 b. Step 1: H0: µ = 26; H1: µ < 26; A left-tailed test. Step 2: Since n > 30, use the normal distribution. Step 3: s x = s / n = 4.3 / 33 = .74853392 z = ( x – μ) / s x = (24.3–26) / .74853392 = –2.27 From the normal distribution table, area to the left of z = –2.27 is .5 –.4884 = .0116 approximately. Hence, p–value = .0116 c. Step 1: H0: µ = 18; H1: µ > 18; A right-tailed test. Step 2: Since n > 30, use the normal distribution. Step 3: s x = s / n = 7.8 / 55 = 1.05175179 z = ( x – μ) / s x = (20.50 – 18) / 1.05175179 = 2.38 From the normal distribution table, area to the right of z = 2.38 is .5 – .4913 = .0087 approximately. Hence, p –value = .0087 TI-83: If you have raw data, first enter your data in a list. Otherwise just select Stat, highlight TESTS, highlight 1: Z Test, and press the ENTER key. If you have raw data choose Data and press the ENTER key, and then using the down arrow to move to enter the requested values and the list name, but be sure to keep Freq set at 1. On the other hand if you have summary statistics, instead of Data highlight Stats and press the ENTER key. Using the down arrow, scroll down to start entering the requested values. In this example we do not have raw data, so will follow the latter method. We select Stat, scroll down enter for part a, µ0 = 46, σ = s = 9.7, x 49.60, n = 40, for µA highlighting µ0, highlighting Calculate, and pressing the ENTER key. The results are shown below where p is the p value we seek. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual 161 Z − Test µ 46 z = 2.347257645 p = .0189121347 x 29.6 n = 40 MINITAB: Select the following Stat, Basic Statistics, and 1-Sample Z to generate a pop-up menu. In the pop-up menu, click beside the words Summarized data as long as you do indeed have summary data , beside the words Sample size enter your n, beside the word Mean enter your x , beside the word standard deviation enter s or if known σ and beside the words Test mean type in µ0. Next click on Options to generate another pop-up menu where you make sure it says the correct Alternative hypothesis. By default µ1 is set to “not equal to”, if your alternative is something else, like greater than or less than, then use the down arrow to highlight the left tailed or right tailed µ1. Once the correct µ1 is in the box, then click on OK. This returns you to the original pop-up menu where you also click on OK and the resulting confidence interval appears in the Session section. The pop-up menus as well as the resulting confidence interval are shown below. The results are shown below where p is the p value we seek. Excel: Open up KADDSTAT and select KADD, Hypothesis Testing, One Sample, Population mean using Z. In the pop-up menu in the first empty box enter µ0 and make sure not equal are highlighted if you are doing a two tailed test. If you have summary statistics and do not have raw data, like in this question, then click beside user input other wise click beside the other choice and enter your range of cells. After clicking beside user input if you, in the boxes enter the n, x , σ if known or s if σ is unknown. In the box beside cell: type in the address on your worksheet where you want the answer to appear. Finally click on OK. For part a, the results and the pop-up menu are shown below. 162 Chapter Nine Caution: For parts b and c, when using technology be sure to select the correct tail for your one tailed test instead the “NOT EQUAL”to used with the two tail test in these instructions which is only appropriate for part a. 9.15 a. Step 1: H0: µ = 72; Hl: µ > 72; A right-tailed test. Step 2: Since n > 30, use the normal distribution. Step 3: sx = s / n = 6 / 36 = 1.0 z = ( x – μ) / s x = (74.07 – 72) / 1.0 = 2.07 The area under the standard normal curve to the right of z = 2.07 is .5 – .4808 = .0192. Hence p–value = .0192 b. For α = .01, do not reject H0, since p–value > .01. c. For α = .025, reject H0, since p–value < .025. 9.17 H0: µ = 16.3 weeks; s x = s / n = 4.2 / H1: µ >16.3 weeks; A right-tailed test. 400 = .21 Test statistic: z = ( x – μ) / s x = (16.9 – 16.3) / .21 = 2.86 The area under the standard normal curve to the right of z = 2.86 is .5 – .4979 = .0021 approximately. Hence, p–value = .0021 For α = .02, reject H0, since p–value < .02. 9.19 H0: µ ≥ 14 hours; s x = s / n = 3.0 / H1 : µ < 14 hours; A left-tailed test. 200 = .21213203 Test statistic: z = ( x – μ) / s x = (13.75 – 14) / .21213203 = –1.18 The area under the standard normal curve to the left of z = –1.18 is .5 – .3810 = .1190 Hence, p–value = .1190. For α = .05, do not reject H0, since p–value > .05. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual 9.21 a. H0: µ = 10 minutes; H1: µ < 10 minutes; s x = s / n = 3.75 / 100 = .375 163 A left-tailed test. z = ( x – μ) / s x = (9.25 – 10) / .375 = –2.00 The area under the standard normal curve to the left of z = –2.00 is .5 – .4772 = .0228. Hence, p–value = .0228 b. Do not reject H0 for α = .02, since p–value > .02. Reject H0 for α = .05, since p–value < .05. 9.23 a. H0: µ = 32 ounces; x / n = .15 / H1: µ ≠ 32 ounces; A two-tailed test. 35 = .02535463 z = ( x – μ) / x (31.90 – 32) / .02535463 = –3.94 The area under the standard normal curve to the left of z = –3.94 is approximately .5 –.5 = 0. Hence, p–value = 2(.00) = .00 approximately b. For α = .01, reject H0, since p–value < .01. For α = .05, reject H0, since p–value < .05. Thus in either case, the inspector will stop the machine. 9.25 The level of significance in a test of hypothesis is the probability of making a Type I error. It is the area under the probability distribution curve where we reject H0. 