Question

The director of Human Resources at a large company wishes to determine if newly instituted training has been effective in reducing work-related injuries. In a random sample of 100 employees taken from the six months before this training began (group 1), she found 15 had suffered a work-related injury. Using a random sample of 150 employees from the six months since the training began (group 2), she found 12 had suffered a work-related injury.
a) Find the 95% confidence interval for this situation.  

b) Does it support the idea that the proportion of work-related injuries has decreased with the new training?

Show work - label all your values first before using technology.

Answer 1

a)

Here, , n1 = 100 , n2 = 150
p1cap = 0.15 , p2cap = 0.08

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.15 * (1-0.15)/100 + 0.08*(1-0.08)/150)
SE = 0.042

For 0.95 CI, z-value = 1.96
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.15 - 0.08 - 1.96*0.042, 0.15 - 0.08 + 1.96*0.042)
CI = (-0.0123 , 0.1523)

b)

As the confidence interval conatin hypothesiswed value 0 so, it does not support the idea

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 < p2

Rejection Region
This is left tailed test, for α = 0.05
Critical value of z is -1.64.
Hence reject H0 if z < -1.64

p1cap = X1/N1 = 15/100 = 0.15
p2cap = X2/N2 = 12/150 = 0.08
pcap = (X1 + X2)/(N1 + N2) = (15+12)/(100+150) = 0.108

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.15-0.08)/sqrt(0.108*(1-0.108)*(1/100 + 1/150))
z = 1.7469

P-value Approach
P-value = 0.9597
As P-value >= 0.05, fail to reject null hypothesis.