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,

\(\displaystyle{x}{1}={192}\)

\(\displaystyle{n}{1}={423}\)

\(\displaystyle{x}{2}={167}\)

\(\displaystyle{n}{2}={192}\)

\(\displaystyle\hat{{p}}_{{1}}=\frac{{{x}{1}}}{{{n}{1}}}=\frac{192}{{423}}={0.4539}\)

\(\displaystyle\hat{{p}}_{{2}}=\frac{{{x}{2}}}{{{x}{2}}}=\frac{167}{{192}}={0.8698}\)

\(\displaystyle\hat{{p}}=\frac{{{x}{1}+{x}{2}}}{{{n}{1}+{n}{2}}}=\frac{{{192}+{167}}}{{{423}+{192}}}={0.5837}\)

Test statistic \(\displaystyle=\frac{{\hat{{p}}_{{1}}-\hat{{p}}_{{2}}}}{\sqrt{{\hat{{p}}{\left({1}-\hat{{p}}\right)}{\left(\frac{1}{{{n}{1}}}+\frac{1}{{{n}{2}}}\right)}}}}\)

\(\displaystyle=\frac{{{0.4359}-{0.8698}}}{\sqrt{{{0.5837}{\left({1}-{0.5837}\right)}{\left(\frac{1}{{423}}+\frac{1}{{192}}\right)}}}}\)

\(\displaystyle=-{9.696}\)

b)

p value:

\(\displaystyle{p}={2}{P}{\left({z}<\text{test statistic}\right)}\)</span>

\(\displaystyle={2}{P}{\left({z}<-{9.696}\right)}\)</span>

\(\displaystyle={2}\cdot{0.0000}\) from the standard normal table

\(\displaystyle={0.0000}\)

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)

\(99\%\) 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 \(99\%\) confidence is 2.576

\(\displaystyle{\left(\hat{{p}}_{{1}}-\hat{{p}}_{{2}}\right)}\pm{\left({z}\sqrt{{\hat{{p}}{\left({1}-\hat{{p}}\right)}{\left(\frac{1}{{{n}{1}}}+\frac{1}{{{n}{2}}}\right)}}}={0.4593}-{0.8698}\right.}\)

\(\displaystyle\pm{\left({2.576}\sqrt{{{0.5837}\cdot{\left({1}-{0.5837}\right)}{\left(\frac{1}{{423}}+\frac{1}{{192}}\right)}}}\right)}\)

\(\displaystyle=-{0.4159}\pm{0.1150}\)

\(\displaystyle={\left(-{0.5264},-{0.3054}\right)}\)

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,

\(\displaystyle{x}{1}={192}\)

\(\displaystyle{n}{1}={423}\)

\(\displaystyle{x}{2}={167}\)

\(\displaystyle{n}{2}={192}\)

\(\displaystyle\hat{{p}}_{{1}}=\frac{{{x}{1}}}{{{n}{1}}}=\frac{192}{{423}}={0.4539}\)

\(\displaystyle\hat{{p}}_{{2}}=\frac{{{x}{2}}}{{{x}{2}}}=\frac{167}{{192}}={0.8698}\)

\(\displaystyle\hat{{p}}=\frac{{{x}{1}+{x}{2}}}{{{n}{1}+{n}{2}}}=\frac{{{192}+{167}}}{{{423}+{192}}}={0.5837}\)

Test statistic \(\displaystyle=\frac{{\hat{{p}}_{{1}}-\hat{{p}}_{{2}}}}{\sqrt{{\hat{{p}}{\left({1}-\hat{{p}}\right)}{\left(\frac{1}{{{n}{1}}}+\frac{1}{{{n}{2}}}\right)}}}}\)

\(\displaystyle=\frac{{{0.4359}-{0.8698}}}{\sqrt{{{0.5837}{\left({1}-{0.5837}\right)}{\left(\frac{1}{{423}}+\frac{1}{{192}}\right)}}}}\)

\(\displaystyle=-{9.696}\)

b)

p value:

\(\displaystyle{p}={2}{P}{\left({z}<\text{test statistic}\right)}\)</span>

\(\displaystyle={2}{P}{\left({z}<-{9.696}\right)}\)</span>

\(\displaystyle={2}\cdot{0.0000}\) from the standard normal table

\(\displaystyle={0.0000}\)

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)

\(99\%\) 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 \(99\%\) confidence is 2.576

\(\displaystyle{\left(\hat{{p}}_{{1}}-\hat{{p}}_{{2}}\right)}\pm{\left({z}\sqrt{{\hat{{p}}{\left({1}-\hat{{p}}\right)}{\left(\frac{1}{{{n}{1}}}+\frac{1}{{{n}{2}}}\right)}}}={0.4593}-{0.8698}\right.}\)

\(\displaystyle\pm{\left({2.576}\sqrt{{{0.5837}\cdot{\left({1}-{0.5837}\right)}{\left(\frac{1}{{423}}+\frac{1}{{192}}\right)}}}\right)}\)

\(\displaystyle=-{0.4159}\pm{0.1150}\)

\(\displaystyle={\left(-{0.5264},-{0.3054}\right)}\)