Statistics for Engineers and Scientists
4th Edition
ISBN: 9780073401331
Author: William Navidi Prof.
Publisher: McGraw-Hill Education
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Chapter 9.2, Problem 15E
To determine
Find the value of MSE so that the F statistic rejects the null hypothesis of no difference at the 5% level, while the Tukey–Kramer method does not find any pair of means to differ at the 5% level.
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Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 36 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 12 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 12,400; SSTR = 4,550.
Chapter 9 Solutions
Statistics for Engineers and Scientists
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- Three different methods for assembling a product were proposed by an industrial engineer.To investigate the number of units assembled correctly with each method, 30 employeeswere randomly selected and randomly assigned to the three proposed methods in such away that each method was used by 10 workers. The number of units assembled correctlywas recorded, and the analysis of variance procedure was applied to the resulting data set.The following results were obtained: SST = 10,800; SSTR = 4560.a. Set up the ANOVA table for this problem.b. Use a = .05 to test for any significant difference in the means for the three assemblymethods.arrow_forwardThree different methods for assembling a product were proposed by an industrial engineer.To investigate the number of units assembled correctly with each method, 30 employeeswere randomly selected and randomly assigned to the three proposed methods in such away that each method was used by 10 workers. The number of units assembled correctlywas recorded, and the analysis of variance procedure was applied to the resulting data set.The following results were obtained: SST = 10,800; SSTR = 4560.a. Set up the ANOVA table for this problemarrow_forwardA nutrition store in the mall is selling "memory booster" which is a concoction of herbs and minerals that is intended to improve memory performance. To test the effectiveness of the herbal mix, a researcher obtains a sample of n=16 participants and has each person take the suggested dosage each day for 4 weeks. At the end of the four week period, each individual takes a standard memory test. The scores from the participants produced a M=26 with a sample variance of s2=64. In the general population, the standardized test is known to have a μμ=20. Do the sample data support the conclusion that memory booster has a significant effect? Use a two-tailed test with αα=.05. Compute Cohen's d to measure the size of the treatment effect.arrow_forward
- Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 10,720; SSTR = 4,510. Set up the ANOVA table for this problem (to 2 decimals, if necessary). Round p-value to four decimal places. Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments Error Total Use = .05 to test for any significant difference in the means for the three assembly methods.The p-value is Selectless than .01between .01 and .025between .025 and .05between .05 and .10greater than .10Item 11 What is your…arrow_forwardA nutrition store in the mall is selling “Memory Booster” which is a concoction of herbs and minerals that is intended to improve memory performance. To test the effectiveness of the herbal mix, a researcher obtains a sample of n = 16 people and has each person take the suggested dosage every day for 4 weeks. At the end of the four week period, each individual takes a standardized memory test. The scores from the sample produced a mean ??� = 24 with a sample variance s2 = 64. In the general population, the standardized test is known to have a mean μ =20. Do the sample data support the conclusion that the Memory Booster has a significant effect? Test with α = .05. is t value of 2 and a rejection zone t value of 2.131 correct?arrow_forwardSnow avalanches can be a real problem for travelers in the western United States and Canada. A very common type of avalanche is called the slab avalanche. These have been studied extensively by David McClung, a professor of civil engineering at the University of British Columbia. Suppose slab avalanches studied in a region of Canada had an average thickness of u = 67 cm. The ski patrol at Vail, Colorado, is studying slab avalanches in its region. A random sample of avalanches in spring gave the following thicknesses (in cm). 59 51 76 38 65 54 49 62 68 55 64 67 63 74 65 79 (i) Use a calculator with sample mean and standard deviation keys to find x and s. (Round your answers to two decimal places.) X= cm S= cm (ii) Assume the slab thickness has an approximately normal distribution. Use a 1% level of significance to test the claim that the mean slab thickness in the Vail region is different from that in the region of Canada. (a) What is the level of significance? State the null and…arrow_forward
- Three different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST=10,800; SSTR =4550. a. Set up the ANOVA table for this problem (to 2 decimals but p-value to 4 decimals, if necessary). Source of Variation Sum of Squares Degrees of Freedom Mean Square Treatments Error Total 1000 000 The p-value is - Select your answer - + What is your conclusion? - Select your answer - F b.Use a = 0.05 to test for any significant difference in the means for the three assembly methods. Calculate the value of the test statistic (to 2 decimals). + p-value Check My Work (5 remaining)arrow_forwardThree different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST = 10,950; SSTR = 4,570. (a) Set up the ANOVA table for this problem. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.) Source Sum Degrees of Freedom Mean F p-value of Variation of Squares Square Treatments Error Total (b) Use a = 0.05 to test for any significant difference in the means for the three assembly methods. State the null and alternative hypotheses. Ho: H1 = H2 = H3 O Ho: H1 = H2 = H3 : Not all the population means are equal. Not all the population means are…arrow_forwardThree different methods for assembling a product were proposed by an industrial engineer. To investigate the number of units assembled correctly with each method, 30 employees were randomly selected and randomly assigned to the three proposed methods in such a way that each method was used by 10 workers. The number of units assembled correctly was recorded, and the analysis of variance procedure was applied to the resulting data set. The following results were obtained: SST =10,870; SSTR =4580. a. Set up the ANOVA table for this problem (to 2 decimals but p-value to 4 decimals, if necessary). Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 4580 Error 6290 Total 10870 b.Use a = 0.05 to test for any significant difference in the means for the three assembly methods. Calculate the value of the test statistic (to 2 decimals). The p-value is - Select your answer - What is your conclusion? Select your answer -arrow_forward
- An airport official wants to prove that the proportion of delayed flights for Airline A (denoted as p1) is less than the proportion of delayed flights for Airline B (denoted as p2). Random samples for both airlines after a storm showed that 51 out of 200 flights for Airline A were delayed, while 60 out of 200 of Airline B's flights were delayed. The test statistic for this problem is -1.00. The p-value for the test statistic for this problem is: p = 0.0228 p = 0.0668 p = 0.3413 p = 0.1587arrow_forwardA nutrition store in the mall is selling "memory booster" which is a concoction of herbs and minerals that is intended to improve memory performance. To test the effectiveness of the herbal mix, a researcher obtains a sample of n=16 participants and has each person take the suggested dosage each day for 4 weeks. At the end of the four week period, each individual takes a standard memory test. The scores from the participants produced a M=26 with a sample variance of s2=64. In the general population, the standardized test is known to have a μμ=20. a. Do the sample data support the conclusion that memory booster has a significant effect? Use a two-tailed test with αα=.05. b. Compute Cohen's d to measure the size of the treatment effect.arrow_forward
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