Question

Every year in Britain there is a No Smoking Day, where many people voluntarily stop smoking for a day. This No Smoking Day occurs on the second Wednesday of March each year. Data are collected about nonfatal injuries on the job, which allows a test of the hypothesis that stopping smoking affects the injury rate (Walters et al., 1998). Many factors affect injury rate, though, such as the year, time of week, etc., so we would like to be able to control some of these factors. One way to do this is to compare the injury rate on the Wednesday of the No Smoking Day to the rate for the previous Wednesday in the same years. Those data for 1987 to 1996 are listed in the following table:

Year	Injuries before No Smoking Day	Injuries on Smoking Day
1987	516	540
1988	610	620
1989	581	599
1990	586	639
1991	554	607
1992	632	603
1993	479	519
1994	583	560
1995	445	515
1996	522	556

a) How many more or fewer injuries are there on No Smoking Day, on average, compared to the normal day? b) What is the 99% confidence interval for this difference? c) In your own words, explain what the 99% confidence interval means. d) Test whether the accident rate changes on No Smoking Day.

Answer 1

a) Average injuries before no smoking day = 550.8

Average injuries on smoking day = 575.8

25 more injuries are there on no smoking day on average compared to normal day

b) for 99% confidence interval

t = 2.8784 with 18 df

M1-M2 = 25

s= 54.84614

therefore 99% confidence interval

= (-45.5991,95.5991)

c) there is 99% chance that the difference in average number of injuries lies between -45.60 to 95.60

d) the null and alternative hypothesis

test statistic

= 1.02

df = 18

p value = 0.3212

the result is not significant

we donot reject H0

there is no significant difference between average injuries in smoking and nonsmoking day

so we cannot say that accident rate changes significantly on No smoking day