Explanation Given info: The table shows the weights in ounces of three different types of digital camera. The level of significance is 0.05. Calculation: The hypotheses are given below: Null hypothesis: H 0 : μ 1 = μ 2 = μ 3 Alternative hypothesis: H 1 : At least one mean is different from the others Here, at least one mean is different from the others is tested. Hence, the claim is that, at least one mean is different from the others. The level of significance is 0.05. The number of samples k is 3, the sample sizes n 1 , n 2 and n 3 are 6, 6 and 6. The degrees of freedom are d .f .N = k − 1 and d .f .D = N − k . Where N = n 1 + n 2 + n 3 = 6 + 6 + 6 = 18 Substitute 3 for k in d .f .N d .f .N = k − 1 =3-1 =2 Substitute 18 for N and 3 for k in d .f .D d .f .D = N − k = 18 − 3 = 15 Critical value: The critical F -value is obtained using the Table H: The F -Distribution with the level of significance α = 0.05 . Procedure: Locate 15 in the degrees of freedom, denominator row of the Table H. Obtain the value in the corresponding degrees of freedom, numerator column below 3. That is, the critical value is 3.67. Rejection region: The null hypothesis would be rejected if F > 3.67 . Software procedure: Step-by-step procedure to obtain the test statistic using the MINITAB software: Choose Stat > ANOVA > One-Way . In Response , enter the Sales . In Factor , enter the Factor . Click OK . Output using the MINITAB software is given below: From the MINITAB output, the test value F is 23.94. From the result of part (c), the test value is 23.94. Here, the F -statistic value is greater than the critical value. That is, 23.94 > 3.67 . Thus, it can be concluding that, the null hypothesis is rejected. Hence, the result concludes that, there is enough evidence to support the claim that at least one mean is different from the others. Consider X ¯ 1 , X ¯ 2 and X ¯ 3 represents the means of 2-3 Megapixels, 4-5 Megapixels, 6-8 Megapixels, s 1 2 , s 2 2 and s 3 2 represents the variances of samples of 2-3 Megapixels, 4-5 Megapixels, 6-8 Megapixels. Software procedure: Step-by-step procedure to obtain the mean and standard deviation using the MINITAB software: Choose Stat > Basic Statistics > Display Descriptive Statistics . In Variables enter the columns 2-3 Megapixels, 4-5 Megapixels, 6-8 Megapixels. Choose option statistics , and select Mean, Variance and N total . Click OK . Output using the MINITAB software is given below: The sample sizes n 1 , n 2 and n 3 are 6, 6 and 6. The means are X ¯ 1 , X ¯ 2 and X ¯ 3 are 7.333, 15.17 and 24.5. The sample variances are s 1 2 , s 2 2 and s 3 2 are 2.338, 5.04 and 4.97. Here, the samples of sizes of three states are equal. So, the test used here is Tukey test. Tukey test: Critical value: Here, k is 3 and degrees of freedom v = N − k Substitute 18 for N and 3 for k in v v = N − k = 18 − 3 = 15 The critical F -value is obtained using the Table N: Critical Values for the Tukey test with the level of significance α = 0.05 . Procedure: Locate 15 in the column of v of the Table H. Obtain the value in the corresponding row below 3. That is, the critical value is 3.67. Comparison of the means: The formula for finding s W 2 is, s W 2 = ∑ ( n i − 1 ) s i 2 ∑ ( n i − 1 ) That is, s W 2 = ( 6 − 1 ) 5.467 + ( 6 − 1 ) 25.37 + ( 6 − 1 ) 24.7 ( 6 − 1 ) + ( 6 − 1 ) + ( 6 − 1 ) = 27.335 + 126.85 + 123.5 15 = 277.685 15 = 18.51 Comparison between the means X ¯ 1 and X ¯ 2 : The hypotheses are given below: Null hypothesis: H 0 : There is no significant difference between X ¯ 1 and X ¯ 2 . Alternative hypothesis: H 1 : There is significant difference between X ¯ 1 and X ¯ 2

Elementary Statistics: A Step-by-Step Approach with Formula Card

9th Edition

ISBN: 9780078136337

Author: Allan G. Bluman

Publisher: McGraw-Hill Education

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Question

Chapter 12.2, Problem 10E

To determine

To analyze: The data using one-way ANOVA.

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Chapter 12 Solutions

Elementary Statistics: A Step-by-Step Approach with Formula Card

Ch. 12.1 - Colors That Make You Smarter The following set of...Ch. 12.1 - What test is used to compare three or more means?Ch. 12.1 - State three reasons why multiple t tests cannot be...Ch. 12.1 - What are the assumptions for ANOVA?Ch. 12.1 - Define between-group variance and within-group...Ch. 12.1 - State the hypotheses used in the ANOVA test.Ch. 12.1 - When there is no significant difference among...Ch. 12.1 - Prob. 7E Ch. 12.1 - For Exercises 7 through 20, assume that all...Ch. 12.1 - For Exercises 7 through 20, assume that all...

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