Question

4. To compare the means of 5 populations, random samples of size 10 are selected from...

4. To compare the means of 5 populations, random samples of size 10 are selected from each population. A partial ANOVA table for this data is as follows: Source df SS MS F Treatment 15.32 Error 0.64 Total a. Fill in the missing entries in the ANOVA table. b. Compute the pooled estimate of the population standard deviation. c. State the null and alternative hypotheses in the experiment. d. Test the hypotheses in part c). Use α=0.05.

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
In order to compare the means of populations, independent random samples of 390 observations are selected...
In order to compare the means of populations, independent random samples of 390 observations are selected from each population, with the results found in the table below x1=5283, x2=5242, s1=143, s2=194 Use a​ 95% confidence interval to estimate the difference between the population means (u1-u2) a) find the confidence interval b) Test the null hypothesis Ho: (u1-u2)=0 vs the alternative hypothesis Ha: (u1-u2)=/=0. Use a=0.05. What is the test statistic? What is the p-value? c) Test the null hypothesis Ho:...
Respond to each of the following questions using this partially completed one-way ANOVA table. Source SS...
Respond to each of the following questions using this partially completed one-way ANOVA table. Source SS DF MS F Between 470 Within 40 Total 1264 44 (a). How many different populations are being compared? (b). Fill in the ANOVA table with the missing (c). State the appropriate null and the alternative (d). Based on the analysis of variance F-test, what conclusion should be reached regarding the null hypothesis? Test Using an α =0.05.
Five population means are being compared. The combined sample size from all the treatment groups is...
Five population means are being compared. The combined sample size from all the treatment groups is 200. Complete the ANOVA table and report the value of the F-test statistic. Source df SS MS F Treatment Error 12 nothing goes here Total 2940 nothing goes here nothing goes here Group of answer choices 23.8 2.11 11.6 12.5
ANOVA summary table. A partially completed ANOVA tale for a competely randomized design is shown here....
ANOVA summary table. A partially completed ANOVA tale for a competely randomized design is shown here. Source df SS MS F Treatments Error 4 - 24.7 Total 34 62.4 (a) Complete the ANOVA table. (b) How many treatments are involved in the experiment? (c) Do the data provide sufficient evidence to indicate a difference among the treatment means? Test using α = .10.
1) A student randomly and independently samples and assigns 9 participants to one of 5 conditions....
1) A student randomly and independently samples and assigns 9 participants to one of 5 conditions. She gives each group a treatment she believes will affect performance on the dependent measure. Higher scores indicate better performance. The ANOVA summary table is from the data that she has collected. Find the dfbetween. If needed, please round to two decimal places. Source SS df MS F Between 78 Within 45 2) A student randomly and independently samples and assigns 9 participants to...
5. Measuring effect size for the repeated-measures ANOVA Alicia F. Lieberman and Patricia Van Horn have...
5. Measuring effect size for the repeated-measures ANOVA Alicia F. Lieberman and Patricia Van Horn have created a psychotherapy model for young children who have witnessed family violence. The therapy focuses on building the parent’s capacity to nurture and protect the child, thereby promoting the child’s emotional health and repairing the parent-child relationship, which has been disrupted by the stress and trauma of family violence. As a clinical psychology intern, you are learning parent-child therapy with Drs. Lieberman and Van...
A researcher used a one-factor ANOVA for independent samples to test the effectiveness of four teaching...
A researcher used a one-factor ANOVA for independent samples to test the effectiveness of four teaching methods for autistic children. The experiment was conducted with four samples of n = 12 autistic children in each group. The results of the analysis are shown in the following summary table. Source SS df MS Between Treatments ____ ____ ____ F = 5.00 Within Treatments 88 ____ ____ Total ____ ____ A. Fill in all missing values in the table. Show your work...
Consider the partial ANOVA table shown below. Let a = .01 Source of Variation DF SS...
Consider the partial ANOVA table shown below. Let a = .01 Source of Variation DF SS MS F Between Treatments 3 180 Within Treatments (Error) Total 19 380 If all the samples have five observations each: there are 10 possible pairs of sample means. the only appropriate comparison test is the Tukey-Kramer. all of the absolute differences will likely exceed their corresponding critical values. there is no need for a comparison test – the null hypothesis is not rejected. 2...
While conducting a one-way ANOVA comparing 4 treatment samples with 5 observations per treatment sample, computed...
While conducting a one-way ANOVA comparing 4 treatment samples with 5 observations per treatment sample, computed value for SS(Total) = 66 and MS(Error) = 3 given. Source of            SS           d.f.              MS                  Test Statistic               Variation                                                                                                                                                      Treatment        .….           ….              …..                      …… Error                  48            ….                  3 Total                 66 What is the value of F-test statistic using ANOVA table? 6 2 3 5
6. (36 pts) The following is an incomplete ANOVA summary table: Source df SS MS F...
6. (36 pts) The following is an incomplete ANOVA summary table: Source df SS MS F Among Groups 3 63 Within Groups 16 97 Total (a) Complete the ANOVA summary table. (b) Determine the number of groups. (b) At the α = 0.05 level of significance, determine whether there is evidence of difference in the population means.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT