Download - Group Statistics[1]
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1.a) find the point estimate and the 95% confidence interval estimate of the mean scores in the pretest of the population of students. Interpret your answer.
Descriptive statistics
N minimum maximum MeanStd.Error
Mean
score 9 38 81 56.44 4.540
One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
score 9 56.4444 13.62086 4.54029
One-Sample Test
Test Value = 0
t df Sig. (2-tailed) Mean Difference
95% Confidence Interval of the Difference
Lower Upper
score 12.432 8 .000 56.444 45.97 66.91
b. Test whether the mean scores of the population is less than 70 at 0.01 level of singnificant.
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One-Sample Statistics
N Mean Std. DeviationStd. Error
Mean
score 9 56.4444 13.62086 4.54029
One-Sample Test
Test Value =70
t df Sig. (2-tailed) Mean Difference
95% Confidence Interval of the Difference
Lower Upper
score -2.986 8 .017 -13.556 -24.03 -3.09
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c. Test at 5% significance level whether this special course makes any statistically significant improvement in the memory retention ability of the population of students.
(5 marks)
d. Construct a 95% confidence interval for mean difference of the scores of the students before and after the memory improvement course
(3 marks)
c)
One- Sample Statistic
N Mean Std. DeviationStd. Error
Mean
SCORE 9 61.3333 15.64449 5.21483
One- Sample Test
Paired Differences
t df Sig. (2-tailed) Mean Difference
95% Confidence Interval of the Difference
Lower Upper
SCORE -1.662 8 .135 08.66667 -20.6921 3.3588
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d)
Paired Sample Statistic
Mean N Std. Deviation Std. Error Mean
Pair 1 BEFORE AFTER
56.444461.3333
99
13.6208615.64449
4.540295.21483
Paired Sample Correlation
N Correlation Sig.
Pair 1 BEFORE AFTER
9 61.3333 .000
Paired Sample Test
Paired differences
T df Sig. (2 tailMean Std. Deviation
Std. Error Mean
95% Confidence IntervalOf the Differences
Lower
UpperPair 1 BEFORE-
AFTER-4.88889 6.05071 2.01690 -9.53988 -.23790 -2.424 8 .0
.
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Another similar study on memory retention ability of a different sample of students is done where the difference in scores between Pretest and Post-Test (PerformanceScore) and the initial level of memory retention (PreTestLevel) are given in File memory.sav. Use the data given in the file to form four improvement levels of memory retention (PerformanceLevel) as follows:
PerformanceLevel 1 if PerformanceScore < -1
PerformanceLevel 2 if -1 PerformanceScore < 2
PerformanceLevel 3 if 2 PerformanceScore < 8
PerformanceLevel 4 if PerformanceScore 8
Where level 1 indicates “Noticeable confusion”, level 2 indicates “No improvement”,
level 3 indicates “Noticeable improvement” and level 4 indicates “Vast improvement”.
e) How many students belong to each initial level of memory retention? Then, present this information using a suitable graphical description.
(4 marks)
f) Is the initial level of memory retention related to the level of improvement of memory retention? Test at 5% significance level.
(5 marks)
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e)
Between –Subject Factors
N
Pre Test Level 1.00 78
2.00 110
3.00 73
4.00 56
Pre TestLevel
Frequency Percent Valid PercentCumulative
Percent
Valid 1.00 78 24.6 24.6 24.6
2.00 110 34.7 34.7 59.3
3.00 73 23.0 23.0 82.3
4.00 56 17.7 17.7 100.0
Total 317 100.0 100.0
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f) Is the initial level of memory retention related to the level of improvement of memory retention? Test at 5% significance level.
(5 marks)
Tests of Between-Subjects Effects
Dependent variable: improvescore
SourceType III Sum of
Squares df Mean Square F Sig.
Corrected ModelInterceptPre TestLevelErrorTotalCorrected Total
682.398(a)9104.011
682.3988558.404
20014.0009240.801
313
313317316
227.4669104.011
227.46627.343
8.319332.954
8.319
.000
.000
.000
a R Squared = .074 (Adjusted R Squared = .065)
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g) Determine whether there is significant difference in the mean of performance scores of the methods at 5% significance level.
(5 marks)
ANOVA
MemoryImprove
SourceType III Sum of
Squares df Mean Square F Sig.
Between GroupsWithin GroupsTotal
305.100585.700890.800
32629
101.70022.527
4.515 .011
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h) Is thereva different effect on performance scores between Method 2 and Method 3? Discuss the effect on performance scores of the methods. (Use 5% significance level)
(5 marks)
Multiple ComparisonsDependent Variable: MemoryImproveBonferroni
(I) time (Binned) (J) time (Binned) Mean Difference (I-J) Std. Error Sig.
95% Confidence Interval
Lower Bound Upper Bound1
1.00 2.00 3.00 4.002.00 1.00 3.00 4.003.00 1.00 2.00 4.004.00 1.00 2.00 3.00
-4.200001.800003.900004.200006.00000
8.10000(*)-1.80000-6.200002.10000
-3.90000-8.10000(*)
-2.10000
2.705782.779122.599632.705782.456422.251352.779122.456422.338982.599632.251352.33898
.7961.000
.874
.796
.130
.0081.000
.1301.000
.874
.0081.000
-11.9264-6.1358-3.5233-3.5264-1.01441.6712
-9.7358-13.0144
-4.5790-11.3233-14.5288
-8.7790
3.52649.7358
11.323311.926413.014414.5288
6.13581.01448.77903.5233
-1.67124.5790
*. The mean difference is significant at the .05 level.
