Chapter Hypothesis Testing Learning Objectives Learn how to formulate and test hypotheses about a population mean and/or a population proportion Understand the types of errors possible when conducting a hypothesis test Be able to determine the probability of making various errors in hypothesis tests Know how to compute and interpret p-values Be able to use critical values to draw hypothesis testing conclusions Be able to determine the size of a simple random sample necessary to keep the probability of hypothesis testing errors within acceptable limits Know the definition of the following terms: null hypothesis alternative hypothesis Type I error Type II error one-tailed test two-tailed test p-value level of significance critical value power curve Solutions: 9-1 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter a H0: µ ≤ 600 Manager’s claim Ha: µ > 600 b We are not able to conclude that the manager’s claim is wrong c The manager’s claim can be rejected We can conclude that µ > 600 a H0: µ ≤ 14 Ha: µ > 14 b There is no statistical evidence that the new bonus plan increases sales volume c The research hypothesis that µ > 14 is supported We can conclude that the new bonus plan increases the mean sales volume a H0: µ = 32 Specified filling weight Ha: µ ≠ 32 Overfilling or underfilling exists b There is no evidence that the production line is not operating properly Allow the production process to continue c Conclude µ ≠ 32 and that overfilling or underfilling exists Shut down and adjust the production line a H0: µ ≥ 220 Ha: µ < 220 Research hypothesis Research hypothesis to see if mean cost is less than $220 b We are unable to conclude that the new method reduces costs c Conclude µ < 220 Consider implementing the new method based on the conclusion that it lowers the mean cost per hour a The Type I error is rejecting H0 when it is true This error occurs if the researcher concludes that young men in Germany spend more than 56.2 minutes per day watching prime-time TV when the national average for Germans is not greater than 56.2 minutes b The Type II error is accepting H0 when it is false This error occurs if the researcher concludes that the national average for German young men is ≤ 56.2 minutes when in fact it is greater than 56.2 minutes a H0: µ ≤ The label claim or assumption Ha: µ > b Claiming µ > when it is not This is the error of rejecting the product’s claim when the claim is true c Concluding µ ≤ when it is not In this case, we miss the fact that the product is not meeting its label specification 9-2 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Hypothesis Testing a H0: µ ≤ 8000 Ha: µ > 8000 Research hypothesis to see if the plan increases average sales b Claiming µ > 8000 when the plan does not increase sales A mistake could be implementing the plan when it does not help c Concluding µ ≤ 8000 when the plan really would increase sales This could lead to not implementing a plan that would increase sales a H0: µ ≥ 220 Ha: µ < 220 b Claiming µ < 220 when the new method does not lower costs A mistake could be implementing the method when it does not help c Concluding µ ≥ 220 when the method really would lower costs This could lead to not implementing a method that would lower costs a z= b Lower tail