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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N Mean Std. DeviationStd. Error

Mean

score 9 56.4444 13.62086 4.54029

One-Sample Test

Test Value =70



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



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



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.


N Mean Std. Deviation Std. Error Mean

blood_sugar 445 5.69 3.202 .152

One-Sample Test

Test Value = 30



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

16

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



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.

group statistics[1]

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