Inferential Statistics 2018: Questions 1-4

ASSIGNMENT 2: INFERENTIAL STATISTICS (2018)
Question 1:
Sample size, n = 26
Sample Mean, X = 23.45
Sample Std. Dev, s = 2.34
(a) = 0.05 => [(1- )*100] % = 95%. The 95% confidence interval for the population
mean is given by: Sample mean ± z* standard error
 Standard error of the mean =

Confidence Interval = 23.45 ± (1.96 * 0.4589) = 23.45 ± 0.899 = [22.55, 24.35]
(b) The confidence interval above means that we can be 95% certain that the true population
mean BMI of all male track athletes lie between the values 22.55 and 24.35
Question 2:
Sample size, n = 145
Sample Mean, X = 22.91
Sample Std. Dev, s = 1.98
(a) = 0.05 => [(1- )*100] % = 95%
 Standard error of the mean =

Confidence Interval = 22.91 ± (1.96 * 0.1644) = 22.91 ± 0.32 = [22.59, 23.23]
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(b) The confidence interval from question 1 has a range of 1.8 (24.35 – 22.55), while the
range of the confidence interval of question 2 is 0.64. This could be attributed to the
difference in size of the wo samples. The sample size in question 1 is 26, which is much
smaller compared to the sample size in question 2 of 145. Given the confidence level is
equal for both questions. Therefore, the larger sample size gives more certainty that the
sample estimates reflects the population. Consequently, the larger sample size will have a
narrower confidence interval.
Question 3:
Mean time per trip = 720 minutes
Claim: mean time was more than 720 minutes
Sample size = 25 trips
Sample mean = 735 minutes
Sample standard deviation = 37.2 minutes
(a) The appropriate distribution to use is the t-distribution because the sample size is small; n
< 30. Moreover, the population variance is unknown.
(b) The hypotheses are stated as:
Null hypothesis, Ho: μ = 720
Alternative hypothesis, H1: μ > 720
(c) This is a one-tailed test. Hence, the critical value, at 0.05 significance level is: Z0.05 =
+1.645
(d) The sample test statistic is computed as:
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