p-value is the area to the left of the test statistic x − µ0 σ/ n = 19.4 − 20 / 50 = −2.12 Using normal table with z = -2.12: p-value =.0170 c p-value ≤ 05, reject H0 d Reject H0 if z ≤ -1.645 -2.12 ≤ -1.645, reject H0 10 a b z= x − µ0 σ/ n = 26.4 − 25 / 40 = 1.48 Upper tail p-value is the area to the right of the test statistic Using normal table with z = 1.48: p-value = 1.0000 - 9306 = 0694 c p-value > 01, not reject H0 d Reject H0 if z ≥ 2.33 1.48 < 2.33, not reject H0 11 a b z= x − µ0 σ/ n = 14.15 − 15 / 50 = −2.00 Because z < 0, p-value is two times the lower tail area Using normal table with z = -2.00: p-value = 2(.0228) = 0456 9-3 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter c p-value ≤ 05, reject H0 d Reject H0 if z ≤ -1.96 or z ≥ 1.96 -2.00 ≤ -1.96, reject H0 12 a z= x − µ0 σ/ n = 78.5 − 80 12 / 100 = −1.25 Lower tail p-value is the area to the left of the test statistic Using normal table with z = -1.25: p-value =.1056 p-value > 01, not reject H0 b z= x − µ0 σ/ n = 77 − 80 12 / 100 = −2.50 Lower tail p-value is the area to the left of the test statistic Using normal table with z = -2.50: p-value =.0062 p-value ≤ 01, reject H0 c z= x − µ0 σ/ n = 75.5 − 80 12 / 100 = −3.75 Lower tail p-value is the area to the left of the test statistic Using normal table with z = -3.75: p-value ≈ p-value ≤ 01, reject H0 d z= x − µ0 81 − 80 = = 83 σ / n 12 / 100 Lower tail p-value is the area to the left of the test statistic Using normal table with z = 83: p-value =.7967 p-value > 01, not reject H0 Reject H0 if z ≥ 1.645 13 a b z= x − µ0 52.5 − 50 = = 2.42 σ/ n / 60 2.42 ≥ 1.645, reject H0 x − µ0 51 − 50 z= = = 97 σ / n / 60 97 < 1.645, not reject H0 9-4 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Hypothesis Testing c z= x − µ0 σ/ n = 51.8 − 50 = 1.74 / 60 1.74 ≥ 1.645, reject H0 14 a z= x − µ0 σ/ n = 23 − 22 10 / 75 = 87 Because z > 0, p-value is two times the upper tail area Using normal table with z = 87: p-value = 2(1 - 8078) = 3844 p-value > 01, not reject H0 b z= x − µ0 σ/ n = 25.1 − 22 = 2.68 10 / 75 Because z > 0, p-value is two times the upper tail area Using normal table with z = 2.68: p-value = 2(1 - 9963) = 0074 p-value ≤ 01, reject H0 c z= x − µ0 σ/ n = 20 − 22 10 / 75 = −1.73 Because z < 0, p-value is two times the lower tail area Using normal table with z = -1.73: p-value = 2(.0418) = 0836 p-value > 01, not reject H0 15 a H0: µ ≥ 1056 Ha: µ < 1056 b z= x − µ0 σ/ n = 910 − 1056 1600 / 400 = −1.83 Lower tail p-value is the area to the left of the test statistic Using normal table with z = -1.83: p-value =.0336 c p-value ≤ 05, reject H0 Conclude the mean refund of “last minute” filers is less than $1056 d Reject H0 if z ≤ -1.645 -1.83 ≤ -1.645, reject H0 16 a H0: µ ≤ 3173 Ha: µ > 3173 9-5 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter b z= x − µ0 σ/ n = 3325 − 3173 1000 / 180 = 2.04 p-value = 1.0000 - 9793 = 0207 c 17 a p-value < 05 Reject H0 The current population mean credit card balance for undergraduate students has increased compared to the previous all-time high of $3173 reported in April 2009 H0: µ = 125,500 Ha: µ ≠ 125,500 b z= x − µ0 σ/ n = 118, 000 − 125,500 30, 000 / 40 = −1.58 Because z < 0, p-value is two times the lower tail area Using normal table with z = -1.58: