I need answer only for questions 6 and 7 using EF2 SPSS FILE, SOME ONE HAS ANSWE
ID: 3350463 • Letter: I
Question
I need answer only for questions 6 and 7 using EF2 SPSS FILE, SOME ONE HAS ANSWERED QUESTIONS 1 TO 5 (USING EF1 SPSS FILE) BUT NOT QUESTIONS 6 AND 7. i am attaching answers for part 1 (from 1 to 5) at the end which someone had solved previously may this will help when draw conculsion for question 6.
Please do not solve it manually as i need respective SPSS output files as well. There are two SPSS file associated with these excercise EF1 (ejection fraction 1), for question number 1-5 and EF2 for used for question number 6 and 7(ejection fraction 2). Please find both SPSS FILES at the end. Thanks.
The data set EF1.SAV contains percentages of fractional fraction, made by 34 subjects in an imaginary clinical study. The efficacy fraction (EF) was calculated before and after treatment, given with the baseline and post variables, while ID is patient number.
1- Check if Baseline is a normally distributed variable.
2- Calculate average and standard deviation for Baseline, and find a 95% confidence interval for the Baseline average.
3- Repeat Points 1 and 2 for the Post variable.
4- Is there a significant difference between the Baseline and Post average? Make an analysis based on both confidence and p-value.
5- Write a short summary of what you found.
A researcher who has not had any course in statistics analyzes the same data using a two-sample method. He considers the data from before and after treatment as if they came from two random sample of patients.
Here you can use the dataset ef2.sav; The data represents the ejaculation fraction before and after a treatment. The measurements are the same as in ef1.sav, but the data structure is slightly different and the ID variable has been removed.
6- Perform the test performed by the researcher. What conclusion do you get? Compare these conclusions you received when analyzing the data in point 4.
7- What's wrong with this method? Why do you get different conclusions when using the two different methods? Consider this based on the effectiveness goals and the uncertainty of the effect goals you get when using the two methods.
EF1 SPSS FILES
ID
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
BASELINE
55
54
57
47
54
57
62
54
51
51
59
47
54
54
53
54
56
59
55
57
58
51
42
58
55
61
54
58
55
57
54
60
48
55
Post baseline
60
54
59
48
54
59
64
53
52
50
61
45
54
55
54
57
57
62
57
57
59
55
42
61
57
64
56
59
57
60
55
59
49
55
THESE THREE SPSS VARIABLES ID, BASELINE AND POST BASELINE, ARE PART OF EF1 FILE
EF2 SPSS FILE
TIME
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
EF (ejection fraction)
55
54
57
47
54
57
62
54
51
51
59
47
54
54
53
54
56
59
55
57
58
51
42
58
55
61
54
58
55
57
54
60
48
55
60
54
59
48
54
59
64
53
52
50
61
45
54
55
54
57
57
62
57
57
59
55
42
61
57
64
56
59
57
60
55
59
49
55
THESE TWO VARIABLES, TIME AND EF(ejection fraction) ARE PART OF EF2 SPSS FILE.
Answered for part 1
The data set EF1.SAV contains percentages of fractional fraction, made by 34 subjects in an imaginary clinical study. The efficacy fraction (EF) was calculated before and after treatment, given with the baseline and post variables, while ID is patient number.
Check if Baseline is a normally distributed variable.
Tests of Normality
Kolmogorov-Smirnova
Shapiro-Wilk
Statistic
df
Sig.
Statistic
df
Sig.
Baseline
0.209
34
0.001
0.935
34
0.044
a. Lilliefors Significance Correction
Kolmogorov-Smirnov test value 0.209, P=0.001 which is < 0.05 level. The test is significant. The data is not normally distributed.
2- Calculate average and standard deviation for Baseline, and find a 95% confidence interval for the Baseline average.
Descriptives
Statistic
Std. Error
Baseline
Mean
54.59
0.723
95% Confidence Interval for Mean
Lower Bound
53.12
Upper Bound
56.06
5% Trimmed Mean
54.78
Median
55.00
Variance
17.765
Std. Deviation
4.215
Minimum
42
Maximum
62
Range
20
Interquartile Range
4
Skewness
-0.911
0.403
Kurtosis
1.429
0.788
95% CI = (53.12, 56.06).
