HELP!!! JUST NEED C) ANSWERED 8) A recent study examined how applicants with a f

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HELP!!! JUST NEED C) ANSWERED

8) A recent study examined how applicants with a facial blemish such as a scar or birthmark fared in job interviews. The results indicate that interviewers recalled less information and gave lower ratings to applicants with a blemish. In a similar study, participants conducted computer-simulated interviews with a series of applicants including one with a facial scar and one with a facial birthmark. The following data represent the ratings given to each applicant.

a. Use a repeated measures ANOVA with ? = .05 to determine whether there are any significant mean differences among the three conditions.

b. Calculate ?2 to measure the size of the treatment effects.

c) Write a sentence demonstrating how a research report would present the results of the hypothesis test and the measure of effect size

Applicant No Person Participant Scar Birthmark Blemish Total T= SS = P=6 4 N=15 G=45 P=15 0 0 P=9 0 6 P=62? = 231 P=9 M=1 M=2 M=6 T=5 T=10 T=30 SS= 6 SS= 10 SS= 10

Explanation / Answer

A recent study examined how applicants with a facial blemish such as a scar or birthmark fared in job interviews. The results indicate that interviewers recalled less information and gave lower ratings to applicants with a blemish. In a similar study, participants conducted computer-simulated interviews with a series of applicants including one with a facial scar and one with a facial birthmark. The following data represent the ratings given to each applicant.

We have given that three conditions.

And we have to test the hypothesis that,

H0 : Three means are equal.

H1 : Atleast one of the mean is differ than 0.

Assume alpha = level of significance = 5% = 0.05

There are 5 values in the three conditions.

We can test three means using ANOVA.

ANOVA in EXCEL :

steps :

ENTER the data in MINITAB sheet --> Data --> Data Analysis --> Anova : Single factor --> ok --> Input : select all data range --> Grouped by : columns --> Alpha : 0.05 --> Output Range : select one empty cell --> ok

Output is :

Test statistic F = 16.15385

P-value = 0.00039

P-value < alpha

Reject H0 at 5% level of significance.

COnclusion : Atleast one of the mean is differ than 0.

2 = SStreatment / SStotal

2 =   70 / 96 = 0.7292 = 72.92%

Treatments accounts for 72.92%.

Anova: Single Factor SUMMARY Groups Count Sum Average Variance Column 1 5 5 1 1.5 Column 2 5 10 2 2.5 Column 3 5 30 6 2.5 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 70 2 35 16.15385 0.000394651 3.885293835 Within Groups 26 12 2.166667 Total 96 14

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