Someone help me with this An independent-measures research study was used to com
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Someone help me with this
An independent-measures research study was used to compare two treatment conditions with n=12 participants in each treatment. The first treatment had a mean of M=55 with a variance of s2=8, and the second treatment had M=52 and s2=4. Do these data indicate a significant difference between the two treatments? use a two-tailed test with a=.05. (Note: Because the two samples are the same size, the pooled variance is simply the average of the two sample variances.) Be sure to show all formulas, steps, processes and calculations for all parts (including pooled variance) of the answers.
Explanation / Answer
Two sample t test
An independent-measures research study was used to compare two treatment conditions with n=12 participants in each treatment. The first treatment had a mean of M=55 with a variance of s2=8, and the second treatment had M=52 and s2=4. Do these data indicate a significant difference between the two treatments? use a two-tailed test with a=.05. (Note: Because the two samples are the same size, the pooled variance is simply the average of the two sample variances.) Be sure to show all formulas, steps, processes and calculations for all parts (including pooled variance) of the answers.
Solution:
Here, we have to check the claim whether there is a significant difference between the average of two treatments or not. For checking this significant difference we have to use the two sample t test for the population mean. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: There is no any significant difference exists between the averages of two treatments mean.
Alternative hypothesis: Ha: There is a significant difference exists between the averages of two treatments mean.
We are given
Sample size n1 = 12 and n2 = 12
M1 = 55
M2 = 52
S1^2 = 8
S2^2 = 4
Level of significance = alpha = 0.05
S1 = sqrt (8) = 2.8284
S2 = sqrt (4) = 2
The test statistic formula is given as below:
t =[ ( M1 – M2) ] / sqrt [(S1^2/n1)+(S2^2/n2)]
The values for test statistic, pooled variance and p-value are given as below:
Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
12
Sample Mean
55
Sample Standard Deviation
2.8284
Population 2 Sample
Sample Size
12
Sample Mean
52
Sample Standard Deviation
2
Intermediate Calculations
Population 1 Sample Degrees of Freedom
11
Population 2 Sample Degrees of Freedom
11
Total Degrees of Freedom
22
Pooled Variance
5.9999
Standard Error
1.0000
Difference in Sample Means
3.0000
t Test Statistic
3.0000
Two-Tail Test
Lower Critical Value
-2.0739
Upper Critical Value
2.0739
p-Value
0.0066
Reject the null hypothesis
Here, we get the p-value as 0.0066 which is less than the given level of significance or alpha value 0.05, so we reject the null hypothesis that there is no any significant difference exists between the averages of two treatments mean.
Pooled-Variance t Test for the Difference Between Two Means
(assumes equal population variances)
Data
Hypothesized Difference
0
Level of Significance
0.05
Population 1 Sample
Sample Size
12
Sample Mean
55
Sample Standard Deviation
2.8284
Population 2 Sample
Sample Size
12
Sample Mean
52
Sample Standard Deviation
2
Intermediate Calculations
Population 1 Sample Degrees of Freedom
11
Population 2 Sample Degrees of Freedom
11
Total Degrees of Freedom
22
Pooled Variance
5.9999
Standard Error
1.0000
Difference in Sample Means
3.0000
t Test Statistic
3.0000
Two-Tail Test
Lower Critical Value
-2.0739
Upper Critical Value
2.0739
p-Value
0.0066
Reject the null hypothesis
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