Statistical Reasoning

Statistical Reasoning
[1]

Answer 1 t-test (as sample standard deviation provided, and population standard
deviation missing)
Answer 2 two-tailed (as difference in sample and population verified)
Answer 3 There is no statistical difference in the amount of empathy expressed
when comparing individuals who were bullied to the general population.
Answer 4 μ bullied = 30.6 (population mean)
Answer 5 Critical Value at α=0 . 05 = ±2.064
Type of test: Two-tailed
Choice of test: t-test
Null hypothesis: μ=30 . 6
Alternate hypothesis: μ≠30 . 6
Level of significance: α=0 . 05
Sample mean = x
¿
=39 .5
Sample size = n=25
Sample standard deviation = s=6 . 6
Calculated t-stat =
x
¿
−μ
s / √ n =39 . 5−30 . 6
6 . 6/ √ 25 =6 .742 > t-critical = 2.064
Decision: Hence, null hypothesis rejected.
Answer 6 Degrees of freedom = n−1=121−1= 120
[2]

Answer 7 t-stat =
x
¿
−μ
s / √ n =60. 41−69 .93
9. 13/ √ 80 = -9.3263 (as sample standard deviation
provided).
Answer 8 Reject the null hypothesis
Type of test: Two-tailed
Choice of test: t-test
Null hypothesis: μ=30 . 6
Alternate hypothesis: μ≠30 . 6
Level of significance: α=0 . 05
Sample mean = x
¿
=39 .5
Sample size = n=25
Sample standard deviation = s=6 . 6
Calculated t-stat =
x
¿
−μ
s / √ n =39 . 5−30 . 6
6 . 6/ √ 25 =6 .742 > t-critical = 2.064
Decision: Hence, null hypothesis rejected.
Answer 9 There is a statistical difference in the amount of empathy expressed by
individuals who were bullied when compared to the general population.
Answer 10 Type I statistical error
Answer 11 37.5 (always the sample mean)
Answer 12 Degrees of freedom = 25 – 1 =24, t-critical for α=0 . 01 = ±2.797
[3]

Answer 13
Steps:
Test: t-test, as sample standard deviation given.
Confidence interval = 99%
Degrees of freedom = n – 1 = 25 – 1 = 24
The t-critical value at α=0 . 01 = 2.797
Sample mean = x
¿
=45 . 55
Sample size = n=25
Sample standard deviation = s=12 .72
Confidence interval:
[ x
¿
−tcrit∗ s
√ n , x
¿
+t crit∗ s
√ n ] =
[ 45 . 55−2 . 797∗12 .72
√ 25 , 45. 55+2 .797∗12 .72
√ 25 ] = [ 38. 43 ,52 .66 ]
_____38.43____ ≤ μ ≤ ___52.6_______,
Below or less than μ=38 . 43
Answer 14 Above or greater than μ=80 . 28
Steps:
Test: t-test, as sample standard deviation given.
Confidence interval = 99%
Degrees of freedom = n – 1 = 25 – 1 = 24
The t-critical value at α=0 . 01 = 2.797
Sample mean = x
¿
=72. 7
Sample size = n=25
Sample standard deviation = s=13 . 55
Confidence interval:
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