Loose Leaf for Statistical Techniques in Business and Economics
17th Edition
ISBN: 9781260152647
Author: Douglas A. Lind
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Question
Chapter 16, Problem 20E
To determine
State the conditions that are necessary to use Kruskal–Wallis test instead of Wilcoxon rank-sum test.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
A chi-square independence test is to be conducted to decide whether an association exists between two variables of a populations. One variable has 4 possible values and the other variable has 2. What are the degrees of freedom for the χ2-statistic?
answer:
A single observation of a random variable having a Beta Distribution with a =2 and B
unknown is used to test the nul hypothesis H,: B = Bo against H1: B - B1 (where B1 < Bo).
Use the Neyman-Pearson Lemma to find the most powerful critical region.
If x, x2 ... Xn is a random sample from N (µ, o²) popula-
tion, find súfficient estimator for µ and ơ.
Chapter 16 Solutions
Loose Leaf for Statistical Techniques in Business and Economics
Ch. 16 - Prob. 1SRCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Calorie Watchers has low-calorie breakfasts,...Ch. 16 - Prob. 4ECh. 16 - Prob. 2SRCh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8E
Ch. 16 - Prob. 3SRCh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 4SRCh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 5SRCh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 6SRCh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 7SRCh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29CECh. 16 - Prob. 30CECh. 16 - Prob. 31CECh. 16 - Prob. 32CECh. 16 - Prob. 33CECh. 16 - Prob. 34CECh. 16 - Prob. 35CECh. 16 - Prob. 36CECh. 16 - Prob. 37CECh. 16 - Prob. 38CECh. 16 - Prob. 39CECh. 16 - Professor Bert Forman believes the students who...Ch. 16 - Prob. 41DACh. 16 - Prob. 42DACh. 16 - Prob. 43DACh. 16 - Prob. 1PCh. 16 - The manufacturer of childrens raincoats wants to...Ch. 16 - Prob. 3PCh. 16 - Prob. 4PCh. 16 - B. Thomas Testing Labs John Thomas, the owner of...Ch. 16 - Prob. 1.1PTCh. 16 - Prob. 1.2PTCh. 16 - Prob. 1.3PTCh. 16 - Prob. 1.4PTCh. 16 - Prob. 1.5PTCh. 16 - Prob. 1.6PTCh. 16 - Prob. 1.7PTCh. 16 - Prob. 1.8PTCh. 16 - Prob. 1.9PTCh. 16 - Prob. 1.10PTCh. 16 - Prob. 2.1PTCh. 16 - Prob. 2.2PTCh. 16 - Prob. 2.3PTCh. 16 - Prob. 2.4PT
Knowledge Booster
Similar questions
- A tire store advertises that the average price of a new set of their tires is only $150. One of their recent customers believes their advertised average is too low – that the true mean price for a set of tires exceeds $150. He plans to carry out a hypothesis test at α = 0.05. In order to perform the test, the customer took an SRS of 8 sets of tires recently sold. The mean of these sets of tires was x̅ = $156.90 and the standard deviation was s = $11.80. a)What are the appropriate null and alternative hypotheses for this test? H0: x̅ = 150 vs. Ha: x̅ < 150 H0: μ = 150 vs. Ha: μ > 150 H0: x̅ = 150 vs. Ha: x̅ > 150 H0: μ = 150 vs. Ha: μ < 150 b)Are the conditions of randomness and normality met for this test? How? c)What are the appropriate degrees of freedom for this test? d)What is the value of the t test statistic for this test? e)Suppose the value for the t test statistic is t = 1.79. What is the p-value for a one-sided test?arrow_forwardPose a null and alternate hypothesis for a two sample t-test using variables: Male and Female, Asthma, number of injuries, or flu shot. Can any of the variables also be used to do a Chi square test or a NOVA test?arrow_forwardThe built-in data set, UKgas, is a time series recording the quarterly UK gas consumption from 1960 to 1986. Using R we can convert this data into the vector x by the assignment x30 we can create a confidence interval for u using a normal critical value. If we want the confidence interval to be at the 94% level and we use a normal critical value, then what critical value should we use? i) Calculate a 94% confidence interval(using a normal critical value) for μ. j) How long is the 94% confidence interval just created in part i?arrow_forward
- A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed gold clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim? Let d= (golf score after using the newly designed golf clubs) - (golf score before using the newly designed golf clubs). Use a significance level of a= 0.1 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs. golfer 1 2 3 4 5 6 7 8 Score (old design ) 91 78 80 91 73 84 90 94 Score (new design) 87 81 79 88 77 80 85 89 Step 1 of 5: State the null and alternative hypothesis for…arrow_forwardA golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim? Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs)d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs. Golfer 1 2 3 4 5 6 7 8 Score (old design) 93 86 84 96 89 81 92 94 Score (new design) 91 90 80 92 91 77 89 87 Step 1 of 5:State the null and…arrow_forwardA golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim? Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs) d = (golf score after using the newly designed golf clubs) − (golf score before using the newly designed golf clubs) . Use a significance level of α=0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs. Step 1 of 5:State the null and alternative hypotheses for the test. Step 2 of 5:Find the value of the standard deviation of the paired…arrow_forward
- A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim? Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs)d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.01 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs. Golfer 1 2 3 4 5 6 7 8 Score (old design) 94 89 87 88 86 90 86 94 Score (new design) 87 95 85 83 88 86 84 88 Step 1 of 5 :State the null…arrow_forwardA golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim? Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs.arrow_forwardA golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim? Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs) d = (golf score after using the newly designed golf clubs) − (golf score before using the newly designed golf clubs) . Use a significance level of α=0.05 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs. Golfer 1 2 3 4 5 6 7 8 Score (old design)88 75 93 77 80 79 76 83 Score (new design)81 78 89 74 82 75 72 76…arrow_forward
- A golf club manufacturer claims that golfers can lower their scores by using the manufacturer's newly designed golf clubs. Eight golfers are randomly selected and each is asked to give his or her most recent score. After using the new clubs for one month, the golfers are asked again to give their most recent score. The scores for each golfer are given in the table below. Is there enough evidence to support the manufacturer's claim? Let d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs)d=(golf score after using the newly designed golf clubs)−(golf score before using the newly designed golf clubs). Use a significance level of α=0.01 for the test. Assume that the scores are normally distributed for the population of golfers both before and after using the newly designed clubs. Golfer 1 2 3 4 5 6 7 8 Score (old design) 94 89 87 88 86 90 86 94 Score (new design) 87 95 85 83 88 86 84 88 Step 2 of 5 : Find the value…arrow_forwarda cereal company claims that the mean weight of the cereal in its packets is at least 14oz. express the null hypothesis and the alternatvie hypothesis in symbolic formarrow_forwardT= DF=arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw Hill
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill