To determine Analyze the data and to draw conclusions at 5% level of significance. Explanation Calculation: The given information is that it is tested the taste of a soft drink beverage on the basis of type of sweetener( A ), ratio of syrup to water ( B ), carbonation level ( C ), and temperature( D ). Each factor has two levels and hence the design is 2 4 factorial design. The response variable is the total point score from 1 to 10. Factor A is Type of sweetener, Factor B is syrup to water, Factor C is carbonation level and Factor D is temperature. State the hypotheses: Main effect of factor A: Null hypothesis: H 0 : Main effect of factor A is not significant. Alternative hypothesis: H 1 : Main effect of factor A is significant. Main effect of factor B: Null hypothesis: H 0 : Main effect of factor B is not significant. Alternative hypothesis: H 1 : Main effect of factor B is significant. Main effect of factor C: Null hypothesis: H 0 : Main effect of factor C is not significant. Alternative hypothesis: H 1 : Main effect of factor C is significant. Main effect of factor D: Null hypothesis: H 0 : Main effect of factor D is not significant. Alternative hypothesis: H 1 : Main effect of factor D is significant. Interaction effect of Factors A and B: Null hypothesis: H 0 : Interaction effect of Factors A and B is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors A and B is significant. Interaction effect of Factors A and C: Null hypothesis: H 0 : Interaction effect of Factors A and C is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors A and C is significant. Interaction effect of Factors A and D: Null hypothesis: H 0 : Interaction effect of Factors A and D is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors A and D is significant. Interaction effect of Factors B and C: Null hypothesis: H 0 : Interaction effect of Factors B and C is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors B and C is significant. Interaction effect of Factors B and D: Null hypothesis: H 0 : Interaction effect of Factors B and D is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors B and D is significant. Interaction effect of Factor C and D: Null hypothesis: H 0 : Interaction effect of Factors C and D is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors C and D is significant. Interaction effect of Factors A, B, and C: Null hypothesis: H 0 : Interaction effect of Factors A, B and C is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors A, B and C is significant. Interaction effect of Factors A,B, and D: Null hypothesis: H 0 : Interaction effect of Factors A, B and D is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors A, B and D is significant. Interaction effect of Factors A,C, and D: Null hypothesis: H 0 : Interaction effect of Factors A, C and D is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors A, C and D is significant. Interaction effect of Factors B,C, and D: Null hypothesis: H 0 : Interaction effect of Factors B, C and D is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors B, C and D are significant. Interaction effect of Factors A,B,C, and D: Null hypothesis: H 0 : Interaction effect of Factors A,B, C and D is not significant. Alternative hypothesis: H 1 : Interaction effect of Factors A,B, C and D is significant. Step-by-step procedure to find the factorial design table is as follows: Software procedure: Choose Stat > DOE > Factorial > Create Factorial Design . Under Type of Design , choose 2 level factorial design . From Number of factors , choose 4 . Click Designs . Select Full factorial . In Number of center pint per block choose 0 . In Number of replicate for corner points choose 2 Click OK . In Printed results, choose “ Summary table , alias table, deign table, defining relation”. Click OK . Enter the corresponding response in the newly created factorial design worksheet. Step-by-step procedure for finding the ANOVA table is as follows: Choose Stat > DOE > Factorial > Analyze Factorial Design . In Response , enter Responses . Click OK in all the dialog boxes. Output obtained by MINITAB procedure is as follows: For Factor A , the F -test statistic is 241.78 and the P -value is 0. For Factor B , the F -test statistic is 2.21 and the P -value is 0.157. For Factor C , the F -test statistic is 98.94 and the P -value is 0. For Factor D , the F -test statistic is 98.94 and the P -value is 0. For interaction of A and B , the F -test statistic is 0.01 and the P -value is 0.91. For interaction of A and C , the F -test statistic is 0.33 and the P -value is 0.575. For interaction of A and D , the F -test statistic is 69.66 and the P -value is 0. For interaction of B and C , the F -test statistic is 0.05 and the P -value is 0.822. For interaction of B and D , the F -test statistic is 1.31 and the P -value is 0.27. For interaction of C and D , the F -test statistic is 1.31 and the P -value is 0.27. For interaction of A,B and C , the F -test statistic is 0.47 and the P -value is 0.503. For interaction of A,B and D , the F -test statistic is 0.21 and the P -value is 0.654. For interaction of A,C and D , the F -test statistic is 0 and the P -value is 1. For interaction of B,C and D , the F -test statistic is 0.01 and the P -value is 0.91. For interaction of A,B,C and D , the F -test statistic is 2.21 and the P -value is 0.157. Decision: If P-value ≤ α , reject the null hypothesis H 0

Applied Statistics and Probability for Engineers

6th Edition

ISBN: 9781118539712

Author: Douglas C. Montgomery

Publisher: WILEY

See similar textbooks

Concept explainers

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.

Videos

Question

Chapter 14.5, Problem 15E

To determine

Analyze the data and to draw conclusions at 5% level of significance.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 14.5, Problem 14E

Chapter 14.5, Problem 16E

Students have asked these similar questions

Set up a free RStudio account at posit.cloud. At the start of your R session, you should enter the following commands. set.seed(1) e<-rnorm(n = 100,mean = 0,sd = 80) x<-seq(1,100,1) y<-10+15*x + e Follow the examples in the Panopto Videos to use R to obtain the values requested below. (As always, if you round your answers, make sure you do so correctly and keep at least three decimal places.) (a) The smallest value of y = (b) The largest value of y = (c) The standard deviation of y = (d) The mean of y = (e) The median of y = (f) The correlation between x and y = : (g) Using the variable e as data, the p-value for a test of Hoμ = 0 vs. HA μ0 is

Does Chi-squared(alpha) always represent the right tail of the the chi-square distribution and does Chi-squared(1-alpha) always represent the left tail of the the chi-square distribution. Also, for a standard normal distribution would Z(1-alpha) be equivalent to -Zalpha in represneting the left tail of the standard normal distribution curve because they both have the same z-score. Furthermore, is my understanding correct in the sense that if we have a critical value,say Zalpha, would the area to the right of it will be alpha and to the left of it would be 1-alpha

One bulb manufacturer claims an average bulb life of 1,600 hours. It is suspected that the actual average is significantly lower. To verify this, a sample of 49 bulbs is selected and the life of each bulb is measured. A sample mean of 1,500 hours and a standard deviation of 120 hours were obtained from them. Can you be sure, at 5% significance, that the mean life is less than what the manufacturer claims?

Chapter 14 Solutions

Applied Statistics and Probability for Engineers

Ch. 14.3 - 14-1. An article in Industrial Quality Control...Ch. 14.3 - Prob. 2E Ch. 14.3 - Prob. 3E Ch. 14.3 - 14-4. An experiment was conducted to determine...Ch. 14.3 - Prob. 5E Ch. 14.3 - 14-6. Johnson and Leone (Statistics and...Ch. 14.3 - 14-7. An article in the IEEE Transactions on...Ch. 14.3 - Prob. 8E Ch. 14.3 - Prob. 9E Ch. 14.3 - Prob. 10E

Ch. 14.3 - Prob. 11E Ch. 14.4 - Prob. 12E Ch. 14.4 - Prob. 13E Ch. 14.5 - Prob. 14E Ch. 14.5 - 14-15. Four factors are thought to influence the...Ch. 14.5 - Prob. 16E Ch. 14.5 - Prob. 17E Ch. 14.5 - Prob. 18E Ch. 14.5 - 14-19. An experiment was run in a semiconductor...Ch. 14.5 - Prob. 20E Ch. 14.5 - Prob. 21E Ch. 14.5 - 14-22. A 24 factorial design was run in a chemical...Ch. 14.5 - Prob. 23E Ch. 14.5 - Prob. 24E Ch. 14.5 - Prob. 25E Ch. 14.5 - Prob. 26E Ch. 14.5 - Prob. 27E Ch. 14.5 - Prob. 28E Ch. 14.5 - Prob. 29E Ch. 14.5 - Prob. 30E Ch. 14.5 - Prob. 31E Ch. 14.5 - Prob. 32E Ch. 14.5 - 14-33. An article in Analytica Chimica Acta...Ch. 14.5 - Prob. 34E Ch. 14.5 - 14-35. The book Using Designed Experiments to...Ch. 14.6 - 14-36. Consider the data from the first replicate...Ch. 14.6 - Prob. 37E Ch. 14.6 - Prob. 38E Ch. 14.6 - Prob. 39E Ch. 14.6 - 14-40. Consider the data from the first replicate...Ch. 14.6 - Prob. 41E Ch. 14.6 - 14-43. An article in Quality Engineering...Ch. 14.6 - Prob. 44E Ch. 14.6 - Prob. 45E Ch. 14.6 - 14-46. An article in Advanced Semiconductor...Ch. 14.6 - Prob. 47E Ch. 14.6 - Prob. 48E Ch. 14.7 - Prob. 49E Ch. 14.7 - Prob. 50E Ch. 14.7 - Prob. 51E Ch. 14.7 - Prob. 52E Ch. 14.7 - 14-54. An article in Quality Engineering [“A...Ch. 14.7 - Prob. 56E Ch. 14.7 - Prob. 57E Ch. 14.7 - Prob. 58E Ch. 14.7 - 14-59. An article in the Journal of...Ch. 14.7 - Prob. 60E Ch. 14.7 - Prob. 61E Ch. 14.7 - Prob. 62E Ch. 14.8 - Prob. 63E Ch. 14.8 - 14-64. An article in Quality Engineering [“Mean...Ch. 14.8 - 14-65. Consider the first-order model where −1 <...Ch. 14.8 - 14-66. A manufacturer of cutting tools has...Ch. 14.8 - Prob. 67E Ch. 14.8 - 14-68. In their book Empirical Model Building and...Ch. 14.8 - 14-69. Consider the first-order model y = 12...Ch. 14.8 - Prob. 73E Ch. 14.8 - Prob. 74E Ch. 14 - Prob. 75SE Ch. 14 - Prob. 76SE Ch. 14 - Prob. 77SE Ch. 14 - Prob. 78SE Ch. 14 - Prob. 79SE Ch. 14 - Prob. 80SE Ch. 14 - Prob. 81SE Ch. 14 - Prob. 82SE Ch. 14 - Prob. 83SE Ch. 14 - 14-84. Construct a design in 16 runs. What are...Ch. 14 - Prob. 85SE Ch. 14 - Prob. 87SE Ch. 14 - Prob. 88SE Ch. 14 - Prob. 89SE Ch. 14 - 14-90. An article in Biotechnology Progress (2001,...Ch. 14 - 14-93. The rework time required for a machine was...Ch. 14 - Prob. 94SE Ch. 14 - Prob. 95SE Ch. 14 - Prob. 97SE Ch. 14 - Prob. 98SE Ch. 14 - Prob. 99SE Ch. 14 - 14-101. Consider an unreplicated 2k factorial, and...Ch. 14 - Prob. 102SE Ch. 14 - Prob. 103SE Ch. 14 - Prob. 104SE Ch. 14 - Prob. 105SE Ch. 14 - Prob. 106SE Ch. 14 - Prob. 107SE Ch. 14 - 14-108. Set up a 27 − 4 design using D = AB, E =...Ch. 14 - Prob. 109SE

Knowledge Booster

$Statistics$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.