9.27 The critical value of z separates the rejection region from the nonrejection region and is found from a table such as the standard normal distribution table. The observed value of z is the value calculated for a sample statistic such as x . 9.29 a. The rejection region lies to the left of z = –2.58 and to the right of z = 2.58. The nonrejection region lies between z = –2.58 and z = 2.58. b. The rejection region lies to the left of z = –2.58. The nonrejection region lies to the right of z = –2.58. c. The rejection region lies to the right of z = 1.96. The nonrejection region lies to the left of z = 1.96. Note: To find the critical value or values of z to create the rejection and nonrejection regions one method is to follow the procedures outlined in Chapter 6 of the text or in problems like 6.53 in Chapter 6 of the solutions manual. 164 Chapter Nine 9.31 If H0 is not rejected, the difference between the hypothesized value of μ and the observed value of x is "statistically not significant". 9.33 a. .10 b. .02 c. .005 9.35 n = 90, x = 15, s = 4, and s x = s / n = 4 / 90 = .42163702 a. Critical value: z = –2.33 Observed value: z = ( x – μ) / s x = (15 – 20) / .42163702 = –11.86 b. Critical value: z = –2.58 and 2.58. Observed value: z = ( x – μ) / s x = (15 – 20) / .42163702 = –11.86 Note: To find the observed value of z and the critical value of z follow the procedures outlined in Chapter 6 of the text or in problems 6.17 and 6.53 of the solutions manual. 9.37 a. The rejection region lies to the left of z = –2.33. The nonrejection region lies to the right of z = –2.33. b. The rejection region lies to the left of z = –2.58 and to the right of z = 2.58. The nonrejection region lies between z = –2.58 and z = 2.58. c. The rejection region lies to the right of z = 2.33. The nonrejection region lies to the left of z = 2.33. 9.39 a. n = 100, x = 43, s = 5, and s x = s / n = 5 / 100 = .50 Critical value: z = –1.96 Test statistic: z = ( x – μ) / s x =(43 – 45) / .50 = –4.00; Reject H0. b. n = 100, x = 43.8, s = 7, and s x = s / n = 7 / 100 = .70 Critical value: z = –1.96 Test statistic: z = ( x – μ) / s x = (43.8 – 45) / .70 = –1.71; Do not reject H0. Comparing parts a and b shows that two samples selected from the same population can yield opposite conclusions on the same test of hypothesis. TI-83: If you have raw data, first enter your data in a list. Otherwise just select Stat, then TESTS, then 1: Z Test, and finally pressing the ENTER key. If you have raw data choose Data and press the ENTER key, and then using the down arrow to move to enter the requested values and the list name, but be sure to keep Freq set at 1. On the other hand if you have summary statistics, instead of Data highlight Stats and press the ENTER key. Using the down arrow, scroll down to start entering the requested values. In this example we do not have raw data, so will follow the latter method. We select Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual 165 Stat, scroll down, and enter for part a, µ0 = 45, σ = s = 5, x 43, n = 100. Then we highlight < 45 for µA, highlight Calculate, and press the ENTER key. The results are shown below where p is the p value we seek to compare with α. If we make out comparison in terms of z, we calculate the critical value or values of z first using the methods describe previously. Next we compare the critical value of z with the observed value of z reported below. Since the observed z is less than the critical value of z we do not reject the null hypothesis for part b, but the observed z is more than the critical value of z so we can reject the null hypothesis in part a. Z − Test µ< 45 z= −4 p = 3.1686035E −5 x 43 n = 100 Z − Test µ< 45 z = − 1.714285714 p = .043238098 x 43.8 n = 100 MINITAB: Select the following Stat, Basic Statistics, and 1-Sample Z to generate a pop-up menu. In the pop-up menu, click beside the words Summarized data as long as you do indeed have summary data , beside the words Sample size enter your n, beside the word Mean enter your x , beside the word standard deviation enter s or if known σ and beside the words Test mean type in µ0. Next click on Options to generate another pop-up menu where you enter in the confidence level or 1− α and then make sure it says the correct Alternative hypothesis. By default µ1 is set to “not equal to”, if your alternative is something else, like greater than or less than, then use the down arrow to highlight a left tailed or right tailed µ1. Once the correct µ1 is in the box, then click on OK. This returns you to the original pop-up menu where you also click on OK and the resulting confidence interval appears in the Session section. The pop-up menus for part a as well as the results of the hypothesis tests for both parts are shown below. Since the observed z is less than the critical value of z we do not reject the null hypothesis for part b, but the observed z is more than the critical value of z so we can reject the null hypothesis for part a. 166 Chapter Nine Excel: Open up KADDSTAT and select KADD, Hypothesis Testing,, One Sample, Population mean using Z. In the pop-up menu in the first empty box enter µ0 and make sure not equal are highlighted if you are doing a two tailed test. If you have summary statistics and do not have raw data, like in this question, then click beside user input other wise click beside the other choice and enter your range of cells. After clicking beside user input if you, in the boxes enter the n, x , σ if known or s if σ is unknown. In the box beside cell: type in the address on your worksheet where you want the answer to appear. For part a, the results and the pop-up menu are shown below, while for part b only the results