Step 1
The sample size is 19, mean is R18500, standard deviation is R1750.
The degrees of freedom is,
\(\displaystyle{d}{f}={n}-{1}\)

\(\displaystyle={19}-{1}\)

\(\displaystyle={18}\) The degrees of freedom is 18. The confidence level is 95%, than the level of significance is 0.05. Computation of critical value: The critical value of t-distribution for 0.05 level of significance at 18 degrees of freedom can be obtained using the excel formula “=T.INV.2T(0.05,18)”. The critical value is 2.1009. Step 2 The 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty is, \(\displaystyle{C}{I}=\overline{{{x}}}\pm{t}_{{{\frac{{\alpha}}{{{2}}}}}}{\left({\frac{{{s}}}{{\sqrt{{{n}}}}}}\right)}\)

\(\displaystyle={18500}\pm{2.1009}{\left({\frac{{{1750}}}{{\sqrt{{{19}}}}}}\right)}\)

\(\displaystyle={18500}\pm{843.464}\)

\(\displaystyle={\left({18500}-{843.464},{18500}+{843.464}\right)}\)

\(\displaystyle={\left({17656.536},{19343.464}\right)}\)

\(\displaystyle\approx{\left({17657},{19343}\right)}\) Thus, the 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty is (17657,19343). Interpretation: When repeated samples of size 19 are taken, about 95% of those confidence intervals would contain the true average monthly salary earned by all employees of People Plus Pty that lies between 17657 and 19343. That is, there is 95% confidence that the true average monthly salary earned by all employees of People Plus Pty that lies between 17657 and 19343.

\(\displaystyle={19}-{1}\)

\(\displaystyle={18}\) The degrees of freedom is 18. The confidence level is 95%, than the level of significance is 0.05. Computation of critical value: The critical value of t-distribution for 0.05 level of significance at 18 degrees of freedom can be obtained using the excel formula “=T.INV.2T(0.05,18)”. The critical value is 2.1009. Step 2 The 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty is, \(\displaystyle{C}{I}=\overline{{{x}}}\pm{t}_{{{\frac{{\alpha}}{{{2}}}}}}{\left({\frac{{{s}}}{{\sqrt{{{n}}}}}}\right)}\)

\(\displaystyle={18500}\pm{2.1009}{\left({\frac{{{1750}}}{{\sqrt{{{19}}}}}}\right)}\)

\(\displaystyle={18500}\pm{843.464}\)

\(\displaystyle={\left({18500}-{843.464},{18500}+{843.464}\right)}\)

\(\displaystyle={\left({17656.536},{19343.464}\right)}\)

\(\displaystyle\approx{\left({17657},{19343}\right)}\) Thus, the 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty is (17657,19343). Interpretation: When repeated samples of size 19 are taken, about 95% of those confidence intervals would contain the true average monthly salary earned by all employees of People Plus Pty that lies between 17657 and 19343. That is, there is 95% confidence that the true average monthly salary earned by all employees of People Plus Pty that lies between 17657 and 19343.