Need some help for this assignment: Report your analyses in APA format and write
ID: 3203541 • Letter: N
Question
Need some help for this assignment: Report your analyses in APA format and write an APA-formatted interpretation of your results for each test.
Move your output into a Microsoft Word document and write your interpretation of each test following the data output for the test in 1 to 2 paragraphs.
2-Sample t: Recall1 by Car
Method
: mean of Recall1 when Car = Blue
: mean of Recall1 when Car = Green
Difference: -
Equal variances are not assumed for this analysis.
Descriptive Statistics: Recall1
Car
N
Mean
StDev
SE Mean
Blue
36
7.5556
3.2815
0.5469
Green
36
7.5000
3.2293
0.5382
Estimation for Difference
Difference
95% CI for Difference
0.0556
(-1.4752, 1.5863)
Test
Null hypothesis
H: - = 0
Alternative hypothesis
H: - 0
T-Value
DF
P-Value
0.07
69
0.9425
2-Sample t: Recall2 by Car
Method
: mean of Recall2 when Car = Blue
: mean of Recall2 when Car = Green
Difference: -
Equal variances are not assumed for this analysis.
Descriptive Statistics: Recall2
Car
N
Mean
StDev
SE Mean
Blue
36
6.3333
2.9472
0.4912
Green
36
6.8056
2.6707
0.4451
Estimation for Difference
Difference
95% CI for Difference
-0.4722
(-1.7946, 0.8502)
Test
Null hypothesis
H: - = 0
Alternative hypothesis
H: - 0
T-Value
DF
P-Value
-0.71
69
0.4786
Method
: mean of Recall1 when Car = Blue
: mean of Recall1 when Car = Green
Difference: -
Equal variances are not assumed for this analysis.
Explanation / Answer
The output for the first test in APA format is as shown below:
""An independent Samples T test is conducted to comapre the means of the two cars. From the data the results obtained state the difference in the two means is 0.056, with a confidence interval CI = (-1.4752, 1.5863). This interval includes the value 0, which indicates that the difference may be equal to Zero as well, thus supporting the null.
Further the T value = 0.07 , with df = 69 , and Pvalue = 0.9425
Since the P value 0.9425 is greater than the significance level 0.05 , we fail to reject the null
And thus we conclude that there is no difference between the means of the two types of cars.There is no significant evidence to support the alternative.""
.
Similarly the output for the second test in APA format is as follows:
""An independent Samples T test is conducted to comapre the means of the two cars. From the data the results obtained state the difference in the two means is -0.4722, with a confidence interval CI = (-1.7946, 0.8502). This interval includes the value 0, which indicates that the difference may be equal to Zero as well, thus supporting the null.
Further the T value = -0.71 , with df = 69 , and Pvalue = 0.4786
Since the P value 0.4786 is greater than the significance level 0.05 , we fail to reject the null
And thus we conclude that there is no difference between the means of the two types of cars.There is no significant evidence to support the alternative.""
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