3- Repeat Points 1 and 2 for the Post variable.
Descriptives
Statistic
Std. Error
Post baseline
Mean
55.88
0.848
95% Confidence Interval for Mean
Lower Bound
54.16
Upper Bound
57.61
5% Trimmed Mean
56.13
Median
57.00
Variance
24.471
Std. Deviation
4.947
Minimum
42
Maximum
64
Range
22
Interquartile Range
5
Skewness
-0.859
0.403
Kurtosis
1.037
0.788
95% CI= (54.16, 57.61)
Tests of Normality
Kolmogorov-Smirnova
Shapiro-Wilk
Statistic
df
Sig.
Statistic
df
Sig.
Post baseline
0.146
34
0.064
0.945
34
0.089
a. Lilliefors Significance Correction
Kolmogorov-Smirnov test value 0.146, P=0.064 which is > 0.05 level. The test is not significant. The data is approximately normally distributed
4- Is there a significant difference between the Baseline and Post average? Make an analysis based on both confidence and p-value.
Paired Samples Statistics
Mean
N
Std. Deviation
Std. Error Mean
Pair 1
Baseline
54.59
34
4.215
0.723
Post baseline
55.88
34
4.947
0.848
Paired Samples Test
Paired Differences
t
df
Sig. (2-tailed)
Mean
Std. Deviation
Std. Error Mean
95% Confidence Interval of the Difference
Lower
Upper
Pair 1
Baseline - Post baseline
-1.294
1.528
0.262
-1.827
-0.761
-4.938
33
0.000
5- Write a short summary of what you found.
Paired sample t test is used. Calculated t=-4.938, P=0.000 which is < 0.05 level. Ho is rejected. We conclude that there is a significant difference between the Baseline and Post average.
A researcher who has not had any course in statistics analyzes the same data using a two-sample method. He considers the data from before and after treatment as if they came from two random sample of patients.
Since data are paired, we have to use dependent sample method. Paired sample t test is appropriate here to compare difference between the Baseline and Post average.
Tests of Normality
Kolmogorov-Smirnova
Shapiro-Wilk
Statistic
df
Sig.
Statistic
df
Sig.
Baseline
0.209
34
0.001
0.935
34
0.044
a. Lilliefors Significance Correction
Histogram 10 Mean = 54.59 Std Dev. = 4.215 N=34 CO 40 45 50 60 BaselineExplanation / Answer
6.
Descriptives
Statistic
Std. Error
EF
Mean
55.23529
.558720
95% Confidence Interval for Mean
Lower Bound
54.12009
Upper Bound
56.35050
5% Trimmed Mean
55.44771
Median
55.00000
Variance
21.227
Std. Deviation
4.607319
Minimum
42.000
Maximum
64.000
Range
22.000
Interquartile Range
4.750
Skewness
-.773
.291
Kurtosis
.953
.574
Tests of Normality
Kolmogorov-Smirnova
Shapiro-Wilk
Statistic
df
Sig.
Statistic
df
Sig.
EF
.174
68
.000
.948
68
.007
a. Lilliefors Significance Correction
Tests of Normality
Kolmogorov-Smirnova
Shapiro-Wilk
Statistic
df
Sig.
Statistic
df
Sig.
EF
.174
68
.000
.948
68
.007
a. Lilliefors Significance Correction
One-Sample Statistics
N
Mean
Std. Deviation
Std. Error Mean
EF
68
55.23529
4.607319
.558720
One-Sample Test
Test Value = 0
t
df
Sig. (2-tailed)
Mean Difference
95% Confidence Interval of the Difference
Lower
Upper
EF
98.861
67
.000
55.235294
54.12009
56.35050
Descriptives
Statistic
Std. Error
EF
Mean
55.23529
.558720
95% Confidence Interval for Mean
Lower Bound
54.12009
Upper Bound
56.35050
5% Trimmed Mean
55.44771
Median
55.00000
Variance
21.227
Std. Deviation
4.607319
Minimum
42.000
Maximum
64.000
Range
22.000
Interquartile Range
4.750
Skewness
-.773
.291
Kurtosis
.953
.574
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