are displayed. Since the observed z is less than the critical value of z we do not reject the null hypothesis for part b, but for part a the observed z is more than the critical value of z so we can reject the null hypothesis. Results for part a: 9.41 a. Step 1: H0: μ = 80; H1: μ ≠ 80; A two-tailed test. Step 2: Since n > 30, use the normal distribution. Step 3: For α = .10, the critical values of z are –1.65 and 1.65. Step 4: x = / n = 15 / 33 = 2.61116484 z = ( x – μ) / x = (76.5 – 80) / 2.61116484 = –1.34 Step 5: b. Step 1: Do not reject H0. H0: μ = 32; H1: μ < 32; A left-tailed test. Results for part b: Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual Step 2: Since n > 30, use the normal distribution. Step 3: For α = .01, the critical value of z is –2.33. Step 4: s x = s / n = 7.4 / 75 = .85447840 167 z = ( x – μ) / s x = (26.5 – 32) / .85447840 = –6.44 Step 5: c. Step 1: Reject H0. H0: μ = 55; H1: μ > 55; A right-tailed test. Step 2: Since n > 30, use the normal distribution. Step 3: For α = .05, the critical value of z is 1.65. Step 4: s x = s / n = 4 / 40 = .63245553 z = ( x – μ) / s x =(60.5 – 55) / .63245553 = 8.70 Step 5: 9.43 Reject H0. Step 1: H0: μ = 16.3 weeks; H1: μ > 16.3 weeks; Step 2: Since n > 30, use the normal distribution. Step 3: For α = .02, the critical value of z is 2.05. Step 4: s x = s / n = 4.2 / 400 = .21 A right-tailed test. z = ( x – μ) / s x =(16.9 – 16.3) / .21 = 2.86 Step 5: 9.45 Reject H0. Conclude the mean duration of unemployment exceeds 16.3 weeks. H0: μ = $2.4 million; H1: μ > $2.4 million; A right-tailed test. For α = .025, the critical value of z is 1.96. s x = s / n = .5 / 32 = ..08838835 Test statistic: z = ( x – μ) / s x = (2.6 – 2.4) / .08838835 = 2.26 Reject H0. Conclude the mean annual cash compensation of CEOs exceeds $2.4 million. 9.47 a. H0: µ = 38.1 years; H1: µ < 38.1 years; A left-tailed test. For α = .01, the critical value of z is –2.33. sx = s / n = 8 / 700 = .30237158 Test statistic: z = ( x – μ) / s x = (37 – 38.1) / .30237158 = –3.64 Reject H0. Conclude that the mean age of motor cycle owners is less than 38.1 years. b. The Type I error in this case would be to conclude that the mean age of motor cycle owners is less than 38.1 years when it is actually equal to 38.1 years . P(Type I error) = α = .01 168 9.49 Chapter Nine a. H0: µ ≥ $35,000; H1: µ < $35,000; A left-tailed test. For α = .01, the critical value of z is –2.33. s x = s / n = 5400 / 150 = $440.90815370 Test statistic: z = ( x – μ) / s x = (33,400 – 35,000) / 440.90815370 = –3.63 Reject H0 and conclude that the company should not open a restaurant in this area. b. If α = 0, there can be no rejection region. Thus, we cannot reject H0. 9.51 a. H0: μ ≥ 8 hours; H1: μ < 8 hours; A left-tailed test For α = .01, the critical value of z is –2.33. s x = s / n = 2.1 / 200 = .14849242 Test statistic: z = ( x – μ) / s x = (7.68 – 8) / .14849242 = –2.15 Do not reject H0. Conclude that the claim is true. b. For α = .025, the critical value of z is –1.96. From part a, the value of the test statistic is –2.15. Hence, we reject H0 and conclude that the claim is false. The decisions in parts a and b are different. The results of this sample are not very conclusive, since raising the significance level from .01 to .025 reverses the decision. 9.53 H1: µ ≠ 32 ounces; H0: µ = 32 ounces; A two-tailed test. For α = .02, the critical values of z are –2.33 and 2.33. x = / n = .15 / 35 = .02535463 Test statistic: z = ( x – μ) / x = (31.90 – 32) / .02535463 = –3.94 Reject H0; the machine needs to be adjusted. 9.55 To make a hypothesis test about the mean textbook costs of college freshmen, first we will take a sample of 30 or more freshmen and collect the information on how much each of them spends on textbooks. Then, using the formulas learned in Chapter 3 for the mean and standard deviation for sample data, we will calculate the sample mean and sample standard deviation. Then we will determine the significance level and make the test of hypothesis. 9.57 a. Area in each tail = α / 2 = .02 / 2 = .01 and df = n – 1 = 20 – 1 = 19 Rejection region: t < –2.539 and t > 2.539 Nonrejection region: –2.539 < t < 2.539 b. Area in the left tail = α = .01 and df = n –1 = 16 – 1 = 15 Rejection region: t < –2.602 Nonrejection region: t > –2.602 c. Area in the right tail = α = .05 and df = n – 1 = 18 – 1 = 17 Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual 169 Rejection region: t > 1.740 Nonrejection region: t < 1.740 Note: you can use technology to solve this problem using the approach similar to the ones describes in problems for the z distribution, only this time selecting the t distribution instead. However, the simplest way to solve this problem is with the t distribution table provided in the textbook inside the back cover. 9.59 a. Area in the right tail = α = .01 and df = n – 1 = 25 – 1 = 24 Critical value: t = 2.492 s x = s / n = 7.5 / 25 = 1.50 Observed value: t = ( x – μ) / s x = (58.5 – 55) / 1.50 = 2.333 b. Area in each tail = α / 2 = .01 / 2 = .005 and df = n–1 = 25–1 = 24 Critical values: t = –2.797 and 2.797 Observed value: t = ( x – μ) / s x = (58.5 – 55) / 1.50 = 2.333 TI-83: If you have raw data, first enter your data in a list. Otherwise just select the following Stat, TESTS, and 2: T Test, followed by pressing the ENTER key. If you have raw data choose Data and press the ENTER key, and then using the down arrow to move to enter the requested values and the list name, but be sure to keep Freq set at 1. On the other hand if you have summary statistics, instead of Data highlight Stats and press the ENTER key. Using the down arrow, scroll down to start entering the