p-value = 2(.0571) = 1142 c p-value > 05, not reject H0 We cannot conclude that the year-end bonuses paid by Jones & Ryan differ significantly from the population mean of $125,500 d Reject H0 if z ≤ -1.96 or z ≥ 1.96 z = -1.58; cannot reject H0 18 a H0: µ = 4.1 Ha: µ ≠ 4.1 b z= x − µ0 σ/ n = 3.4 − 4.1 / 40 = −2.21 Because z < 0, p-value is two times the lower tail area Using normal table with z = -2.21: p-value = 2(.0136) = 0272 c p-value = 0272 < 05 Reject H0 and conclude that the return for Mid-Cap Growth Funds differs significantly from that for U.S Diversified funds 19 H0: µ ≤ 14.32 Ha: µ > 14.32 z= x − µ0 14.68 − 14.32 = = 2.15 σ/ n 1.47 75 Upper tail p-value is the area to the right of the test statistic 9-6 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Hypothesis Testing Using normal table with z = 2.15: p-value = 1.0000 - 9842 = 0158 p-value ≤ 05, reject H0 Conclude that there has been an increase in the mean hourly wage of production workers 20 a H0: µ ≥ 32.79 Ha: µ < 32.79 x − µ0 z= c Lower tail p-value is area to left of the test statistic σ/ n = 30.63 − 32.79 b 5.6 50 = −2.73 Using normal table with z = -2.73: p-value = 0032 d p-value ≤ 01; reject H Conclude that the mean monthly internet bill is less in the southern state 21 a H0: µ ≤ 15 Ha: µ > 15 x−µ z= c Upper tail p-value is the area to the right of the test statistic σ/ n = 17 − 15 b / 35 = 2.96 Using normal table with z = 2.96: p-value = 1.0000 - 9985 = 0015 d 22 a p-value ≤ 01; reject H0; the premium rate should be charged H0: µ = Ha: µ ≠ b Because z > 0, p-value is two times the upper tail area Using normal table with z = 1.37: p-value = 2(1 - 9147) = 1706 c Do not reject H0 Cannot conclude that the population mean waiting time differs from minutes d x ± z.025 (σ / n ) 8.4 ± 1.96 (3.2 / 120) 8.4 ± 57 (7.83 to 8.97) Yes; µ = is in the interval Do not reject H0 9-7 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter 23 a b t= x − µ0 s/ n = 14 − 12 4.32 / 25 = 2.31 Degrees of freedom = n – = 24 Upper tail p-value is the area to the right of the test statistic Using t table: p-value is between 01 and 025 Exact p-value corresponding to t = 2.31 is 0149 c p-value ≤ 05, reject H0 d With df = 24, t.05 = 1.711 Reject H0 if t ≥ 1.711 2.31 > 1.711, reject H0 24 a b t= x − µ0 s/ n = 17 − 18 4.5 / 48 = −1.54 Degrees of freedom = n – = 47 Because t < 0, p-value is two times the lower tail area Using t table: area in lower tail is between 05 and 10; therefore, p-value is between 10 and 20 Exact p-value corresponding to t = -1.54 is 1303 c p-value > 05, not reject H0 d With df = 47, t.025 = 2.012 Reject H0 if t ≤ -2.012 or t ≥ 2.012 t = -1.54; not reject H0 25 a t= x − µ0 s/ n = 44 − 45 5.2 / 36 = −1.15 Degrees of freedom = n – = 35 Lower tail p-value is the area to the left of the test statistic Using t table: p-value is between 10 and 20 Exact p-value corresponding to t = -1.15 is 1290 p-value > 01, not reject H0 b t= x − µ0 s/ n = 43 − 45 4.6 / 36 = −2.61 9-8 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Hypothesis Testing Lower tail p-value is the area to the left of the test statistic Using t table: p-value is between 005 and 01 Exact p-value corresponding to t = -2.61 is 0066 p-value ≤ 01, reject H0 c t= x − µ0 s/ n = 46 − 45 / 36 = 1.20 