Stats Help!! (Here is the data) time/ group 47.2 1 22 1 20.4 1 19.7 1 17.4 1 14.

Question

Stats Help!! (Here is the data)

time/ group
47.2 1
22 1
20.4 1
19.7 1
17.4 1
14.7 1
13.4 1
13 1
12.3 1
12.2 1
10.3 1
9.7 1
9.7 1
9.5 1
9.1 1
8.9 1
8.9 1
8.4 1
8.1 1
7.9 1
7.8 1
6.9 1
6.3 1
6.1 1
5.6 1

4.7 1

4.7 1

4.3 1
4.2 1
3.9 1
3.4 1
3.1 1
3.1 1
2.7 1
2.4 1
2.3 1
2.3 1
2.1 1
2.1 1
2 1
1.9 1
1.7 1
1.7 1
19.7 2
16.2 2
15.9 2
15.4 2
9.7 2
8.9 2
8.6 2
8.6 2
7.4 2
6.3 2
6.1 2
6 2
6 2
5.9 2
4.9 2
4.6 2
3.8 2
3.6 2
3.5 2
3.3 2
3.3 2
2.9 2
2.8 2
2.7 2
2.4 2
2.3 2
2 2
1.8 2
1.7 2
1.7 2
1.6 2
1.4 2
1.2 2
1.1 2
1 2

Question!

In theory, the ANOVA F-test can be used when there are only two samples if thenormality and homoscedasticity assumptions are met. However, it’s better to use Welch’st-test because it has weaker assumptions (e.g. variances may be dierent) while the F-test issensitive to violations of its assumptions when there are only two samples.
Recall for the untransformed stereogram data on Canvas (stereograms.txt in the Datafolder of FIles):

> stereograms = read.table(file.choose(), header=TRUE)

> treatment = stereograms$time[stereograms$group==2]

> control = stereograms$time[stereograms$group==1]

> t.test(treatment, control)

Welch Two Sample t-test
data: treatment and controlt = -2.0384, df = 70.039, p-value = 0.04529alternative hypothesis: true difference in means is not equal to 0
> t.test(treatment, control, var.equal=T)Two Sample t-testdata: treatment and control
t = -1.9395, df = 76, p-value = 0.05615alternative hypothesis: true difference in means is not equal to 0

1

So Welch’s test gives a P-value of 0.04529, while Student’s test gives a P-value of 0.05615.
Find a P-value for the F-test of the null hypothesis that the treatment and control populationshave the same mean. How does this compare to the Welch and Student P-values?

Explanation / Answer

Sol)

Null hypothesis: All the means are equal

Alternative hypothesis: All the means are not equal

Test Statistics:

Using SPSS

Conclusion: F cal=3.762 and P value=0.056

and alpha=0.05

P value > alpha, we accept the null hypothesis.

Descriptives time N Mean Std. Deviation Std. Error 95% Confidence Interval for Mean Minimum Maximum Lower Bound Upper Bound 1.00 43 8.5605 8.08541 1.23301 6.0721 11.0488 1.70 47.20 2.00 35 5.5514 4.80174 .81164 3.9020 7.2009 1.00 19.70 Total 78 7.2103 6.93601 .78535 5.6464 8.7741 1.00 47.20

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