requested values. In this example we do not have raw data, so will follow the latter method. We select Stat, scroll down, and enter µ0 = 55, σ = s =7.5, x 58.5, n = 25. Next we highlight > 55 for part a and ≠ 55 for part b, highlight Calculate, and press the ENTER key. The results are shown below where p is the p value we seek to compare with α or we can compare the critical value of t with the observed value of t reported below. T − Test µ> 55 t = 2.333333333 p = .0141686084 x 58.5 Sx = 7.5 n = 25 T − Test µ ≠ 55 t = − 1.714285714 p = .0283372167 x 58.5 Sx = 7.5 n = 25 MINITAB: Select the following Stat, Basic Statistics, and 1-Sample t to generate a pop-up menu. In the pop-up menu, click beside the words Summarized data as long as you do indeed have summary data , beside the words Sample size enter your n, beside the word Mean enter your x , beside the word standard deviation enter s or if known σ and beside the words Test mean type in µ0. Next click on Options to generate another pop-up menu where you make sure it says the correct Alternative 170 Chapter Nine hypothesis. By default µ1 is set to “not equal to”, if your alternative is something else, like greater than or less than, then use the down arrow to highlight left a tailed or right tailed µ1. Once the correct µ1 is in the box, then click on OK. This returns you to the original pop-up menu where you also click on OK and the resulting confidence interval appears in the Session section. The pop-up menus for part a as well as the results for both parts of this problem are shown below. Excel: Open up KADDSTAT and select KADD, Hypothesis Testing,, One Sample, Population mean using t. In the pop-up menu in the first empty box enter µ0 and make sure not equal are highlighted if you are doing a two tailed test. If you have summary statistics and do not have raw data, like in this question, then click beside user input other wise click beside the other choice and enter your range of cells. After clicking beside user input if you, in the boxes enter the n, x , σ if known or s if σ is unknown. In the box beside cell: type in the address on your worksheet where you want the answer to appear. For part a, the results and the pop-up menu are shown below, while for part b only the results are displayed. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual Results for part a: 9.61 171 Results for part b: a. The rejection region lies to the left of t = –2.539. The nonrejection region lies to the right of t = –2.539. b. The rejection region lies to the left of t = –2.861 and to the right of t = 2.861. The nonrejection region lies between t = –2.861 and t = 2.861 c. The rejection region lies to the right of t = 2.539. The nonrejection region lies to the left of t = 2.539. 9.63 a. Step 1: Step 2: H0: µ = 80; H1: µ ≠ 80; A two-tailed test. The sample size is small (n < 30), the population is normally distributed and the population standard deviation σ, is unknown. Hence, we use the t distribution to make the test. Step 3: df = n – 1 = 25 – 1 = 24 and α / 2 = .01 / 2 = .005 Critical values: t = –2.797 and 2.797 Step 4: sx = s / n = 8 / 25 = 1.60 t = ( x – μ) / s x = (77 – 80) / 1.60 = –1.875 Step 5: Do not reject H0. b. Steps 1, 2, and 3 are the same as for part a. Step 4: sx = s / n = 6 / 25 = 1.20 t = ( x – μ) / s x = (86 – 80) / 1.20 = 5.000 Step 5: Reject H0. Comparing parts a and b shows that two samples selected from the same population can yield opposite conclusions on the same test of hypothesis. Note: Using technology follow the instructions outline in 9.59, and then compare your α with the 172 Chapter Nine p value or the critical value of t with its observed value. This will allow you to determine whether to reject the null hypothesis or not. 9.65 Hl: µ ≠ 24; a. H0: µ = 24; A two-tailed test. df = n– 1 = 25 – 1 = 24 and α / 2 = .01 / 2 = .005 The critical values of t are –2.797 and 2.797. s x = s / n = 4.9 / 25 = .98 t = ( x – μ) / s x = (28.5–24) / .98 = 4.592 b. H0 :µ = 30; Reject H0. H1 : µ < 30: A left-tailed test. df = n–1 = 16–1 = 15 and α = .025 The critical value of t is –2.131. s x = s / n = 6.6 / 16 = 1.65 c. t = ( x – μ) / s x = (27.5 – 30) / 1.65 = –1.515 Do not reject H0. H0: µ = 18, A right tailed test. H1: µ > 18; df = n – 1 = 20 – 1 = 19 and α = .10 The critical value of t is 1.328. sx = s / n = 8 / 20 = 1.78885438 t = ( x – μ) / s x = (22.5 – 18) / 1.78885438 = 2.516 9.67 H0: µ = 69.5 inches; H1: µ ≠ 69.5 inches; Reject H0. A two-tailed test. df = n –1 = 25 –1 = 24 and α / 2 = .01 / 2 = .005 The critical values of t are –2.797 and 2.797. s x = s / n = 2.1 / 25 = .420 t = ( x – μ) / s x =(70.25–69.5) / .420 = 1.786 9.69 H0: µ ≤ 7 hours; H1: µ > 7 hours; Do not reject H0. A right-tailed test. df = n – 1 = 20 – 1 = 19 and α = .025 The critical value of t is 2.093. s x = s / n = 2.3 / 9.71 20 = .51429563 t = ( x – μ) / s x = (10.5 – 7) / .51429563 = 6.805 Reject H0. The president's claim is not true. H0: µ ≤ 30 calories; A right-tailed test. H1: µ > 30 calories; df = n – 1 = 16 – 1 = 15 and α = .05 The critical value of t is 1.753. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual 173 s x = s / n = 3 / 16 = .750 9.73 t = ( x – μ) / s x = (32–30) / .750 = 2.667 Reject H0. The manufacturer's claim is false. a. H0: µ ≤ 45 minutes; A right-tailed test. H1: µ > 45 minutes; df = n –1 = 20 –1 = 19 and α = .01 The critical value of t is 2.539. sx = s / n = 3 / 20 = .67082039 t = ( x – μ) / s x = (49.50 – 45) / .67082039 = 6.708 Reject H0. The mean drying time for these paints is more than 45 minutes. b. The Type I error would occur if the mean drying time for these paints is 45 minutes or less, but we conclude otherwise. The probability of such an error is .01 here. 9.75 a. If α = 0, there is no rejection region. Hence, the decision must be: “Do not reject H0”. b. H0: µ ≥ 1200 words; H1: µ < 1200 words; A left-tailed test. df = n – 1 = 25 – 1 = 24 and α = .05 Critical value: t = –1.711 s x = s / n = 85 / 25 = 17.00 t = ( x – μ) / s x = (1125 – 1200) / 17.00 = –4.412 Reject H0. Conclude that the claim of the business school is false. 