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Statistics Solution

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Statistics
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Question 1:
Here is the SPSS output for single sample t test applied to given data:
One-Sample Statistics
Mean
3.8875
Std. Deviation
2.6743
Std. Error Mean
0.6686
N
16
df
15
t
1.6266
P-value (one-tailed)
0.06232
t Critical (one-tailed)
1.75305
Question 2:
According to the above SPSS result, the mean average score of the children in the given area is
3.89. Since the test wants to determine if the children in the given area have higher scores than
children in the general population, who normally score of 2.3, the one-tailed test should be used
in this case. Since the population of sample is less than 30, we applied t-test statistics in this case,
and the t-test value can be computed according to the formula:
/
X
t
n
=

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3.89 2.3
2.67 / 16
1.62
t
t
=
=
The p-value for one-tailed t test with level of significance = 0.05 is 0.062. Since this p-value is
greater than the level of significance = 0.05, we do not have evidence to reject the null
hypothesis. That is, we are not able to conclude from the data that the children in this specific
area indeed have higher scores than children in the general population. The difference in the test
score which is higher than the average of the general population is due to chance.
Question 3:
Here is the SPSS output for the paired sample t test applied to the data:
Mean
N
Std.
Deviation
Std. Error
Mean
t
df
P-value
(2-tailed)
Before Geometry Test
84.25
16
12.39
3.0975
-1.872
15
0.081
After Geometry Test
86.19
16
10.86
2.715
According to the above SPSS output, the mean average score of 16 students before the geometry
test is 84.25, with the standard deviation of 12.39. The mean average score of those students after
the geometry test is 86.19, with the standard deviation of 10.86.
By conducting the paired-sampled t test for the two above data, the t test statistics is found to be

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Question 1: Here is the SPSS output for single sample t test applied to given data: One-Sample Statistics Mean 3.8875 Std. Deviation 2.6743 Std. Error Mean 0.6686 N 16 df 15 t 1.6266 P-value (one-tailed) 0.06232 t Critical (one-tailed) 1.75305 Question 2: According to the above SPSS result, the mean average score of the children in the given area is 3.89. Since the test wants to determine if the children in the given area have higher scores than children in the general population, who normally score of 2.3, the one-tailed test should be used in this case. Since the population of sample is less than 30, we applied t-test statistics in this case, and the t-test value can be computed according to the formula: t= X − / n 3.89 − 2.3 2.67 / 16 t = 1.62 t= The p-value for one-tailed t test with level of significance = 0.05 is 0.062. Since this p-value is greater t ...
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