Anova: Single Factor SUMMARY Groups Count Sum Average Variance Control No Exercise 15 1061 70.7333333 107.638095...
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Anova: Single Factor SUMMARY Groups Count Sum Average Variance Control No Exercise 15 1061 70.7333333 107.638095 Group 2 Cardio 15 948 63.2 63.1714286 Group 3 Cardio + Lifting 15 936 62.4 72.5428571 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 634.177778 2 317.088889 3.90900908 0.02774085 3.21994229 Within Groups 3406.93333 42 81.1174603 Total 4041.11111 44 - Based on whether or not you rejected or accepted your null hypothesis, what can you state about the resting heart rates of the students you sampled? Is there a difference between the different group heart rates?
Homework Answers
Solution-A:
Ho:mu1=mu2=mu3
Ha :atleast one of the group means are different
alpha=0.05
F=3.90900908
p=0.02774085
p<0.05
Reject null hypothesis
Accept alternative Hypothesis.
There is suffcient statisitical evidence at 5% level of significance to conclude that there are
differences in mean resting heart rates of the 3 groups
Add Answer to:
Anova: Single Factor
SUMMARY
Groups
Count
Sum
Average
Variance
Control No Exercise
15
1061
70.7333333
107.638095...
Question Completion Status QUESTION 4 Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 4...
Question Completion Status QUESTION 4 Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 4 108 27 32.66666667 B 4 96 24 13.33333333 4 120 30 56 ANOVA Source of Variation SS df F crit MS P-value F Treatments 72 36 1.058823529 0.386396621 4.256494729 Error 306 9 34 Total 378 11 Based on the Results above of Single Factor ANOVA: the MSTR O 36 O 378 O 72 O 34 Click Save and Submit to save and submit. Chick...
Anova: Single Factor SUMMARY Groups Count CAR1 CAR2 CAR3 CAR4 SumAverage Variance 178.9 19.87778 ...
T test statistic
Anova: Single Factor SUMMARY Groups Count CAR1 CAR2 CAR3 CAR4 SumAverage Variance 178.9 19.87778 9.309444 11119.4 10.85455 10.15873 11241.1 21.91818 13.06164 14.49 7.894333 9 10 144.9 ANOVA Source of Varid Ss MS P-value F crit Between ( 815.10793 271.7026 26.61437 2.39E-09 2.8587916 Within Gr 377.7282 37 10.20887 Total 1192.836 40 Conduct a test of hypothesis that the mean mileage of makes 2 and 3 do not differ. Use the method for single means when σ is not...
SUMMARY Groups Count Sum Average Variance 6 195 32.5 8 226 28.25 19.64285714 4 77 19.25...
SUMMARY Groups Count Sum Average Variance 6 195 32.5 8 226 28.25 19.64285714 4 77 19.25 2.916666667 43.5 ANOVA Source of Variation s df MS Between Groups 426.25 2 213.125 8.788659794 Within Groups 363.75 15 24.25 Total 790 17 Our conclusion at a-0.01 should be: O No conclusion can be made from this experiment O The average alertness level is not the same for all three dosages O The average alertness level is the same for all three dosages
Anova: Single Factor SUMMARY Count Av Variance Sum 165.000000000 27.500000000 8.300000000 174.000000000 29.000000000 2.000000000 155.000000000 25.833333333...
Anova: Single Factor SUMMARY Count Av Variance Sum 165.000000000 27.500000000 8.300000000 174.000000000 29.000000000 2.000000000 155.000000000 25.833333333 1.366666667 188.000000000 31.333333333 0.666666667 ANOVA P.value Fcrit Between Groups 98.166666667 3 Within Groups 61.666666667 20 MS 32.722222222 10.612612613 0.000217836 3.098391196 3.083333333 Total 159.833333333 23 TUKEY MULTIPLE COMPARISON TEST Distance 2.838767338 3.96 0.05 Means joined by a double line are not significantly different. 25.833333969 27.500000000 29.000000000 31.333333969 19. Is there a statistically significant difference in the average processing time between the four products? A. Yes,...
What is the ANOVA F test statistic value? A researcher wishes to see if there is...
What is the ANOVA F test statistic value?
A researcher wishes to see if there is a difference in the number of stories in the tall buildings of Chicago, Houston, and New York City. The researcher randomly selects six buildings in each city and records the number of stories in each building. At a = 0.05, can it be concluded that there is a significant difference in the mean number of stories in the tall buildings in each city? Use...
End Of Game Anova: Two-Factor Without Replication SUMMARY Count Sum Average Variance 1 5 11322.6 ...
End Of Game Anova: Two-Factor Without Replication SUMMARY Count Sum Average Variance 1 5 11322.6 22.3 | 23 | 10.51 5 115 3 5 12224.412.8 5 127254 21.8 25.4 132 26.4 32.3 64.3 5 128 5110 25.6 Mon Tue Wed Thurs Fri 7 175 7 168 25 40 24 17 7 14721 3.333333 22 6.666667 7203 29 13.33333 71 154 ANOVA Source of Variation MSF P-value F crit 0.863774 0.535234 2.508189 Rows Columns Error 85.6 271.6 67.9 24 Total 753.634...
Anova: Single Factor SUMMARY Groups Count Average Variance A 30 4.967 5.344 B 30 4.567 3.220...
Anova: Single Factor SUMMARY Groups Count Average Variance A 30 4.967 5.344 B 30 4.567 3.220 C 30 7.267 7.375 D 30 6.500 7.638 E 30 9.400 12.248 Compare Group E to Group A What is the t-value for this test? What is the P-value for this test? What is the confidence interval for mean difference in this test? Keeping in mind that the Margin of Error is the difference from the mean difference to either value above, what is...
I am working on a statistical problem where I have to interpret a single factor ANOVA...
I am working on a statistical problem where I have to interpret
a single factor ANOVA test that has a 95% confidence Interval. The
ANOVA analysis was made up of this data:
This is the ANOVA data analysis generated from the data
above:
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1. What is the null hypothesis for an ANOVA test?
2. Do the 4 box designs produce the same mean compressive
strength?
3. Provide 2 specific statistical comparisons and describe how
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For Questions 11-15, consider the following: The Human Resource Manager of a large insurance company wishes...
For Questions 11-15, consider the following: The Human Resource Manager of a large insurance company wishes to evaluate the leadership qualities of 3 manager groups-supervisors, mid-level managers and upper-level managers. 12 people from each management level were sampled. Is there a difference on the average of the leadership scores for the three groups? Part of the leadership scores (higher is better) and ANOVA table are as follows: Supervisor Mid-Manager Upper-Manager 33 29 32 43 36 39 34 32 40 29...
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Please help with b and c, thanks! A one-way analysis of variance experiment produced the following ANOVA table. (You may find it useful to reference the q table). SUMMARY Groups Count Average Column 1 6 0.89 Column 2 6 1.31 Column 3 6 2.35 Source of Variation SS df MS F p-value Between Groups 8.65 2 4.33 16.65 0.0002 Within Groups 3.83 15 0.26 Total 12.48 17 b. Calculate 99% confidence interval estimates of μ1 − μ2,μ1 − μ3, and...
Question Completion Status QUESTION 4 Anova: Single Factor SUMMARY Groups Count Sum Average Variance A 4 108 27 32.66666667 B 4 96 24 13.33333333 4 120 30 56 ANOVA Source of Variation SS df F crit MS P-value F Treatments 72 36 1.058823529 0.386396621 4.256494729 Error 306 9 34 Total 378 11 Based on the Results above of Single Factor ANOVA: the MSTR O 36 O 378 O 72 O 34 Click Save and Submit to save and submit. Chick...
T test statistic
Anova: Single Factor SUMMARY Groups Count CAR1 CAR2 CAR3 CAR4 SumAverage Variance 178.9 19.87778 9.309444 11119.4 10.85455 10.15873 11241.1 21.91818 13.06164 14.49 7.894333 9 10 144.9 ANOVA Source of Varid Ss MS P-value F crit Between ( 815.10793 271.7026 26.61437 2.39E-09 2.8587916 Within Gr 377.7282 37 10.20887 Total 1192.836 40 Conduct a test of hypothesis that the mean mileage of makes 2 and 3 do not differ. Use the method for single means when σ is not...
SUMMARY Groups Count Sum Average Variance 6 195 32.5 8 226 28.25 19.64285714 4 77 19.25 2.916666667 43.5 ANOVA Source of Variation s df MS Between Groups 426.25 2 213.125 8.788659794 Within Groups 363.75 15 24.25 Total 790 17 Our conclusion at a-0.01 should be: O No conclusion can be made from this experiment O The average alertness level is not the same for all three dosages O The average alertness level is the same for all three dosages
Anova: Single Factor SUMMARY Count Av Variance Sum 165.000000000 27.500000000 8.300000000 174.000000000 29.000000000 2.000000000 155.000000000 25.833333333 1.366666667 188.000000000 31.333333333 0.666666667 ANOVA P.value Fcrit Between Groups 98.166666667 3 Within Groups 61.666666667 20 MS 32.722222222 10.612612613 0.000217836 3.098391196 3.083333333 Total 159.833333333 23 TUKEY MULTIPLE COMPARISON TEST Distance 2.838767338 3.96 0.05 Means joined by a double line are not significantly different. 25.833333969 27.500000000 29.000000000 31.333333969 19. Is there a statistically significant difference in the average processing time between the four products? A. Yes,...
What is the ANOVA F test statistic value?
A researcher wishes to see if there is a difference in the number of stories in the tall buildings of Chicago, Houston, and New York City. The researcher randomly selects six buildings in each city and records the number of stories in each building. At a = 0.05, can it be concluded that there is a significant difference in the mean number of stories in the tall buildings in each city? Use...
End Of Game Anova: Two-Factor Without Replication SUMMARY Count Sum Average Variance 1 5 11322.6 22.3 | 23 | 10.51 5 115 3 5 12224.412.8 5 127254 21.8 25.4 132 26.4 32.3 64.3 5 128 5110 25.6 Mon Tue Wed Thurs Fri 7 175 7 168 25 40 24 17 7 14721 3.333333 22 6.666667 7203 29 13.33333 71 154 ANOVA Source of Variation MSF P-value F crit 0.863774 0.535234 2.508189 Rows Columns Error 85.6 271.6 67.9 24 Total 753.634...
I am working on a statistical problem where I have to interpret
a single factor ANOVA test that has a 95% confidence Interval. The
ANOVA analysis was made up of this data:
This is the ANOVA data analysis generated from the data
above:
Questions:
1. What is the null hypothesis for an ANOVA test?
2. Do the 4 box designs produce the same mean compressive
strength?
3. Provide 2 specific statistical comparisons and describe how
you established your conclusion. Be...
For Questions 11-15, consider the following: The Human Resource Manager of a large insurance company wishes to evaluate the leadership qualities of 3 manager groups-supervisors, mid-level managers and upper-level managers. 12 people from each management level were sampled. Is there a difference on the average of the leadership scores for the three groups? Part of the leadership scores (higher is better) and ANOVA table are as follows: Supervisor Mid-Manager Upper-Manager 33 29 32 43 36 39 34 32 40 29...
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