9.77 From the given data: n = 10, x = 257, and x 2 = 7341 x = x / n = 257 / 10 = 25.70 hours. s 2 x ( x) 2 / n = n 1 H0: µ = 18 hours; 7341 (257 ) 2 / 10 = 9.04372097 10 1 H1: µ ≠ 18 hours; A two-tailed test. df = n – 1 = 10 – 1 = 9 and α / 2 = .05 / 2 = .025 Critical values: t = –2.262 and t = 2.262 s x = s / n = 9.04372097 / 10 = 2.85987568 t = ( x – μ) / s x = (25.70 – 18) / 2.85987568 = 2.692 Reject H0. Conclude that the claim of the earlier study is false. 9.79 To make a hypothesis test about the mean amount spent on gas by all customers at the given gas station, first we will take a sample of less than 30 customers and collect the information on how much they spent on gas at this gas station. Then, using the formulas learned in Chapter 3 for the mean and 174 Chapter Nine standard deviation for sample data, we will calculate the sample mean and sample standard deviation. Assuming the amounts spent on gas by this station’s customers are normally distributed, we will determine the significance level and make the test. Our hypotheses would be: H0: µ = $10.90 versus H1: µ ≠ $10.90 9.81 In order to use the normal distribution in a test of hypothesis about a population proportion, both np and nq must be greater than 5, where p is the value of the population proportion in the null hypothesis and q = 1 – p. 9.83 a. Yes; np = 30(.65) = 19.5 > 5; and nq = 30(.35) = 10.5 > 5 b. No; np = 70(.05) = 3.5 < 5 c. No; np = 60(.06) = 3.6 < 5 d. Yes; np = 900(.17) = 153 > 5, and nq = 900(.83) = 747 > 5 9.85 a. The rejection region lies to the left of z = –1.96 and to the right of z = 1.96. The nonrejection region lies between z = –1.96 and z = 1.96. b. The rejection region lies to the left of z = –2.05. The nonrejection region lies to the right of z = –2.05. c. The rejection region lies to the right of z = 1.96. The nonrejection region lies to the left of z = 1.96. 9.87 a. Area in the left tail = α = .01 Critical value: z = –2.33 pˆ pq = n .63(.37 ) = .03413942 200 Observed value of z = pˆ p pˆ = .60 .63 = –.88 .03413942 b. Area in each tail = α / 2 = .01 / 2 = .005 Critical values: z = –2.58 and 2.58 Observed value of z is –.88 as in part a. 9.89 a. The rejection region lies to the left of z = –2.33. The nonrejection region lies to the right of z = –2.33. b. The rejection region lies to the left of z = –2.58 and to the right of z = 2.58. The nonrejection region lies between z = –2.58 and z = 2.58. c. The rejection region lies to the right of z = 2.33. The nonrejection region lies to the left of z = 2.33. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual 9.91 a. Step 1: Step 2: H0: p = .45; H1: p < .45; 175 A left-tailed test. np = 400(.45) = 180 and nq = 400(.55) = 220 Since np > 5 and nq > 5, use the normal distribution. Step 3: For α = .025, the critical value of z is –1.96. Step 4: pˆ z= Step 5: pq = n pˆ p pˆ = .45 (. 55 ) = .02487469 400 .42 .45 = –1.21 .02487469 Do not reject H0. b. Steps 1, 2, and 3 are identical to part a. pˆ p Step 4: z= Step 5: Reject H0. pˆ = .39 .45 = –2.41 .02487469 The results of parts a and b show that two different samples from the same population can yield opposite decisions on a test of the same hypothesis. TI-83: If you have raw data, first enter your data in a list. Otherwise just select the following Stat, TESTS, and 5: 1−PropZTest, followed by pressing the ENTER key. Now enter the requested values. In this example p = .45. x is the number of successes in the sample or the sample mean of n* p̂ . If it is not yet calculated you can enter the values in parenthesis and the calculation will be done for you, so in the first part of this problem we enter (400*.42) followed by pressing the ENTER key and the calculator returns a value of 168 for x. Next we enter n = 400, highlight the < symbol, highlight Calculate, and press the ENTER key. The results are shown below where p is the p value we seek to compare with α or we can compare the critical value of z with its observed value reported below. The results show we can not reject the null hypothesis for part a, while for part b we can reject the null hypothesis. 1 −PropZTest prop < .45 z = -1.206045378 p = .1139000591 p̂ = .42 n = 400 1 −PropZTest prop < .45 z = -2.412090757 p = .007930662 p̂ = .39 n = 400 MINITAB: Select the following Stat, Basic Statistics, and 1 Proportion to generate a pop-up menu. In the pop-up menu, click beside the words Summarized data and enter n in the first box, and enter µ or the value of n* p̂ in the 2nd box. Next click on Options to generate another pop-up menu where you 176 Chapter Nine enter the proportion from the null hypothesis beside the words Test proportion and make sure it says the correct Alternative hypothesis. By default µ1 is set to “not equal to”, if your alternative is something else, like greater than or less than, then use the down arrow to highlight left a tailed or right tailed µ1. Once the correct µ1 is in the box, click on the empty box below the alternative hypothesis, and then click on OK. This returns you to the original pop-up menu where you also click on OK and the resulting confidence interval appears in the Session section. The pop-up menus for part a as well as the results for both parts of this problem are shown below. Excel: Open up KADDSTAT and select KADD, Hypothesis Testing, One Sample, Population mean using t. In the pop-up menu in the first empty box enter µ0 and make sure not equal are highlighted if you are doing a two tailed test. If you have summary statistics and do not have raw data, like in this question, then click beside user input other wise click beside the other choice and enter your range of cells. After clicking beside user input if you enter the n in the first box and in the next box enter x which is number of successes in the sample which is n* p̂ . In the box beside cell: type in the address on your worksheet where you want the answer to appear. For part a, the results and the pop-up menu are shown below, while for part b only the results are displayed. The results show we can not reject the null hypothesis for part a, while for part b we can reject the null hypothesis. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual Results for part a: 9.93 a. Step 1: Step 2: H1: p ≠ .57; H0: p = .57; A two-tailed test. np = 800(.57) = 456 and nq = 800(l – .57) = 344 Since np > 5 and nq > 5, use the normal distribution. Step 3: For α = .05, the critical values of z are –1.96 and 1.96. Step 4: p = .57 and q = 1 – p = 1 – .57 = .43 pˆ z= Step 5: b. Step 1: Step 2: pq = n pˆ p pˆ = .57 (.43) = .01750357 800 .50 .57 = –4.00 .01750357 Reject the null hypothesis. H0: p = .26; H1: p < .26; A left-tailed test. np = 400(.26) = 104 and nq = 400(l –.26) = 296 Since np > 5 and nq > 5, use the normal distribution. Step 3: For α = .01, the critical value of z is –2.33. Step 4: p = .26 and q = 1 – p = 1 –.26 = .74 pˆ z= Step 5: c. Step l: Step 2: pq = n pˆ p pˆ = .26 (. 74 ) = .02193171 400 .23 .26 = –1.37 .02193171 Do not reject the null hypothesis. H0: p = .84; Hl: p > .84; A right-tailed test. np = 250(.84) = 210 and nq = 250(l –.84) = 40 Since np > 5, and nq > 5, use the normal distribution. Step 3: For α = .025, the critical value of is 1.96. Step 4: p = .84 and q = 1 –p = 1 –.84 = .16 177 Results for part b: 178 Chapter Nine pˆ z= Step 5: 9.95 pq = n pˆ p pˆ = .84 (. 16 ) = .02318620 250 .85 .84 = .43 .02318620 Do not reject the null hypothesis. H0: p = .49; H1: p > .49; A right-tailed test. For α = .025, the critical value of z is 1.96. pˆ pq / n = (. 49 )(. 51) / 200 = .03534827 p̂ = .52 z= pˆ p pˆ = .52 .49 = .85 .03534827 Do not reject H0. Do not conclude that the current percentage of management and professional jobs held by women exceeds 49%. 9.97 H0: p = .45; H1: p > .45; A right-tailed test. For α = .01, the critical value of z is 2.33. p̂ = 248 / 500 = .496 pˆ pq / n = z= pˆ p pˆ = (. 45 )(. 55 ) / 500 = .02224860 .496 .45 = 2.07 .02224860 Do not reject H0. Do not conclude that the proportion of employers monitoring their employees use of company phones exceeds 45%. 9.99 a. H0: p = .32; H1: p > .32; A right-tailed test. For α = .025, the critical value of z is 1.96. p̂ = 396 / 1100 = .36 pˆ pq / n = z= pˆ p pˆ = (.32 )(. 68 ) / 1100 = .01406479 .36 .32 = 2.84 .01406479 Reject H0. Conclude that more than 32% of households earning $75,000 or more would struggle to pay an unexpected bill of $5000. b. The Type I error would occur if we concluded that more than 32% of households earning $75,000 or more would struggle to pay an unexpected bill of $5000, when actually it is not true. The probability of making this error is α = .025. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual 9.101 a. H0: p ≥ .60; H1: p < .60; 179 A left-tailed test. For α = .01, the critical value of z is –2.33. p̂ = 208 / 400 = .52 pˆ z= pq = n pˆ p pˆ = .60 (. 40 ) = .02449490 400 .52 .60 = –3.27 .02449490 Reject H0. Conclude that the company's claim is not true. b. If α = 0, there is no rejection region, so we could not reject H0. Thus, we would conclude that the company's claim is true. 9.103 a. H0: p ≤ .07; H1: p > .07; A right-tailed test. For α = .02, the critical value of z = 2.05. p̂ = 22 / 200 = .11 pˆ z= pq = n pˆ p pˆ = .07 (.93) = .01804162 200 .11 .07 = 2.22 .01804162 Reject H0. Conclude that the machine should be stopped. b. If α = .01, the critical value of z is 2.33. Since the value of the test statistic is 2.22, we do not reject H0. Thus, our decision differs from that of part a; we conclude that the machine should not be stopped. The decisions in parts a and b are different. The results of this sample are not very conclusive, since lowering the significance level from 2% to 1% reverses the decision. 9.105 Select a random sample of 40 students from your school and ask them whether or not they hold off– campus jobs. From this data calculate p̂ . Using hypotheses H0: p = .65 and H1: p ≠ .65, choose a level of significance and find the critical values from the table of the normal distribution. Then compute pˆ 9.107 H0: µ = 40; (.65 )(. 35 ) pˆ .65 and z = and make a decision. 40 pˆ H1: µ ≠ 40 sx = s / n = 6 / 64 = .75 z = ( x – μ) / s x = (38.4 – 40) / .75 = –2.13 180 Chapter Nine a. For α = .02, the critical values of z are –2.33 and 2.33. Hence, do not reject H0. b. P(Type I error) = α = .02 c. Area to the left of z = –2.13 is .5 – .4834 = .0166. Hence, p–value = 2(.0166) = .0332 If α = .01, do not reject H0 since p–value > .01. If α = .05, reject H0 since p–value < .05. 9.109 H0: p = .44; pˆ z= H1: p < .44 pq = n pˆ p pˆ = .44 (. 56 ) = .02339991 450 .39 .44 = –2.14 .02339991 a. For α = .02, the critical value of z is –2.05. Reject H0. b. P(Type I error) = α = .02 c. Area to the left of –2.14 is .5 –.4838 = .0162. Hence, p–value = .0162 If α = .01, do not reject H0, since p–value > .01. If α = .025, reject H0, since p–value < .025. 9.111 H0: µ = 1245 cubic feet; H1: µ < 1245 cubic feet s x = s / n = 250 / 100 = 25.00 z = ( x – μ) / s x = (1175 – 1245) / 25.00 = –2.80 Area to the left of z = –2.80 is .5 – .4974 = .0026. Hence, p–value = .0026 Reject H0 for α = .025, since p–value < .025. 9.113 a. H0: µ = 180 months; H1: µ < 180 months; A left-tailed test. For α = .02, the critical value of z is –2.05. s x = s / n = 27 / 60 = 3.48568501 z = ( x – μ) / s x = (171 – 180) / 3.48568501 = –2.58 Reject H0. The sample supports the alternative hypothesis that the current mean sentence for such crimes is less than 180 months. b. The Type I error would be to conclude that the current mean sentence for such crimes is less than 180 months when in fact it is not. P(Type I error) = .02 c. If α = 0, there is no rejection region and, hence, we cannot reject H0. Thus, our conclusion would change. 9.115 a. H0: µ ≤ 2400 square feet; H1: µ > 2400 square feet; For α = .05, the critical value of z is 1.65. A right-tailed test. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual s x = s / n = 472 / 181 50 = 66.75088014 square feet z = ( x – μ) / s x = (2540 – 2400) / 66.75088014 = 2.10 Reject H0. The sample supports the alternative hypothesis that the real estate agents’ claim is false. b. For α = .01, the critical value of z is 2.33. From part a, the observed value of z is 2.10. Hence, do not reject H0. The results of parts a and b show that the sample does not support the alternative hypothesis very strongly, since lowering the significance level from .05 to .01 reverses the conclusion. 9.117 H0: µ = 8 minutes; H1: µ < 8 minutes; A left-tailed test. For α = .025, the critical value of z is –1.96. s x = s / n = 2.1 / 32 = .37123106 minute z = ( x – μ) / s x = (7.5 – 8) / .37123106 = –1.35 Do not reject H0. At the .025 level of significance, the difference between the sample mean and the hypothesized value of the population mean is small enough to attribute to chance. Thus, the manager's claim is not justified. 9.119 H0: µ = 25 minutes; H1: µ≠ 25 minutes; A two-tailed test. df = n – 1 = 16 – 1 = 15and α / 2 = .01 / 2 = .005 The critical values of t are –2.947 and 2.947. s x = s / n = 4.8 / 16 = 1.20 t = ( x – μ) / s x = (27.5 – 25) / 1.20 = 2.083 Do not reject H0. 9.121 a. H0: µ = 114 minutes; H1: µ < 114 minutes; A left-tailed test. df = n – 1 = 25 –1 = 24 and α = .01 The critical value of t is –2.492. s x = s / n = 11 / 25 = 2.20 minutes t = ( x – μ) / s x = (109 – 114) / 2.20 = –2.273 Do not reject H0. The mean time currently spent by all adults with their families is not less than 114 minutes a day. b. If α = 0, there is no rejection region, so there is no need to go through the five steps of hypothesis testing. We cannot reject H0. 9.123 H0: µ ≤ 150 calories; H1:µ > 150 calories df = n – 1 = 10 – 1 = 9 and α = .025 182 Chapter Nine The critical value of t is 2.262. x = 1527, and x 2 = 233,663 x = x / n = 1527 / 10 = 152.7 2 x ( x) 2 / n = s n 1 233 ,663 (1527 ) 2 / 10 = 7.37940076 10 1 s x = s / n = 7.37940076 / 10 = 2.33357142 t = ( x – μ) / s x = (152.70 – 150) / 2.33357142 = 1.157 9.125 a. H0: p = .32; Hl: p > .32; Do not reject H0. A right-tailed test. For α = .02, the critical value of z = 2.05. pq = n pˆ z= pˆ p .35 .32 = 1.88 .016 = pˆ .32 (. 68 ) = .016 850 Do not reject H0. Do not conclude that the current percentage of faculty who hold this opinion exceeds 32%. b. The Type I error would occur if we concluded that the current percentage of the faculty who hold this opinion exceeds 32%, when in fact it does not. P(Type I error) = .02. 9.127 H0: p = .56; H1: p < .56; A left-tailed test. For α = .02, the critical value of z is –2.05. p̂ = .50 pˆ z= pq = n pˆ p pˆ = .56 (. 44 ) = .04052982 150 .50 .56 = –1.48 .04052982 Do not reject H0. Do not conclude that the current percentage of attorneys who take work on vacation is less than 56%. 9.129 a. H0: p ≥ .90; H1: p < .90; For α = .02, the critical value of z is –2.05. pˆ pq = n .90 (. 10 ) = .03162278 90 p̂ = 75 / 90 = .833 A left-tailed test. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual z= pˆ p = pˆ 183 .833 .90 = –2.12 .03162278 Reject H0. Conclude that the company's policy is not maintained. b. If α = 0, there is no rejection region; thus, we cannot reject H0 and cannot conclude that the company's policy is not maintained. 9.131 H0: p ≥ .90; pˆ H1: p < .90; pq = n A left-tailed test. .90 (. 10 ) = .03162278 90 p̂ = 75 / 90 = .833 z= pˆ p = pˆ .833 .90 = –2.12 .03162278 The area to the left of z = –2.12 under the normal curve is .5 – .4830 = .0170. Hence, p–value = .0170 If α = .05, reject H0 since p–value < .05. If α = .01, do not reject H0 since p–value > .01. 9.133 Since np > 5 and nq > 5, the distribution of p̂ is approximately normal. For α = .05, the critical value of z is 1.65. pˆ Now z = pq = n .04 (. 96 ) = .01718676 130 pˆ p pˆ , so 1.65 = pˆ .04 .01718676 Solving for p̂ yields p̂ = 1.65(.01718676) +.04 = .0684 Thus, we would reject H0 if p̂ > .0684. Hence, .0684 = c c n 130 Then, c = 130(.0684) = 8.89 9 Therefore, reject H0 and shut down the machine if the number of defectives in a sample of 130 parts is 9 or more. 9.135 First, we must be sure that the cure rate of the new therapy is not lower than that of the old therapy. Let p be the proportion of all patients cured with the new therapy. We must test the hypotheses: H0: p = .60; H1: p < .60 For α = .01, the critical value of z is –2.33. 184 Chapter Nine pˆ pq = n .60 (. 