The sample means and sums of squares of the scores for each of the groups are presented in the following table. Group Sample Mean Sum of Squares Away-facing 93.6 2,343.75 Parent-facing 107.2 2,574.15 You decide to use an ANOVA at α = 0.01 to test the null hypothesis that there is no difference between the groups. The ANOVA table follows. Calculate the F-ratio and enter it into the table. Source of Variation ANOVA Table Sum of Squares Degrees of Freedom Mean Square F Between Treatments 1,479.68 1 1,479.68 Within Treatments 4,917.90 30 163.93 Total 6,397.58 31 Use the Distributions tool that follows to find Fcriticalcritical at a significance level of α = 0.01, the value of F that bounds the critical region for the F-ratio test statistic. (Note: Do not use the t-tables. In some cases, when the degrees of freedom are not exact, they do not provide answers with enough precision for this exercise.) F Distribution Numerator Degrees of Freedom = 26 Denominator Degrees of Freedom = 26 0.001.002.003.004.005.006.007.008.00F.5000.50001.000 Fcriticalcritical at α = 0.01 is . Now evaluate the null hypothesis that the population means for all treatments are equal. At a significance level of α = 0.01, the null hypothesis is . You find that you conclude that the direction that the stroller faces influences a child’s expressive vocabulary at 18 months. Now you decide to use a t test to test the hypothesis that there is no difference between the groups. The estimated standard error (sM1 – M2M1 – M2) is 4.53, so the t test statistic is . Use the following tool to find the critical regions for α = 0.01. t Distribution Degrees of Freedom = 21 -3.0-2.0-1.00.01.02.03.0t.2500.5000.2500.5000.5000-0.6860.6860.000 The critical t-scores (the values for t-scores that separate the tails from the main body of the distribution, forming the critical regions) are . Now use the tool to evaluate the null hypothesis. At a significance level of α = 0.01, the null hypothesis is . You find that you conclude that the direction the stroller faces influences a child’s expressive vocabulary at 18 months. When you evaluated the mean difference from the independent-measures study comparing only two samples, the ANOVA and the t test resulted in conclusion.

The sample means and sums of squares of the scores for each of the groups are presented in the following table. Group Sample Mean Sum of Squares Away-facing 93.6 2,343.75 Parent-facing 107.2 2,574.15 You decide to use an ANOVA at α = 0.01 to test the null hypothesis that there is no difference between the groups. The ANOVA table follows. Calculate the F-ratio and enter it into the table. Source of Variation ANOVA Table Sum of Squares Degrees of Freedom Mean Square F Between Treatments 1,479.68 1 1,479.68 Within Treatments 4,917.90 30 163.93 Total 6,397.58 31 Use the Distributions tool that follows to find Fcriticalcritical at a significance level of α = 0.01, the value of F that bounds the critical region for the F-ratio test statistic. (Note: Do not use the t-tables. In some cases, when the degrees of freedom are not exact, they do not provide answers with enough precision for this exercise.) F Distribution Numerator Degrees of Freedom = 26 Denominator Degrees of Freedom = 26 0.001.002.003.004.005.006.007.008.00F.5000.50001.000 Fcriticalcritical at α = 0.01 is . Now evaluate the null hypothesis that the population means for all treatments are equal. At a significance level of α = 0.01, the null hypothesis is . You find that you conclude that the direction that the stroller faces influences a child’s expressive vocabulary at 18 months. Now you decide to use a t test to test the hypothesis that there is no difference between the groups. The estimated standard error (sM1 – M2M1 – M2) is 4.53, so the t test statistic is . Use the following tool to find the critical regions for α = 0.01. t Distribution Degrees of Freedom = 21 -3.0-2.0-1.00.01.02.03.0t.2500.5000.2500.5000.5000-0.6860.6860.000 The critical t-scores (the values for t-scores that separate the tails from the main body of the distribution, forming the critical regions) are . Now use the tool to evaluate the null hypothesis. At a significance level of α = 0.01, the null hypothesis is . You find that you conclude that the direction the stroller faces influences a child’s expressive vocabulary at 18 months. When you evaluated the mean difference from the independent-measures study comparing only two samples, the ANOVA and the t test resulted in conclusion.

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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12. Comparing ANOVA and the t test for an independent-measures hypothesis test

Suzanne Zeedyk, a developmental psychologist at Dundee University’s School of Psychology, conducted a pilot study in which parents started a half-hour walk with their infants in a parent-facing or an away-facing stroller and then switched to the other type of stroller midway. Her results suggest that parents talked less to the babies, the babies had higher heart rates, and they were less likely to fall asleep in away-facing strollers than in parent-facing strollers.

You are interested in testing the hypothesis that babies who travel in parent-facing strollers have different expressive vocabularies than babies who travel in away-facing strollers. You randomly assign 16 newborns to parent-facing strollers and 16 newborns to away-facing strollers. You then test the babies’ expressive vocabularies at age 18 months using the Expressive Vocabulary Test (EVT), which is designed primarily to assess children’s expressive vocabulary.

The sample means and sums of squares of the scores for each of the groups are presented in the following table.

Group	Sample Mean	Sum of Squares
Away-facing	93.6	2,343.75
Parent-facing	107.2	2,574.15

You decide to use an ANOVA at α = 0.01 to test the null hypothesis that there is no difference between the groups. The ANOVA table follows. Calculate the F-ratio and enter it into the table.

Source of Variation	ANOVA Table
Sum of Squares	Degrees of Freedom	Mean Square	F
Between Treatments	1,479.68	1	1,479.68
Within Treatments	4,917.90	30	163.93
Total	6,397.58	31

Use the Distributions tool that follows to find Fcriticalcritical at a significance level of α = 0.01, the value of F that bounds the critical region for the F-ratio test statistic. (Note: Do not use the t-tables. In some cases, when the degrees of freedom are not exact, they do not provide answers with enough precision for this exercise.)

F Distribution

Numerator Degrees of Freedom = 26

Denominator Degrees of Freedom = 26

0.001.002.003.004.005.006.007.008.00F.5000.50001.000

Fcriticalcritical at α = 0.01 is .

Now evaluate the null hypothesis that the population means for all treatments are equal. At a significance level of α = 0.01, the null hypothesis is . You find that you conclude that the direction that the stroller faces influences a child’s expressive vocabulary at 18 months.

Now you decide to use a t test to test the hypothesis that there is no difference between the groups. The estimated standard error (sM1 – M2M1 – M2) is 4.53, so the t test statistic is .

Use the following tool to find the critical regions for α = 0.01.

t Distribution

Degrees of Freedom = 21

-3.0-2.0-1.00.01.02.03.0t.2500.5000.2500.5000.5000-0.6860.6860.000

The critical t-scores (the values for t-scores that separate the tails from the main body of the distribution, forming the critical regions) are .

Now use the tool to evaluate the null hypothesis. At a significance level of α = 0.01, the null hypothesis is . You find that you conclude that the direction the stroller faces influences a child’s expressive vocabulary at 18 months.

When you evaluated the mean difference from the independent-measures study comparing only two samples, the ANOVA and the t test resulted in conclusion.

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 + ...

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