Lower tail p-value is the area to the left of the test statistic Using t table: p-value is between 80 and 90 Exact p-value corresponding to t = 1.20 is 8809 p-value > 01, not reject H0 26 a t= x − µ0 s/ n = 103 − 100 11.5 / 65 = 2.10 Degrees of freedom = n – = 64 Because t > 0, p-value is two times the upper tail area Using t table; area in upper tail is between 01 and 025; therefore, p-value is between 02 and 05 Exact p-value corresponding to t = 2.10 is 0397 p-value ≤ 05, reject H0 b t= x − µ0 s/ n = 96.5 − 100 11/ 65 = −2.57 Because t < 0, p-value is two times the lower tail area Using t table: area in lower tail is between 005 and 01; therefore, p-value is between 01 and 02 Exact p-value corresponding to t = -2.57 is 0125 p-value ≤ 05, reject H0 c t= x − µ0 s/ n = 102 − 100 10.5 / 65 = 1.54 Because t > 0, p-value is two times the upper tail area Using t table: area in upper tail is between 05 and 10; therefore, p-value is between 10 and 20 Exact p-value corresponding to t = 1.54 is 1285 p-value > 05, not reject H0 27 a H0: 238 9-9 â 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter Ha: µ < 238 b t= x − µ0 s/ n = 231 − 238 80 / 100 = −.88 Degrees of freedom = n – = 99 Lower tail p-value is the area to the left of the test statistic Using t table: p-value is between 10 and 20 Exact p-value corresponding to t = -.88 is 1905 c p-value > 05; not reject H0 Cannot conclude mean weekly benefit in Virginia is less than the national mean d df = 99 t.05 = -1.66 Reject H0 if t ≤ -1.66 -.88 > -1.66; not reject H0 28 a H0: µ ≥ Ha: µ < b t= x − µ0 s/ n = 7.27 − 6.38 / 85 = −2.50 Degrees of freedom = n – = 84 Lower tail p-value is P(t ≤ -2.50) Using t table: p-value is between 005 and 01 Exact p-value corresponding to t = -2.50 is 0072 c 29 a p-value ≤ 01; reject H0 The mean tenure of a CEO is significantly lower than years The claim of the shareholders group is not valid H0: µ = 5600 Ha: µ ≠ 5600 b t= x − µ0 s/ n = 5835 − 5600 520 / 25 = 2.26 Degrees of freedom = n – = 24 Because t < 0, p-value is two times the upper tail area Using t table: area in lower tail is between 01 and 025; therefore, p-value is between 02 and 05 - 10 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter β = 0853 c Power = - β at µ = 26, Power = - 0853 = 9147 When µ = 26, there is a 9147 probability that the test will correctly reject the null hypothesis that µ = 28 51 a b Accepting H0 and letting the process continue to run when actually over - filling or under - filling exists Decision Rule: Reject H0 if z ≤ -1.96 or if z ≥ 1.96 indicates Accept H0 if 15.71 < x < 16.29 Reject H0 if x ≤ 15.71 or if x ≥ 16.29 For µ = 16.5 z= 16.29 − 16.5 / 30 = −1.44 β = 0749 c β x c Power = - 0749 = 9251 d The power curve shows the probability of rejecting H0 for various possible values of µ In particular, it shows the probability of stopping and adjusting the machine under a variety of underfilling and overfilling situations The general shape of the power curve for this case is - 22 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Hypothesis Testing 1.00 75 50 Power 25 00 15.6 15.8 16.0 16.2 16.4 Possible Values of u c = µ0 + z.01 52 σ = 15 + 2.33 = 16.32 n 50 At µ = 17 z = 16.32 − 17 / 50 = −1.20 β = 1151 At µ = 18 z = 16.32 − 18 / 50 = −2.97 β = 