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CHAPTER 8 INTRODUCTION TO HYPOTHESIS TESTING 8.1. a. H o : μ ≤ 20 H A : μ > 20 b. H o : μ = 50 H A : μ ≠ 50 c. H o : μ ≥ 35 H A : μ < 35 d. H o : μ ≤ 87 H A : μ > 87 e Ho : μ ≤ 6 HA :μ > 6 8.3. a. If x > 205.2344 reject Ho If x < 205.2344 do not reject Ho x α = 200 + 1.645(45/ 200 ); x α = 205.2344 If z > 1.645 reject Ho If z < 1.645 do not reject Ho b. z = (204.50 – 200)/(45/ 200 ) = 1.41; Since 1.41 < 1.645 do not reject Ho Since 204.5 < 205.2344 do not reject Ho c. The alternative hypothesis. The burden of proof is always to on the alternative hypothesis. 8.4. a. If z > 1.88 reject Ho If z < 1.88 do not reject Ho If p-value < 0.03 reject Ho b. z = (24.85 – 24.78)/(9/ 50 ) = 0.0550; Since 0.0550 < 1.88 do not reject Ho p-value = 0.4781 Instructor’s Solutions Manual 265 8.5. a. Even though the population standard deviation is unknown, since n = 100 is large, we can use the standard normal distribution to obtain the critical value. x α = 4,000 - 1.645(205/ 100 ); x α = 3966.2775 If x < 3966.2775 reject Ho If x > 3966.2775 do not reject Ho If p-value < .05, reject Ho If p-value > .05, do not reject Ho b. Since 3980 > 3966.2775 do not reject Ho p-value = P(z < -.9756) = .50 - .3365 = .1635 Since p-value = .1635 > 0.05 do not reject Ho c. The two research hypotheses that could have produced the null and alternative hypotheses are: The population mean is less than 4,000. The population mean is at least 4,000. 8.8. a. If: z > 1.96 reject Ho If: z < -1.96 reject Ho Otherwise, do not reject Ho b. z = (1,338 – 1,346)/(90/ 64 ) = -0.7111 Since z = -0.7111 > -1.96, do not reject the null hypothesis. c. Since the null hypothesis is not rejected, a Type II error may have been committed. 8.14. a. Ho: μ < 240 seconds Ha: μ > 240 seconds b. Students can use Excel’s AVERAGE and STDEV functions to determine the sample mean and standard deviation or compute them manually. x = 219.6667 s = 57.5884 t = (219.6667 – 240)/(57.5884/ 12 ) = -1.2231 The critical t value for a one tailed test with alpha = .10 and 11 degrees of freedom is 1.3634. Since t = -1.2231 < 1.3634, do not reject the null hypothesis. There is not enough evidence to conclude that the average time exceeds 4 minutes (240 seconds). 266 Chapter 8: Introduction to Hypothesis Testing Also could use Excel’s TDIST function to determine the p-value = 0.8766. Since p-value = 0.8766 > 0.1, do not reject Ho c. x α = 240 + 1.3634(57.5884/ 12 ); x α = 262.6656 Since x = 219.6667 < 262.6656, do not reject the null hypothesis. 8.16. a. Ho: μ = 24 ounces Ha: μ ≠ 24 ounces b. t = (24.32 – 24)/(0.7/ 16 ) = 1.83 t.05/2 = + 2.1315 Since –2.1315 < 1.83 < 2.1315 do not reject Ho and conclude that the filling machine remains all right to operate. c. Because the production control manager does not want the boxes under-filled or overfilled. d. Using Excel’s TDIST function, p-value = 0.0872 > 0.025; therefore do not reject Ho e. Since the null hypothesis was “accepted”, a Type II error may have been committed.