In the early 1990s, researchers in the UK collected data on traffic accident rel
ID: 3216823 • Letter: I
Question
In the early 1990s, researchers in the UK collected data on traffic accident related emergency room admissions at a particular hospital on Friday the 13th and on the previous Friday, Friday the 6th. The data is shown to the right. (a) Perform a hypothesis test to determine if the average number of accident related emergency room admissions is greater on Friday the 13th than on Friday the 6th. Let alpha = .05. Structure your test so that a conclusion that Friday the 13th is more dangerous than Friday the 6th (if in fact that is the conclusion) is the strongest statistical argument that you could make. (b) You cannot directly compute the p-value associated with the hypothesis test in 6a; however, by using the t-table you can provide bounds on the p-value. What are the bounds? (c) Construct a 95% confidence interval on the difference in mean accident related emergency room admissions between the two Fridays. (d) Does your confidence interval in 6c allow you to reach the same (overall) conclusion as your hypothesis test in 6a? Briefly state why (or why not)?Explanation / Answer
H0 = There is no significant difference between mean number of accidents related emergency room admission on Friday the 13th and Friday the 6th That is µ1 = µ2
H1 = The mean number of accidents related emergency room admission is greater on Friday the 13th than on Friday the 6th That is µ1 > µ2
Means:
x1 = 10.8333
x2 = 7.5
Sample variance:
s12 = 12.96667
s22= 11.1
Assuming the population variances are equal, the test statistics is given by:
t = x1 - x2 / sp(1/n1 + 1/n2)
where sp is the pooled standard deviation.
Sp2 = (n1-1)s12 + (n2-1)s22 / (n1 + n2 -2)
= (5*12.96667)+(5*11.1)/10
=12.03334
Sp = 3.46891
t = 10.83333-7.5/3.46891*0.57735
= 1.665316
= 1-0.95 = 0.05
d.f = 6+6-2 = 10
t critical value = t0.05,10 = 1.812
p-value = 0.0635
since, tcal = 1.665316 < ttab = 1.812 and p-value > =0.05, we accept the null hypothesis and conclude that There is no significant difference between mean number of accidents related emergency room admission on Friday the 13th and Friday the 6th .
b) The df= 10 and the t statistic is t=1.665
From the t table, we notice that
1.372 < 1.665 < 1.812
Converting to P-values
0.10 > P( t > 1.665) > 0.05
Therefore, 0.05 < P-value < 0.10
c) 95% confidence interval is given by:
( x1 - x2 ) ± (t/2, n1+n2 -2) Sp(1/n1 + 1/n2)
(10.8333-7.5) ± t0.025,10 * 3.46891 *0.57735
From t table, we notice that
t0.025,10 =2.228
95% confidence interval:
(10.8333-7.5) ± 2.228 * 3.46891 *0.57735
3.333 ± 4.462183
The 95% confidence interval is : (-1.12885,7.795516)
d) We can be 95% confidence that the true difference in mean lies between -1.12885 and 7.795516.
Since the interval contains 0, we can conclude that the difference between the two means is insignificant that is there is no difference between the two means.