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Statistics
Statistics
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Statistics
TEST I
Number 1
Use a paired sample t-test to determine the difference at a given alpha value then compare the
means of the two.
Number 2
The chicken dish rated 10 on the ordinal scale was the best.
Number 3
A paired sample t-test will be applied.
Number 4
An independent samples t-test will be applied.
Number 5
An independent samples t-test will be applied
Number 6
A chi-square test of association is used.
Number 7
A paired sample t-test will be applied to determine whether the difference is significant. If the
difference is significant, then compare the means.
Number 8
Pearson’s correlation test
Number 9
A chi-square test of association is used.
Number 10
Analysis of variance is the most appropriate test.
Number 11
Analysis of variance is the most appropriate test since the treatments are more than two.
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Number 12
A chi-square test for independence is appropriate
Number 13
Pearson’s correlation test
Number 14
A chi-square test of association is used.
Number 15
An independent samples t-test will be applied.
TEST II
PROBLEM 1
Research question: Does review type affect maths score?
Null hypothesis: Review type has got no effect on math score
Statistical tool: Analysis of variance (ANOVA)
Table of results
ANOVA
Math_test
Sum of Squares df Mean Square F Sig.
Between Groups 2420.281 2 1210.141 70.236 .000
Within Groups 809.799 47 17.230
Total 3230.080 49
Table 1
Since the p-value calculated (0.00) is less than the level of significance (0.05), the null
hypothesis is rejected. It is concluded that review type has got no effect on math score.
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PROBLEM 2
Research question: Is there association between gender and exam ranks?
Null hypothesis: There is no association between gender and exam ranks.
Statistical tool: Chi-square test
Table of results
Chi-Square Tests
Value df Asymp. Sig. (2-
sided)
Pearson Chi-Square 50.000a 48 .394
Likelihood Ratio 68.593 48 .027
Linear-by-Linear Association .823 1 .364
N of Valid Cases 50
a. 98 cells (100.0%) have expected count less than 5. The minimum
expected count is .44.
Table 2
Since the p-value calculated (0.39) is greater than the level of significance (0.05), the null
hypothesis is not rejected. It is concluded that there is no association between gender and exam
ranks.
PROBLEM 3
Research question: Is there is a significant difference on the interview scores of those who
practice and those who did not practice?
Null hypothesis: There is no difference on the interview scores of those who practice and those
who did not practice.
Statistical tool: Independent samples t-test
Table of results
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Independent Samples Test
Levene's Test for
Equality of
Variances
t-test for Equality of Means
F Sig. t df Sig. (2-
tailed)
Mean
Difference
Std. Error
Difference
95% Confidence
Interval of the
Difference
Lower Upper
Interview_s
core
Equal variances
assumed
15.769 .000 -8.702 48 .000 -17.49919 2.01098 -21.54255 -13.45584
Equal variances not
assumed
-8.185 27.374 .000 -17.49919 2.13807 -21.88335 -13.11504
Table 3
Since the p-value calculated (0.00) is less than the level of significance (0.05), the null
hypothesis is rejected. It is concluded that there is a significant difference on the interview scores
of those who practice and those who did not practice.
PROBLEM 4
Research question: Is there a relationship between socio economic status and opinion on same
sex marriage?
Null hypothesis: There is no relationship between socio economic status and opinion on same
sex marriage.
Statistical tool: Chi-square test
Table of results
Chi-Square Tests
Value df Asymp. Sig. (2-
sided)
Pearson Chi-Square 13.190a 2 .001
Likelihood Ratio 13.760 2 .001
Linear-by-Linear Association 11.832 1 .001
N of Valid Cases 50
a. 0 cells (0.0%) have expected count less than 5. The minimum
expected count is 6.72.
Table 4
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Since the p-value calculated (0.01) is less than the level of significance (0.05), the null
hypothesis is rejected. It is concluded that there is a significant relationship between socio
economic status and opinion on same sex marriage.
PROBLEM 5
Research question: Are the mean English tests scores of the students from Week 1 to Week 3
equal?
Null hypothesis: mean for Week 1= Week 2 = Week 3
Statistical tool: Anova
Table of results
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
english test (Week 1) 50 180 3.6 19.42857
english test (week2) 50 200 4 14.28571
english test (week3) 50 185 3.7 17.15306
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 4.333333 2 2.166667 0.127783 0.880142 3.057621
Within Groups 2492.5 147 16.95578
Total 2496.833 149
Table 5
Since the p-value calculated (0.88) is greater than the level of significance (0.05), the null
hypothesis is rejected. It is concluded that at least one week has a different test score.
PROBLEM 6
Research question: Are predictors significant?
Null hypothesis: The model does not significantly predict the dependent variable
Statistical tool: Multiple analysis of variance
Table of results
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ANOVAa
Model Sum of Squares df Mean Square F Sig.
1
Regression 5419.366 5 1083.873 60.002 .000b
Residual 794.814 44 18.064
Total 6214.180 49
a. Dependent Variable: Interview score
b. Predictors: (Constant), Anxiety, School type, Math test, Science, Gender
Table 6
Coefficientsa
Model Unstandardized Coefficients Standardized
Coefficients
t Sig.
B Std. Error Beta
1
(Constant) 40.100 13.281 3.019 .004
Gender .939 2.588 .042 .363 .718
School type 3.075 1.403 .137 2.191 .034
Math test .170 .132 .123 1.289 .204
Science .376 .130 .318 2.897 .006
Anxiety -4.644 1.001 -.454 -4.638 .000
a. Dependent Variable: Interview score
Table 7
Since the p-value calculated (0.00) is less than the level of significance (0.05), the null
hypothesis is rejected. It is concluded that the model does significantly predict the dependent
variable.
PROBLEM 7
Research question: Does intervention program have an effect on performance of students in
Chemistry?
Null hypothesis: Intervention program have no effect on performance of students in Chemistry
Statistical tool: Paired samples t-test
Table of results
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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 Pretest chemistry –
Posttest chemistry
-
1.86000
5.67130 .80204 -3.47177 -.24823 -2.319 49 .025
Table 8
Since the p-value calculated (0.025) is less than the level of significance (0.05), the null
hypothesis is rejected. It is concluded that the intervention program have a significant effect on
performance of students in Chemistry.
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