Statistical Reasoning Assignment: t-tests, Errors, and Confidence

This document presents a comprehensive solution to a statistical reasoning assignment, focusing on the application of t-tests for hypothesis testing and confidence interval estimation. The assignment explores concepts such as one-tailed and two-tailed tests, null and alternative hypotheses, and the calculation of t-statistics. It covers the interpretation of critical values, degrees of freedom, and the identification of Type I and Type II errors. The solution provides step-by-step calculations, including standard error determination and the construction of confidence intervals. Furthermore, the assignment assesses the student's ability to determine statistical significance, make informed decisions based on the results, and understand the implications of these statistical findings in practical scenarios, such as determining the effect of a drug on blood pressure. The assignment also examines the central limit theorem and its application to determine standard error and the effects of sample size. The provided solution offers a detailed explanation of the methods and concepts used to solve the problems.

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
¿
+tcrit∗ 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:
[4]

[ x
¿
−tcrit∗ s
√ n , x
¿
+tcrit∗ s
√ n ] =
[ 72. 7−2. 797∗13 .55
√ 25 , 72. 7+2 . 797∗13 . 55
√ 25 ] = [ 65. 12 , 80. 28 ]
_____65.12____ ≤ μ ≤ ___80.28_______,
Above or greater than μ=80 . 28
Answer 15 Type I error = level of significance = 0.05
Answer 16 Fail to reject null, when it is really true = 1−α=1−0 .08=0 . 92
Answer 17 Rejecting the null hypothesis, when it is false = Power of the test = β =
0.14
Answer 18 Type II error = 1- Power of the test (β) = 1 – 0.13 = 0.87
Answer 19 Sample standard deviation given, so appropriate statistical procedure for one
sample comparison with population = t-test
Answer 20 To determine whether people with hypertension taking Atenolol have
significantly lower systolic pressure than people who do not consume Atenolol, the test will
be = one-tailed
Answer 21 People taking Atenolol will experience a significantly lower systolic blood
pressure when compared to the general population of people who do not take Atenolol.
Steps: Type of test: One-tailed
Choice of test: t-test
Null hypothesis: μ=165
Alternate hypothesis: μ<165
Level of significance: α=0 . 05
[5]

Sample mean = x
¿
=147
Sample size = n=30
Sample standard deviation = s=6
Calculated t-stat =
x
¿
−μ
s / √ n =147−165
6 / √ 30 =−16 . 43 < t-critical = -1.699
Decision: Hence, null hypothesis rejected.
Answer 22 μAtenolol < 165
Answer 23 This is a left-tailed test. Degrees of freedom = 147 – 1 = 146. Level of
significance = 0.01. The t-critical = -2.462
Answer 24 Standard Error =
s
√ n =20. 57
√ 124 =
1.847
Steps: Type of test: One-tailed
Choice of test: t-test
Null hypothesis: μ=165
Alternate hypothesis: μ<165
Level of significance: α=0 . 05
Sample mean = x
¿
=147
Sample size = n=124
Sample standard deviation = s=20 . 57
Calculated t-stat =
x
¿
−μ
s / √ n =147−165
20. 57/ √ 124 =−9 . 744 < t-critical = -1.657
Decision: Hence, null hypothesis rejected.
[6]

Answer 25 t-value = -31.426, t-critical = -1.65, Null hypothesis to be rejected.
Steps: Type of test: One-tailed
Choice of test: t-test
Null hypothesis: μ=175
Alternate hypothesis: μ<175
Level of significance: α=0 . 05
Sample mean = x
¿
=130
Sample size = n=258
Sample standard deviation = s=23
Calculated t-stat =
x
¿
−μ
s / √ n =130−175
20. 57/ √ 124 =−31 . 426 < t-critical = -1.657
Decision: Hence, null hypothesis rejected. Atenolol significantly lowers systolic blood
pressure.
Answer 26 Reject the null hypothesis
Steps: Type of test: One-tailed
Choice of test: t-test
Null hypothesis: μ=165
Alternate hypothesis: μ<165
Level of significance: α=0 . 01
Sample mean = x
¿
=147
Sample size = n=30
[7]

Sample standard deviation = s=6
Calculated t-stat =
x
¿
−μ
s / √ n =147−165
6 / √ 30 =−16 . 43 < t-critical = -2.46
Decision: Hence, null hypothesis rejected.
Answer 27 There is a significant difference in systolic blood pressure when
comparing people who take Atenolol to the general population of people who do not
take Atenolol.
Answer 28 Type I statistical error (as null hypothesis rejected)
Answer 29 Mean used = 146
Test: t-test, as sample standard deviation given.
Confidence interval = 95%
Degrees of freedom = n – 1 = 30 – 1 = 29
The t-critical value at α=0 . 05 = 2.045
Sample mean = x
¿
=146
Sample size = n=30
Sample standard deviation = s=6
Confidence interval:
[ x
¿
−tcrit∗ s
√ n , x
¿
+tcrit∗ s
√ n ] =
[ 146−2. 045∗ 6
√ 30 ,146+ 2. 045∗ 6
√ 30 ]= [ 143 .76 , 148 .24 ]
Population mean μ=165 is outside the 95% confidence interval. Hence, null hypothesis
rejected.
Answer 30 The t-critical value = +/- 2.045
Test: t-test, as sample standard deviation given.
[8]

Confidence interval = 95%
Degrees of freedom = n – 1 = 30 – 1 = 29
The t-critical value at α=0 . 05 = 2.045
Sample mean = x
¿
=147
Sample size = n=30
Sample standard deviation = s=6
Confidence interval:
[ x
¿
−tcrit∗ s
√ n , x
¿
+tcrit∗ s
√ n ] =
[ 147−2. 045∗ 6
√ 30 ,147+ 2. 045∗ 6
√ 30 ]= [ 144 . 76 , 149 . 24 ]
Population mean μ=165 is outside the 95% confidence interval. Hence, null hypothesis
rejected.
Answer 31
Test: t-test, as sample standard deviation given.
Confidence interval = 95%
Degrees of freedom = n – 1 = 30 – 1 = 29
The t-critical value at α=0 . 05 = 2.045
Sample mean = x
¿
=164
Sample size = n=30
Sample standard deviation = s=16
Confidence interval:
[ x
¿
−tcrit∗ s
√ n , x
¿
+t crit∗ s
√ n ] =
[ 164−2. 045∗16
√ 30 ,164 +2 .045∗16
√ 30 ] = [ 158. 03 , 169 . 97 ]
Population mean μ=157 is outside the 95% confidence interval. Hence, null hypothesis
rejected.
___158.27______ ≤ μ ≤ ___169.73____
Value below or less than μ= 158.27
[9]

Answer 32
Test: t-test, as sample standard deviation given.
Confidence interval = 95%
Degrees of freedom = n – 1 = 30 – 1 = 29
The t-critical value at α=0 . 05 = 2.045
Sample mean = x
¿
=173
Sample size = n=30
Sample standard deviation = s=14
Confidence interval:
[ x
¿
−tcrit∗ s
√ n , x
¿
+t crit∗ s
√ n ] =
[ 173−2. 045∗14
√ 30 , 173+2 . 045∗14
√ 30 ]= [ 167 .77 , 178 .23 ]
Population mean μ=152 is outside the 95% confidence interval. Hence, null hypothesis
rejected.
____167.77_____ ≤ μ ≤ ___178.23_______
Value greater than μ=152 = 178.23
Answer 33
Mean = 11 mint
Standard deviation = 14 mint
Sample size = 112
By Central Limit Theorem, standard error =
s tan dard deviation
√ sample size =14
√ 112 =1.323
Answer 34 By central limit theorem, mean of sampling distribution is equal to the
population mean = 12 minutes
[10]

Answer 35 Increase the number of subjects
Answer 36 Increases
Answer 37 Increases
Answer 38 Typically the population standard deviation is unknown.
Answer 39 Could the relationship observed in the sample have occurred by chance?
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