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Problem #4

Problem #4

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Problem #4

Suppose you are interested in studying whether lighting brightness affects spatial reasoning abilities. You decide to test spatial reasoning using completion time scores for the paper-folding test with five people, repeating the test on each person with three different lighting levels (800, 1,000, and 1,200 lux). In this experiment, the null hypothesis is that: O The completion time mean for at least one lighting brightness is different from another o There are no individual differences in the completion time means o There are no differences in the mean completion times among the lighting levels compared The factor of interest is the the The results of the study are presented in the following data table. All scores are times necessary to complete the paper-folding task, recorded in seconds. Participant 800 lux A B с D E Lighting Brightness 1,000 lux 3.42 3.19 3.50 3.75 3.90 T = 17.76 SS=0.3115 ANOVA Table Source Between treatments Within treatments Between subjects Error Total SS 3.40 3.52 3.70 4.12 4.16 ▼ I and the dependent variable is T = 18.90 T = 17.84 SS = 0.4784 SS = 0.1803 df 14 1,200 lux 3.50 3.27 3.54 3.83 3.70 The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population means are equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for the missing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on the order in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between your answers and the answer choices.) MS Participant Totals 0.0809 0.2157 P = 10.32 P = 9.98 P = 10.74 P = 11.70 P = 11.76 n = k = N G ΣΧ2 = F = = 5 3 15 54.5( 199.14
Transcribed Image Text:Suppose you are interested in studying whether lighting brightness affects spatial reasoning abilities. You decide to test spatial reasoning using completion time scores for the paper-folding test with five people, repeating the test on each person with three different lighting levels (800, 1,000, and 1,200 lux). In this experiment, the null hypothesis is that: O The completion time mean for at least one lighting brightness is different from another o There are no individual differences in the completion time means o There are no differences in the mean completion times among the lighting levels compared The factor of interest is the the The results of the study are presented in the following data table. All scores are times necessary to complete the paper-folding task, recorded in seconds. Participant 800 lux A B с D E Lighting Brightness 1,000 lux 3.42 3.19 3.50 3.75 3.90 T = 17.76 SS=0.3115 ANOVA Table Source Between treatments Within treatments Between subjects Error Total SS 3.40 3.52 3.70 4.12 4.16 ▼ I and the dependent variable is T = 18.90 T = 17.84 SS = 0.4784 SS = 0.1803 df 14 1,200 lux 3.50 3.27 3.54 3.83 3.70 The three treatments in the experiment define three populations of interest. You use an ANOVA to test the hypothesis that the three population means are equal. Present the results of your analysis in the following ANOVA table by entering the missing df values and selecting the correct values for the missing SS, MS, and F entries. (Note: For best results, retain at least two additional decimal points throughout your calculations. Depending on the order in which you do these calculations and the number of digits you retain, you may find slight rounding differences in the last digit between your answers and the answer choices.) MS Participant Totals 0.0809 0.2157 P = 10.32 P = 9.98 P = 10.74 P = 11.70 P = 11.76 n = k = N G ΣΧ2 = F = = 5 3 15 54.5( 199.14
F Distribution Numerator Degrees of Freedom = 26 Denominator Degrees of Freedom = 26 0 .5000 1 1.00 .5000 2 3 4 6 8 F Use the Distributions tool to find the critical value for a = 0.05. The critical value is F = At a significance level of a = 0.05, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis is You conclude that lighting brightness affects spatial reasoning abilities.
Transcribed Image Text:F Distribution Numerator Degrees of Freedom = 26 Denominator Degrees of Freedom = 26 0 .5000 1 1.00 .5000 2 3 4 6 8 F Use the Distributions tool to find the critical value for a = 0.05. The critical value is F = At a significance level of a = 0.05, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis is You conclude that lighting brightness affects spatial reasoning abilities.
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