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: Cola Black Tea Coffee Caffeine Content (mg/oz) 2.9 5.9 13.4 85 85 92 ΣX² = 147,641 86 89 87 G = 1,755 82 82 80 N = 21 75 75 89 k = 3 66 88 96 78 76 83 87 82 92 T₁ = 559 T₂ = 577 T₃ = 619 SS₁ = 338.86 SS₂ = 177.71 SS₃ = 185.71 n₁ = 7 n₂ = 7 n₃ = 7 M₁ = 79.8571 M₂ = 82.4286 M₃ = 88.4286 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? The population mean test scores for all three treatments are equal. The population mean test scores for all three treatments are different. The population mean test scores for all three treatments are not all equal. The population mean test score for the cola population is different from the population mean test score for the black tea population. Calculate the degrees of freedom and the variances for the following ANOVA table: Source SS df MS Between Within 702.28 Total 973.14 The formula for the F-ratio is: FF = = MSbetweenMSbetween / / MSwithinMSwithin Using words (chosen from the dropdown menu), the formula for the F-ratio can be written as: FF = = / / Using the data from the ANOVA table given, the F-ratio can be written as: FF = = / / Thus: FF = = Use the Distributions tool to find the critical region for α = 0.05. F Distribution Numerator Degrees of Freedom = 26 Denominator Degrees of Freedom = 26 012345678F At the α = 0.05 level of significance, what is your conclusion? You can reject the null hypothesis; caffeine does appear to affect test performance. You can reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance. You cannot reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance. You cannot reject the null hypothesis; caffeine does appear to affect test performance.

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: Cola Black Tea Coffee Caffeine Content (mg/oz) 2.9 5.9 13.4 85 85 92 ΣX² = 147,641 86 89 87 G = 1,755 82 82 80 N = 21 75 75 89 k = 3 66 88 96 78 76 83 87 82 92 T₁ = 559 T₂ = 577 T₃ = 619 SS₁ = 338.86 SS₂ = 177.71 SS₃ = 185.71 n₁ = 7 n₂ = 7 n₃ = 7 M₁ = 79.8571 M₂ = 82.4286 M₃ = 88.4286 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? The population mean test scores for all three treatments are equal. The population mean test scores for all three treatments are different. The population mean test scores for all three treatments are not all equal. The population mean test score for the cola population is different from the population mean test score for the black tea population. Calculate the degrees of freedom and the variances for the following ANOVA table: Source SS df MS Between Within 702.28 Total 973.14 The formula for the F-ratio is: FF = = MSbetweenMSbetween / / MSwithinMSwithin Using words (chosen from the dropdown menu), the formula for the F-ratio can be written as: FF = = / / Using the data from the ANOVA table given, the F-ratio can be written as: FF = = / / Thus: FF = = Use the Distributions tool to find the critical region for α = 0.05. F Distribution Numerator Degrees of Freedom = 26 Denominator Degrees of Freedom = 26 012345678F At the α = 0.05 level of significance, what is your conclusion? You can reject the null hypothesis; caffeine does appear to affect test performance. You can reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance. You cannot reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance. You cannot reject the null hypothesis; caffeine does appear to affect test performance.

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Description: 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

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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The students receive the following arithmetic test scores:

	Cola	Black Tea	Coffee
Caffeine Content (mg/oz)	2.9	5.9	13.4
	85	85	92	ΣX² = 147,641
	86	89	87	G = 1,755
	82	82	80	N = 21
	75	75	89	k = 3
	66	88	96
	78	76	83
	87	82	92
	T₁ = 559	T₂ = 577	T₃ = 619
	SS₁ = 338.86	SS₂ = 177.71	SS₃ = 185.71
	n₁ = 7	n₂ = 7	n₃ = 7
	M₁ = 79.8571	M₂ = 82.4286	M₃ = 88.4286

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?

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

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

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

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

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

Source	SS	df	MS
Between
Within	702.28
Total	973.14

The formula for the F-ratio is:

= =

MSbetweenMSbetween

/ /

MSwithinMSwithin

Using words (chosen from the dropdown menu), the formula for the F-ratio can be written as:

= =

/ /

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

= =

/ /

Thus:

= =

Use the Distributions tool to find the critical region for α = 0.05.

F Distribution

Numerator Degrees of Freedom = 26

Denominator Degrees of Freedom = 26

012345678F

At the α = 0.05 level of significance, what is your conclusion?

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

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

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

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

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Expert Solution

Step 1

Solution:

Given information:

N= 21 observation

k =3 treatments

Sum of square of within =SSwithin =702.28

Sum of square of total = SStot = 973.14

Sum of square of total = Sum of square of between + Sum of square of within

Sum of square of between=Sum of square of total - Sum of square of within

Sum of square of between= SSbetween = 973.14-702.28 = 270.86

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