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ADM 2304 Study Guide - Railways Act 1921, Parametric Statistics, John Tukey

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University of Ottawa
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Administration
Course
ADM 2304
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All

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Document Summary

S = 0. 002569 r-sq = 51. 01% r-sq(adj) = 41. 21% Since the p(0. 005) < (0. 05), and f-stat(5. 21)< f-crit(4. 60), we reject the null hypothesis. 0. 0050 l i a u d s e. Also, no outliers seem to exist in the boxplots of pin sizes. Normal probability plot (response is pin size) t n e c r e. E: using tukeys method, the differences in mean performance are as follows: Means that do not share a letter are significantly different. By observing this information, the mean of a and b are different, as well as the mean of b and d. H = 12. 02 df = 4 p = 0. 017. H = 12. 15 df = 4 p = 0. 016 (adjusted for ties) The non parametric test used is the kruskal-wallis test. Since the p-value (0. 017)< (0. 05), we reject the hypothesis and confirm our conclusion from part a).

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Related Questions

Hypothesis testing with ANOVA opinions about whether caffeine enhances test performance differ. You design a study to test the impact of drinks with different caffeine contents on students' test-taking abilities. You choose 21 students at random from your introductory psychology course to participate in your study. You randomly assign each student to one of three drinks, each with a different caffeine concentration, such that there are seven students assigned to each drink. You then give each of them a plain capsule containing the precise quantity of caffeine that would be consumed in their designated drink and have them take an arithmetic test 15 minutes later.

The students receive the following arithmetic test scores:

Ā  Cola Black Tea Coffee Ā 
Caffeine Content (mg/oz) 2.9 5.9 13.4 Ā 
Ā  85 85 92 āˆ‘X2 = 147,641
Ā  86 89 87 G = 1,755
Ā  82 82 80 N = 21
Ā  75 75 89 k = 3
Ā  66 88 96 Ā 
Ā  78 76 83 Ā 
Ā  87 82 92 Ā 
Ā  T1 = 559 T2 = 559 T3 = 559 Ā 
Ā  SS1 = 338.86 SS2 = 338.86 SS3 = 338.86 Ā 
Ā  n1 = 7 n2 = 7 n3 = 7 Ā 
Ā  M1 = 79.8571 M2 = 79.8571 M3 = 79.8571 Ā 

1.) You plan to use an ANOVA to test the impact of drinks with different caffeine contents on students' test-taking abilities. What is the null hypothesis?

(a) The population mean test score for the cola population is different from the population mean test score for the black tea population.

(b) The population mean test scores for all three treatments are equal.

(c) The population mean test scores for all three treatments are different.

(d) The population mean test scores for all three treatments are not all equal.

Ā 

2.) Calculate the degrees of freedom and the variances for the following ANOVA table:

Source SS df MS
Between - - -
Within 702.28 - -
Total 973.14 - Ā 

Ā 

The formula for the F-ratio is:

Ā 

Ā 

3.) Using words, the formula of the F-ratio can be written as:

Ā 

Ā 

4.) Using the data from the ANOVA table given, the F-ratio can be written as:

Ā 

5.) Calculate for F-ratio:

Ā 

Ā 

6.)Ā At the level of significance, what is your conclusion?

(a)You can reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance.

(b)You cannot reject the null hypothesis; caffeine does appear to affect test performance.

(c)You cannot reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance.

(d)You can reject the null hypothesis; caffeine does appear to affect test performance.