When would you use the Tukey-Kramer procedure?   Question 1 options:   to test independence of errors   to test for normality   to test for differences in pairs of means   to test for homogeneity of variance

An experiment has been conducted for four treatments with eight blocks. Complete the following analysis of variance table (to 2 decimals but p-value to 4 decimals, if necessary). If answer is zero enter "0". Source Sum Degrees Mean of Variation of Squares of Freedom Square F p-value Treatments 900 Blocks 400 Error Total 1800 Use a = 0.05 to test for any significant differences. The p-value is less than 0.01 What is your conclusion? Conclude not all treatment means are equal

A test statistic for one of the tests in this unit is given, along with information about degrees of freedom. Give the p-value and indicate whether the results are significant at the 5% level. An analysis of variance test with F-statistic 8.71 and df -numerator 3 and df -denominator = 17 Round your answer to four decimal places. p-value

In a completely randomized design, 12 experimental units were used for the first treatment, 15 for the second treatment, and 20 for the third treatment. Complete the following analysis of variance (to 2 decimals, if necessary). If the answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square 8 8 Treatments Error 1000 700 2 24 * Ⓡ 650 The p-value is less than 0.01 What is your conclusion? Cannot conclude that the means of the three treatments are equal v 29.17 Total 1800 44 At a 0.05 level of significance, is there a significant difference between the treatments? F 47.65 * p-value 0 ✔ 4

In a completely randomized design, 12 experimental units were used for the first treatment, 15 for the second treatment, and 20 for the third treatment. Complete the following analysis of variance (to 2 decimals, if necessary). If the answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 1,100         Error           Total 1,500         At a=.05  level of significance, is there a significant difference between the treatments?

Could please help interpret this SPSS output of a Independent sample t-test?

Start A sales manager for a large department store believes that customer spending per visit with a sale is lower than customer spending without a sale, and would like to test that claim. A simple random sample of customer spending is taken from without a sale and with a sale. The results are shown below. Without sale with sale 82.904 71.316 Mean Variance 1984.96 1826.41 Observations 150 Hypothesized Mean Difference 328 PT<=t) one-tail 0.0071 Critical one-tail -1.65 PT<=t) two-tail Critical two-tail -1.967 6 Confidence Level 3 Ex: 1.234 Samples from without sale: n₁ = Ex 9 Samples from with sale: n₂ Point estimate for spending without sale: 1 Point estimate for spending with sale: 2 = 0 J. Ex: 1.234 EN P -7 T P= = t= DE UNDE IS-

When comparing more than two treatment means, why should you use an analysis of variance instead of using multiple t tests?   Question 31 options:   Using multiple t tests increases the risk of a Type II error   The analysis of variance is more likely to detect a treatment effect   Using multiple t tests increases the risk of a Type I error   There is no advantage to using an analysis of variance instead of multiple t tests

In a completely randomized design, 12 experimental units were used for the first treatment, 15 for the second treatment, and 20 for the third treatment. Complete the following analysis of variance (to 2 decimals, if necessary). If the answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Treatments 1,100         Error           Total 1,700         At a .05 level of significance, is there a significant difference between the treatments? The -value is - Select your answer - less than .01 between .01 and .025 between .025 and .05 between .05 and .10 greater than .10 What is your conclusion? - Select your answer - Cannot conclude that the means of the three treatments are equal Conclude that the means of the three treatments are equal