Question

you use critical range = 31.369 for all (the sample sizes are equal). Conduct the Tukey-...

you use critical range = 31.369 for all (the sample sizes are equal).

Conduct the Tukey- Kramer procedure with alpha= 0.05 level, q0.95 = 3.38 (from Stundentized range table).

The sample means(average) are given in the table below.

Summary

Groups Count Average Variance
City 1 30 236.2393 334.11
City 2 30 344.0701 2201.818
City 3 30 201.503

2215

What is your conclusion

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1
Groups Count Average Variance SSB SSE
City 1 30 236.2393 334.11 17809.35 9689.19
City 2 30 344.0701 2201.818 208997 63852.722
City 3 30 201.503 2215 104788.3 64235
Grand Mean 260.6041 SSB SSE
331594.7 137776.912
Source SS df MS F p-value
Bet 331594.7 2 165797.3 104.6937 0
Wit 137776.9 87 1583.643
Total 469371.6 89
Q 3.38
MSE 1583.643
Groups Mean Difference Lower Bound Upper Bound Significant Difference
City 1 - 2 -107.831 -132.388 -83.2733 Yes
City 1 - 3 34.7363 10.17877 59.29383 Yes
City 2 - 3 142.5671 118.0096 167.1246 Yes

There is a significant difference between all the groups.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Item Group 1 Group 2 Group 3 1 14 17 17 2 13 16 14 3...
Item Group 1 Group 2 Group 3 1 14 17 17 2 13 16 14 3 12 16 15 4 15 18 16 5 16 14 6 16 a. Conduct a one-way analysis of variance on the data assuming the populations have equal variances and the populations are normally distributed. Use alpha = 0.05. b. If warranted, use the Tukey-Kramer procedure to determine which populations have different means. Use an alpha level of 0.05.
Consider the data in the table collected from four independent populations. The conclusion of a​ one-way...
Consider the data in the table collected from four independent populations. The conclusion of a​ one-way ANOVA test using alphaαequals=0.05 is that the population means are not all the same. Determine which means are different using alphaαequals=0.05 Sample 1 Sample 2 Sample 3 Sample 4 6 13 23 9 7 16 13 8 8 19 19 10 9 22 Click here to view the ANOVA summary table. Find the​ Tukey-Kramer critical range CR for each of these differences. CR​1,2 equals=...
Conduct an independent sample t-test (alpha= .05) in Excel to determine whether there is a significant...
Conduct an independent sample t-test (alpha= .05) in Excel to determine whether there is a significant difference in the age of offenders at prison admissions between White and Black offenders. For this t-test, assume that the variances of the two groups are equal. In your summary, please discuss the null hypothesis, alternative hypothesis, the result of this hypothesis test, and interpret the result. t-Test: Two-Sample Assuming Equal Variances White Black Mean 235.5714286 199.164557 Variance 44025.35714 87049.21616 Observations 21 79 Pooled...
The conclusion of a​ one-way ANOVA procedure for the data shown in the table is to...
The conclusion of a​ one-way ANOVA procedure for the data shown in the table is to reject the null hypothesis that the means are all equal. Determine which means are different using alpha α equals=.05 Sample 1 Sample 2 Sample 3 88 1515 2020 1515 1414 2424 77 1717 2222 99 1313 1919 LOADING... Click the icon to view the ANOVA summary table. LOADING... Click the icon to view a studentized range table for alphaαequals=0.050.05. Let x overbarx1​, x overbarx2​,...
Test the Hypotheses Below Null Hypothesis: Mean Student Debt in 2011 is equal to Mean Student...
Test the Hypotheses Below Null Hypothesis: Mean Student Debt in 2011 is equal to Mean Student Debt in 2007 Alternative Hypothesis: Mean Student Debt in 2011 is not equal to Mean Student Debt in 2007 Alpha Level = 0.05 t-Test: Two-Sample Assuming Equal Variances Variable 1 Variable 2 Mean 3925.76 2876.82 Variance 222129.8188 140278.3547 Observations 50 50 Pooled Variance 181204.0867 Hypothesized Mean Difference 0 df 98 t Stat 12.32073615 P(T<=t) one-tail 6.27467E-22 t Critical one-tail 1.660551217 P(T<=t) two-tail 1.25493E-21 t...
This requires you to show ALL work needed to conduct the two tests of hypotheses below....
This requires you to show ALL work needed to conduct the two tests of hypotheses below. Determine if your tests are one tailed or two-tailed according to your alternative hypothesis. You will have to use a z table as well. 1) A random sample of 20 observations selected from a normal population produced x bar = 72.6 and s^2 (variance) = 19.4 a) Test the null hypothesis: mean = 80 against the alternative hypothesis: mean < 80. Use alpha =...
Test the Hypotheses Below Null Hypothesis: Mean Student Debt in 2011 is equal to Mean Student...
Test the Hypotheses Below Null Hypothesis: Mean Student Debt in 2011 is equal to Mean Student Debt in 2007 Alternative Hypothesis: Mean Student Debt in 2011 is not equal to Mean Student Debt in 2007 Alpha Level = 0.05 Variable 1 Variable 2 Mean 3925.76 2876.82 Variance 222129.8188 140278.3547 Observations 50 50 Pooled Variance 181204.0867 Hypothesized Mean Difference 0 df 98 t Stat 12.32073615 P(T<=t) one-tail 6.27467E-22 t Critical one-tail 1.660551217 P(T<=t) two-tail 1.25493E-21 t Critical two-tail 1.984467455 Run the...
Question #6 Beth’s Ltd manufactures screws which are used to manufacture tables. The table manufacturers, M...
Question #6 Beth’s Ltd manufactures screws which are used to manufacture tables. The table manufacturers, M & M Ltd, require that the screws have a length of 150 mm, with a small tolerance for differences in length. Beth’s Ltd randomly sampled and measured the average length of screws for 25 production runs. Due to the fact that the table manufacturers require a small tolerance for differences in the length of the screws, a 5% (0.05) significance level for evaluation was...
The following is the sample information. Test the hypothesis at the 5% level of significance that...
The following is the sample information. Test the hypothesis at the 5% level of significance that the treatment means are equal. Present your solution by the p value method      Treatment 1 Treatment 2 Treatment 3 9 13 10 7 20 9 11 14 15 9 13 14 12 15 10 Anova: Single Factor SUMMARY Groups Count Sum Average Variance Treatment 1 6 58 9.66666667 3.066667 Treatment 2 4 60 15 11.33333 Treatment 3 5 63 12.6 8.3 ANOVA Source...
You want to compare the means of several different groups to determine if all are equal...
You want to compare the means of several different groups to determine if all are equal or if any are significantly different from the rest. For instance on the Insurance file you may wish to see if there is any significant differences in satisfaction among individuals with different levels of education. You have rearranged the Insurance File to give the following: College degree Graduate degree Some College 5 3 4 3 4 1 5 5 4 3 5 2 3...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT