Transport company PlusBus run a rush-hour bus service to and from the city centr
ID: 3235866 • Letter: T
Question
Transport company PlusBus run a rush-hour bus service to and from the city centre train station to the out of town business park. To improve their service they are investigating two possible routes (labelled A and B). A random sample of 30 journeys by each route were collected with the following sample means and standard deviations. a) Carry out an appropriate parametric statistical test (i.e. Z or t) of whether or not there is a difference between the mean travel times by the two routes. What are the assumptions required to carry out this test? b) Suppose that the travel time data above is paired, i.e. the first observation for route A is paired with the first observation for route B, and so on. In addition, the mean and standard deviation of the differences are -5 and 12 mins respectively. Carry out an appropriate statistical test (i.e. Z or t) of whether or not there is a difference between the mean travel times by the two routes. What are the assumptions required to carry out this test? c) Provide examples of how the paired data referred to in part (b) might have been collected in practice. d) Explain how your answers to parts (a) and (b) would have differed if the sample means and sample standard deviations had been based on only 15 journeys rather than 30 journeys for each of the two routes. There is no need to do any further calculations.Explanation / Answer
Solution:-
a) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 = 2
Alternative hypothesis: 1 2
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = sqrt[(s12/n1) + (s22/n2)]
SE = 3.383
DF = 58
t = [ (x1 - x2) - d ] / SE
t = - 1.18
where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 58 degrees of freedom is more extreme than -1.18; that is, less than -1.18 or greater than 1.18.
Thus, the P-value = 0.2682
Interpret results. Since the P-value (0.268) is greater than the significance level (0.05), we have to accept the null hypothesis.
From the above test we do not have sufficient evidence in the favor of the claim that mean of two route is significantly different.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: d = 0
Alternative hypothesis: d 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).
s = sqrt [ ((di - d)2 / (n - 1) ]
s = 12
SE = s / sqrt(n)
S.E = 2.191
DF = n - 1 = 30 -1
D.F = 29
t = [ (x1 - x2) - D ] / SE
t = - 2.28
where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 29 degrees of freedom is more extreme than - 2.28; that is, less than - 2.28 or greater than - 2.28.
Thus, the P-value = 0.0302
Interpret results. Since the P-value (0.0302) is less than the significance level (0.05), we have to reject the null hypothesis.
d) If the sample size is 15 for each route than in both the cases the value of t test statistics would decrease due to increase in standard error.
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