# The International Business Society conducts a series of managerial surveys to detect challenges that international managers encounter and what strateg

The International Business Society conducts a series of managerial surveys to detect challenges that international managers encounter and what strategies help them overcome those challenges. A previous study showed that employee retention is the most critical organizational challenge currently confronted by international managers. One strategy that may impact employee retention is a successful employee recognition program. The International Business Society randomly selected 423 small firms and 192 large firms. The study showed that 326 of the 423 small firms have employee retention programs compared to 167 of the 192 large organizations.
a) At the 0.01 level of significance, is there evidence of a significant difference between small firms and large firms concerning the proportion that has employee recognition program?
b) Find the p-value and interpret its meaning?
c) Construct and interpret a $99\mathrm{%}$ confidence interval estimate for the difference between small firms and large firms concerning the proportion of employee recognition program.
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a)
Null Hypothesis:
H0: There is no evidence of a significant difference between small firms and large firms concerning the proportion that has employee recognition program.
Alternative Hypothesis:
H1: There is a evidence of a significant difference between small firms and large firms concerning the proportion that has employee recognition program.
From the given information,
$x1=192$
$n1=423$
$x2=167$
$n2=192$
${\stackrel{^}{p}}_{1}=\frac{x1}{n1}=\frac{192}{423}=0.4539$
${\stackrel{^}{p}}_{2}=\frac{x2}{x2}=\frac{167}{192}=0.8698$
$\stackrel{^}{p}=\frac{x1+x2}{n1+n2}=\frac{192+167}{423+192}=0.5837$
Test statistic $=\frac{{\stackrel{^}{p}}_{1}-{\stackrel{^}{p}}_{2}}{\sqrt{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)\left(\frac{1}{n1}+\frac{1}{n2}\right)}}$
$=\frac{0.4359-0.8698}{\sqrt{0.5837\left(1-0.5837\right)\left(\frac{1}{423}+\frac{1}{192}\right)}}$
$=-9.696$
b)
p value:
$p=2P\left(z<\text{test statistic}\right)$
$=2P\left(z<-9.696\right)$
$=2\cdot 0.0000$ from the standard normal table
$=0.0000$
Interpretation:
The p value is less than 0.05, hence the results are statistically significant.
There is a evidence of a significant difference between small firms and large firms concerning the proportion that has employee recognition program.
c)
$99\mathrm{%}$ confidence interval estimate for the difference between small firms and large firms concerning the proportion of employee recognition program:
The z critical value corresponding with $99\mathrm{%}$ confidence is 2.576
$\left({\stackrel{^}{p}}_{1}-{\stackrel{^}{p}}_{2}\right)±\left(z\sqrt{\stackrel{^}{p}\left(1-\stackrel{^}{p}\right)\left(\frac{1}{n1}+\frac{1}{n2}\right)}=0.4593-0.8698$
$±\left(2.576\sqrt{0.5837\cdot \left(1-0.5837\right)\left(\frac{1}{423}+\frac{1}{192}\right)}\right)$
$=-0.4159±0.1150$
$=\left(-0.5264,-0.3054\right)$