0015 Increasing the sample size reduces the probability of making a Type II error 53 a b Accept µ ≤ 100 when it is false Critical value for test: c = µ0 + z.05 σ 75 = 100 + 1.645 = 119.51 n 40 At µ = 120 z= 119.51 − 120 75 / 40 = −.04 β = 4840 c At µ = 130 z = 119.51 − 130 75 / 40 = −.88 β = 1894 d Critical value for test: - 23 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter c = µ0 + z.05 σ 75 = 100 + 1.645 = 113.79 n 80 At µ = 120 z = 113.79 − 120 75 / 80 = −.74 β = 2296 At µ = 130 z = 113.79 − 130 75 / 80 = −1.93 β = 0268 Increasing the sample size from 40 to 80 reduces the probability of making a Type II error ( zα + zβ ) σ 54 n= 55 n= 56 At µ0 = 3, (µ0 − µa ) ( zα + z β ) σ (µ0 − µa )2 = (1.645 + 1.28)2 (5)2 = 214 (10 − 9) = (1.96 + 1.645) (10) = 325 (20 − 22) α = 01 z.01 = 2.33 At µa = 2.9375, β = 10 z.10 = 1.28 σ = 18 n= 57 ( zα + z β ) σ (µ0 − µa )2 = (2.33 + 1.28) (.18) = 108.09 Use 109 (3 − 2.9375) At µ0 = 400, α = 02 z.02 = 2.05 At µa = 385, β = 10 z.10 = 1.28 σ = 30 n= 58 ( zα + z β ) σ (µ0 − µa )2 = (2.05 + 1.28)2 (30) = 44.4 Use 45 (400 − 385) At µ0 = 28, α = 05 Note however for this two - tailed test, zα / = z.025 = 1.96 At µa = 29, β = 15 z.15 = 1.04 σ =6 n= 59 ( zα / + z β ) σ ( µ0 − µ a ) At µ0 = 25, = (1.96 + 1.04) (6) = 324 (28 − 29) α = 02 z.02 = 2.05 - 24 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Hypothesis Testing At µa = 24, β = 20 z.20 = 84 σ =3 n= 60 a ( zα + z β ) σ (µ0 − µa ) = (2.05 + 84) (3) = 75.2 Use 76 (25 − 24) H0: µ = 16 Ha: µ ≠ 16 b z= x − µ0 σ/ n = 16.32 − 16 / 30 = 2.19 Because z > 0, p-value is two times the upper tail area Using normal table with z = 2.19: p-value = 2(.0143) = 0286 p-value ≤ 05; reject H0 Readjust production line c z= x − µ0 σ/ n = 15.82 − 16 / 30 = −1.23 Because z < 0, p-value is two times the lower tail area Using normal table with z = -1.23: p-value = 2(.1093) = 2186 p-value > 05; not reject H0 Continue the production line d Reject H0 if z ≤ -1.96 or z ≥ 1.96 For x = 16.32, z = 2.19; reject H0 For x = 15.82, z = -1.23; not reject H0 Yes, same conclusion 61 a H0: µ = 900 Ha: µ ≠ 900 b x ± z.025 σ n 935 ± 1.96 935 ± 25 c 180 200 (910 to 960) Reject H0 because µ = 900 is not in the interval - 25 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter d z= x − µ0 σ/ n = 935 − 900 180 / 200 = 2.75 Because z > 0, p-value is two times the upper tail area Using normal table with z = 2.75: p-value = 2(.0030) = 0060 62 a H0: µ ≤ 119,155 Ha: µ > 119,155 b z= x − µ0 σ/ n = 126,100 − 119,155 20, 700 / 60 = 2.60 Upper tail p-value is the area to the right of the test statistic Using normal table with z = 2.60: p-value = 1.0000 - 9953 = 0047 c 63 p-value ≤ 01, reject H0 We can conclude that the mean annual household income for theater goers in the San Francisco Bay area is higher than the mean for all Playbill readers The hypothesis test that will allow us to conclude that the consensus estimate has increased is given below H0: µ ≤ 250,000 Ha: µ > 250,000 t= x − µ0 s/ n = 266, 000 − 250, 000 24, 000 / 20 = 2.981 Degrees of freedom = n – = 19 Upper tail p-value is the area to the right of the test statistic Using t table: p-value is less than 005 Exact p-value corresponding to t = 2.981 is 0038 p-value ≤ 01; reject H0 The consensus estimate has increased 