40 ) = .03464102 200 From the data: p̂ = x / n = 108 / 200 = .54 Test statistic: z = pˆ p pˆ = .54 .60 = –1.73 .03464102 Do not reject H0. Thus, we cannot conclude that the new therapy has a lower cure rate. Next, we must see if the new therapy is effective in reducing the number of visits. Let µ be the mean number of visits required for all patients using the new therapy regime. We must test the hypotheses: H0: µ = 140 visits; H1: µ < 140 visits For α = .01, the critical value of z is –2.33. s x = s / n = 38 / 200 = 2.68700577 Test statistic: z = ( x – μ) / s x = (132 – 140) / 2.68700577 = –2.98 Reject H0. Conclude that the new therapy regime requires fewer visits on average. Based on the results of these two hypothesis tests, the health care provider should support the new therapy regime. 9.137 a. Let p be the proportion of all people receiving the new vaccine who contract the disease within a year. Then the appropriate hypotheses are: H0: p = .30; H1: p < .30 b. Let x be the number of people in a sample of 100 inoculated with the new vaccine who contract the disease within a year. Then, under H0, x is a binomial random variable with n = 100 and p = .30. Hence, σ= µ = np = 100(.30) = 30 npq 100 (.30 )(. 70 ) = 4.58257569 If 84 or more of the 100 people in the sample do not contract the disease, then x < 16. Using a normal approximation, and correcting for continuity, we need P(x < 16.5). For x = 16.5: z = 16 .5 30 = –2.95 4.58257569 Thus, P( x 16.5 p .30) P( z 2.95) .5 .4984 .0016 c. Let x be the number of people in a sample of 20 inoculated with the new vaccine who contract the disease within a year. Then, under H0, x is a binomial random variable with n = 20 and p = .30. Using Table IV, Appendix C: P( x 3 p .30) = P(0) + P(l) + P(2) = .0008 +.0068 +.0278 = .0354 9.139 The following are two possible experiments we might conduct to investigate the effectiveness of middle taillights. Mann – Introductory Statistics, Fifth Edition, Students Solutions Manual 185 I. Let: p1 = proportion of all collisions involving cars built since 1984 that were rear–end collisions. Let: p2 = proportion of all collisions involving cars built before 1984 that were rear–end collisions. (p2 would be known) We would test H0: p1 = p2 versus H1: p1 < p2. We would take a random sample of collisions involving cars built since 1984 and determine the number that were rear–end collisions. We would have to assume the following: i. The only change in cars built since 1984 that would reduce rear–end collisions is the middle taillight. ii. None of the cars built before 1984 had middle taillights. iii. People's driving habits, traffic volume, and other variables that might affect rear–end collisions have not changed appreciably since 1984. II. Let: µ1 = mean number of rear–end collisions per 1000 cars built since 1984 µ2 = mean number of rear–end collisions per 1000 cars built before 1984 (µ2 would be known) We would test H0: µ1 = µ2 versus H1: µ1 < µ2. We could take several random samples of 1000 cars built since 1984. We would determine the number of rear–end collisions in each sample of 1000 cars. We would find the mean and standard deviation of these numbers and use them to form the test statistic. We would require the same assumptions as those listed for the test in part 1. Also, if we took less than 30 samples, we would have to assume that the number of rear–end collisions per 1000 cars has a normal distribution. Self–Review Test for Chapter Nine 1. a 2. b 3. a 4. b 5. a 6. a 7. a 8. b 9. c 10. a 11. c 12. b 13. d 14. c 15. a 16. b 17. a. Step 1: H0: µ = $85,900; H1: µ >85,900; Step 2: Since n > 30, use the normal distribution. Step 3: s x = s / n = 27,000 / A right-tailed test. 36 = $4500 z = ( x – μ) / s x = (95,000 – 85,900) / 4500 = 2.02 The area to the right of z = 2.02 under the normal curve is .5 – .4783 = .0217. Hence, p–value = .0217 b. If α = .01, do not reject H0, since p–value > .01. If α = .05, reject H0, since p–value < .05. 186 Chapter Nine 18. a. Step 1: H0: µ =185 minutes; H1: µ < 185 minutes Step 2: Since n > 30, use the normal distribution. Step 3: For α = .01, the critical value of z is –2.33. Step 4 s x = s / n = 12 / 36 = 2 z = ( x – μ) / s x = (179 – 185) / 2 = –3.00 Step 5: Reject H0. Conclude that the mean durations of games have decreased after the meeting. b. The Type I error would be to conclude that the mean durations of games have decreased after the meeting when they are actually equal to the duration of games before the meeting. P(Type I error) = α = .01 c. If α = 0, there is no rejection region, so do not reject H0. d. From part a, z = –3.00. The area to the left of z = –3.00 under the normal curve is .5 – .4987 = .0013. 19. a. Hence, p–value =.0013. For α = .01, reject H0, since p–value < .01. H0: µ ≥ 31 months; H1: µ < 31 months The critical value of t is: –2.131 s x = s / n = 7.2 / 16 = 1.80 t = ( x – μ) / s x = (25–31) / 1.80 = –3.333 Reject H0. Conclude that the editor's claim is false. b. The Type I error would be to conclude that the editor's claim is false when it is actually true. P(Type I error) = α = .025 c. For α = .001, the critical value of t is –3.733. Since the observed value of t = –2.778 is greater than –3.733, we would not reject H0. 20. a. H0: p = .50; H1: p < .50; A left-tailed test. For α = .05, the critical value of z is –1.65. pˆ p̂ = 450 / 1000 = .45 z= pˆ p pˆ = .45 .50 = –3.16 .015811388 pq = n .50 (. 50 ) = .015811388 1000 Reject H0. Conclude that the less than 50% have a will. b. The Type I error would be to conclude that the percentage of adults with wills was less than 50 % when it is actually 50%. P(Type I error) = α = .05 c. If α = 0, there is no rejection region, so do not reject H0. 21. a. Referring to Problem 20. The area to the left of z = –3.16 under the normal curve is .5 –.5000 = .0000. Hence, p–value = .0000 b. If α = .05, reject H0, since p–value < .05 If α = .01, reject H0, since p–value <.01