The manager of a factory has devised a detailed plan for evacuating the building
ID: 3230469 • Letter: T
Question
The manager of a factory has devised a detailed plan for evacuating the building as quickly as possible in the event of a fire or other emergency. An industrial psychologist believes that workers actually leave the factory faster at closing time without following any system. The company holds fire drills periodically in which a bell sounds and workers leave the building according to the system. The evacuation time for each drill is recorded. For comparison, the psychologist also records the evacuation time when the bell sounds for closing time each day. A random sample of 37 fire drills showed a mean evacuation time of 5.3 minutes with a standard deviation of 1.1 minutes. A random sample of 37 days at closing time showed a mean evacuation time of 4.5 minutes with a standard deviation of 1.0 minutes. a. Construct a 95% confidence interval for the difference between the two population means. b. Test at the 1% significance level whether the mean evacuation time is smaller at closing time than during fire drills.Explanation / Answer
Given,
Fire drill values
x1=5.3
s1=1.1
n=37
closing time values
x2=4.5
s2=1
n=37
A.a) 95% confidence interval is given as
x1-x2 ± t(n1+n2-2)*SE
SE=sqrt{(n1-1)*s1^2 + (n2-1)*s2^2/n1+n2-2}
SE=sqrt(36*1.1*1.1+36*1*1/72
solving we get
SE=1.025
t(n1+n2-2)=t-value at df=72
t(72)=1.994
Confidence interval is
CI=(5.3-4.5)±1.994*1.025
CI=0.8±2.04
Confidence interval is (-1.24,2.84)
A.b) H0: µ1-µ2>0
H1: µ1-µ20
t-statistic is given as
t=(x1-x2)-(µ1-µ2)/sqrt[{s1^2*(n1-1)+s2^2*(n2-1)}/n1+n2-2]
=0.8-0/sqrt(1.1*1.1*36+1*1*36/72)
solving we get
t-value=0.76
t-value from table is
t(0.01,72)=2.38(at 0.01 significance level)
as observed value is less than table value, we reject the null hypothesis
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