64 H0: µ = 25 Ha: µ ≠ 25 t= x − µ0 s/ n = 24.0476 − 25.0 5.8849 / 42 = −1.05 Degrees of freedom = n – = 41 Because t < 0, p-value is two times the lower tail area Using t table: area in lower tail is between 10 and 20; therefore, p-value is between - 26 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Hypothesis Testing 20 and 40 Exact p-value corresponding to t = -1.05 is 2999 Because p-value > α = 05, not reject H0 There is no evidence to conclude that the mean age at which women had their first child has changed 65 a H0: µ ≥ 6883 Ha: µ < 6883 b t= x − µ0 s/ n = 6130 − 6883 2518 / 40 = −1.89 Degrees of freedom = n – = 39 Lower tail p-value is the area to the left of the test statistic Using t table: p-value is between 05 and 025 Exact p-value corresponding to t = -1.89 is 0331 c We should conclude that Medicare spending per enrollee in Indianapolis is less than the national average d Using the critical value approach we would: Reject H0 if t ≤ −t.05 = -1.685 Since t = -1.89 ≤ -1.685, we reject H0 66 H0: µ ≤ 125,000 Ha: µ > 125,000 t= x − µ0 s/ n = 130,000 − 125, 000 12,500 / 32 = 2.26 Degrees of freedom = 32 – = 31 Upper tail p-value is the area to the right of the test statistic Using t table: p-value is between 01 and 025 Exact p-value corresponding to t = 2.26 is 0155 p-value ≤ 05; reject H0 Conclude that the mean cost is greater than $125,000 per lot 67 H0: µ = 2.357 Ha: µ ≠ 2.357 - 27 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter x= s= t= Σxi = 2.3496 n Σ ( xi − x ) = 0444 n −1 x − µ0 s/ n = 2.3496 − 2.3570 0444 / 50 = −1.18 Degrees of freedom = 50 - = 49 Because t < 0, p-value is two times the lower tail area Using t table: area in lower tail is between 10 and 20; therefore, p-value is between 20 and 40 Exact p-value corresponding to t = -1.18 is 2437 p-value > 05; not reject H0 There is not a statistically significant difference between the National mean price per gallon and the mean price per gallon in the Lower Atlantic states 68 a H0: p ≤ 50 Ha: p > 50 b c p= z= 64 = 64 100 p − p0 p0 (1 − p0 ) n = 64 − 50 50(1 − 50) 100 = 2.80 Upper tail p-value is the area to the right of the test statistic Using normal table with z = 2.80: p-value = 1.0000 - 9974 = 0026 p-value ≤ 01; reject H0 College graduates have a greater stop-smoking success rate 69 a H0: p = 6667 Ha: p ≠ 6667 b c p= z= 355 = 6502 546 p − p0 p0 (1 − p0 ) n = 6502 − 6667 6667(1 − 6667) 546 = −.82 Because z < 0, p-value is two times the lower tail area - 28 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Hypothesis Testing Using normal table with z = -.82: p-value = 2(.2061) = 4122 p-value > 05; not reject H0; Cannot conclude that the population proportion differs from 2/3 70 a H0: p ≤ 80 Ha: p > 80 b c p= z= 252 = 84 (84%) 300 p − p0 p0 (1 − p0 ) n = 84 − 80 80(1 − 80) 300 = 1.73 Upper tail p-value is the area to the right of the test statistic Using normal table with z = 1.73: p-value = 1.0000 - 9582 = 0418 d 71 a b p-value ≤ 05; reject H0 Conclude that more than 80% of the customers are satisfied with the service provided by the home agents Regional Airways should consider implementing the home agent system p= 503 = 553 910 H0: p ≤ 50 Ha: p > 50 c z= p − p0 p0 (1 − p0 ) n = 553 − 500 (.5)(.5) 910 = 3.19 Upper tail p-value