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2a) List down the value of the mean and standard deviation of time-length of self-treatment for age, where age can be group asCompare the means of age group.
Group Statistics
age group N Mean Std. Deviation Std. Error Mean
time 1.00 107 23.25 13.062 1.263
2.00 338 29.11 19.961 1.086
b. is there significant difference in time-length of self-treatment between age group?
Independent Samples Test
Levene's Test for Equality of Variances t-test for Equality of Means
F Sig. t dfSig. (2-tailed)
Mean Difference
Std. Error Difference
95% Confidence Interval of the
Difference
Lower Upper
time Equal variances assumed
45.666 .000 -2.849 443 .005 -5.860 2.057 -9.903 -1.817
Equal variances not assumed
-3.519 273.617 .001 -5.860 1.665 -9.139 -2.582
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c. Give a comment on the means regarding to your answer in (a) and (b).
d. obtain the 95% confidence interval for the means difference in time-length of self-treatment between age group. Interpret the obtained value.
Test of homogeneity of variances age
ANOVA
age
Sum of Squares df Mean Square F Sig.
Between GroupsWithin GroupsTotal
3055.76028362.56831418.328
2442444
1527.88064.169
23.810 .000
Levene statistic df1 df2 Sig.
2.996 2 442 .051
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e. Related your result in (b) and (d). Does you result in (d) support the answer in (b)? What is your reason?
f. Code the time-length of self-treatment (timegroup) as
Thus, determine whether there is enough evidence to conclude that there is significant difference in the mean age between different categories of time-length of self-treatment. Check the equality of the variances assumption and do the necessary multiple comparison tests if applicable.
Multiple ComparisonsDependent Variable: AgeBonferroni
(I) time group (J)timegroup Mean Difference (I-J) Std. Error Sig.
95% Confidence Interval
Lower Bound Upper Bound
1.00dimension3
11 - 50 2.801 1.312 .100 -.35 5.95
>50 -4.357* 1.563 .017 -8.11 -.60
2.00dimension3
<=10 -2.801 1.312 .100 -5.95 .35
>50 -7.158* 1.053 .000 -9.69 -4.63
3.00dimension3
<=10 4.357* 1.563 .017 .60 8.11
11 - 50 7.158* 1.053 .000 4.63 9.69
*. The mean difference is significant at the 0.05 level.
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g. From the past studies, the research believes that the average of blood sugar reading was at most 5 mmo1/L. Test the claim.
One-Sample Statistics
N Mean Std. Deviation Std. Error Mean
blood_sugar 445 5.69 3.202 .152
One-Sample Test
Test Value = 30
t df Sig. (2-tailed) Mean Difference
95% Confidence Interval of the Difference
Lower Upper
Blood_sugar 4.545 444 .000 .690 .39 .99
h. Recode the blood sugar reading as followa:
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the blood sugar reading (blood_sugar group) as
Test whether the blood sugar reading (blood_sugar group) are equally likely distributed.
bsugargrp
Observed N Expected N Residual
1.002.003.00Total
290135
20445
148.3148.3148.3
141.7-13.3
-128.3
Test Statistic
bsugargrp
Chi- SquareªDfAsymp.Sig.
247.5282
.000
a. 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 148.3
I. Is there enough evidence to conclude that the blood pressure (bpressgrp) is related to the
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blood sugar reading (bsugargrp)?
Case Processing Summary
Cases
Valid Missing Total
N Percent N Percent N Percent
bpressgrp* bsugargrp 445 100.0% 0 .0% 445 100.0%
Bpressgrp* bsugargrp Crosstabulation
count
bsugargrp
total1.00 2.00 3.00
bpressgrp normal
prehypertension
hypertension
Total
290
0
0
290
0
135
0
135
0
0
20
20
290
135
20
445
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Chi- Square Tests
a. 1 cells (11.1%) have expected count less than 5. The minimum expected count is .90.
Value
df
Asymp. Sig. (2-sided)
Pearson Chi-squareLikelihood RatioLinear-by-Linear AssociationN of Valid Cases
890.000ª 694.502 444.000
445
4 4 1
.000 .000 .000
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j. Another experiment using completely Random Block design to study the effect of time length of self-treatment on systolic blood pressure with the existence of gender as block has been done. The result of systolic reading are given in the following table:
Systolic pressure Reading
Time-length of self-treatment (in minutes/ day)
0-31
31-60
> 30
Men 130 125 115
Women 125 122 110
Draw your conclusion based on the analysis.
Between-Subject Factors
Test of Between- Subjects Effects
Dependent Variable: pressuread
SourceType III Sum of
Squares df Mean Square F Sig.
Corrected ModelInterceptGenderTimelenghtErrorTotalCorrected Total
253.500ª87604.167
20.167233.333
5.33387863.000
258.833
3112265
84.50087604.167
20.167116.667
2.667
31.68832851.563
7.56343.750
.031
.000
.111
.022
NGender 1.00 2.00Timelenght 1.00 2.00 3.00
3 3 2 2 2
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REFERENCE
Fathilah Binti Mohd Alipiah , Rohaiza Binti Zakaria (2010). Applied Statistic. By The McGraw-Hill Companies
Yuhaniz Haji Ahmad, Shamshuritawati Sharif, Zahayu Md. Yusof (2010). SQQS 2013 Applied Statistics.