is the area to the right of the test statistic Using normal table with z = 3.19: p-value ≈ You can tell the manager that the observed level of significance is very close to zero and that this means the results are highly significant Any reasonable person would reject the null hypotheses and conclude that the proportion of adults who are optimistic about the national outlook is greater than 50 72 H0: p ≥ 90 Ha: p < 90 p= 49 = 8448 58 - 29 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter z= p − p0 p0 (1 − p0 ) n = 8448 − 90 90(1 − 90) 58 = −1.40 Lower tail p-value is the area to the left of the test statistic Using normal table with z = -1.40: p-value =.0808 p-value > 05; not reject H0 Claim of at least 90% cannot be rejected 73 a H0: p ≥ 24 Ha: p < 24 b c p= z= 81 = 2025 400 p − p0 p0 (1 − p0 ) n = 2025 − 24 24(1 − 24) 400 = −1.76 Lower tail p-value is the area to the left of the test statistic Using normal table with z = -1.76: p-value =.0392 p-value ≤ 05; reject H0 The proportion of workers not required to contribute to their company sponsored health care plan has declined There seems to be a trend toward companies requiring employees to share the cost of health care benefits 74 a H0: µ ≤ 72 Ha: µ > 72 Reject H0 if z ≥ 1.645 z= x − µ0 σ/ n = x − 72 20 / 30 = 1.645 Solve for x = 78 Decision Rule: Accept H0 if x < 78 Reject H0 if x ≥ 78 b For µ = 80 - 30 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Hypothesis Testing z= 78 − 80 20 / 30 = −.55 β = 2912 c For µ = 75, z= 78 − 75 20 / 30 = 82 β = 7939 d For µ = 70, H0 is true In this case the Type II error cannot be made e Power = - β 1.0 P o w e r 72 75 76 78 80 74 Possible Values of µ Ho False 82 84 H0: µ ≥ 15,000 Ha: µ < 15,000 At µ0 = 15,000, α = 02 z.02 = 2.05 At µa = 14,000, β = 05 z.10 = 1.645 n= 76 ( zα + z β ) σ (µ0 − µa ) = (2.05 + 1.645)2 (4, 000) = 218.5 Use 219 (15, 000 − 14,000) H0: µ = 120 Ha: µ ≠ 120 At µ0 = 120, α = 05 With a two - tailed test, zα / = z.025 = 1.96 At µa = 117, β = 02 z.02 = 2.05 - 31 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part Chapter n= b ( zα / + z β ) σ ( µ0 − µ a ) = (1.96 + 2.05) (5) = 44.7 Use 45 (120 − 117)2 Example calculation for µ = 118 Reject H0 if z ≤ -1.96 or if z ≥ 1.96 z= x − µ0 x − 120 = σ / n / 45 Solve for x At z = -1.96, x = 118.54 At z = +1.96, x = 121.46 Decision Rule: Accept H0 if 118.54 < x < 121.46 Reject H0 if x ≤ 118.54 or if x ≥ 121.46 For µ = 118, z= 118.54 − 118 / 45 = 72 β = 2358 Other Results: If µ is 117 118 119 121 122 123 z 2.07 72 -.62 +.62 +.72 -2.07 - 32 β 0192 2358 7291 7291 2358 0192 © 2010 Cengage Learning All Rights Reserved May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part ... continue c Conclude µ ≠ 32 and that overfilling or underfilling exists Shut down and adjust the production line a H0: µ ≥ 220 Ha: µ < 220 Research hypothesis Research hypothesis to see if mean... in part Chapter z= x − µ0 σ/ n = x − 15 / 35 = 2.33 Solve for x = 16.58 Decision Rule: Accept H0 if x < 16.58 Reject H0 if x ≥ 16.58 For µ = 17, z= 16.58 − 17 = −.62 / 35 β = 2676 c For µ = 18,... publicly accessible website, in whole or in part Hypothesis Testing z= 78 − 80 20 / 30 = −.55 β = 2912 c For µ = 75, z= 78 − 75 20 / 30 = 82 β = 7939 d For µ = 70, H0 is true In this case the Type