Construct a 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty, if it was found that the average monthly salary earned by a sample of 19 employees of the company was R18 500, with a standard deviation of R1 750. Interpret your answer

Step 1 The sample size is 19, mean is R18500, standard deviation is R1750. The degrees of freedom is, ${\displaystyle df=n-1}$
${\displaystyle =19-1}$
${\displaystyle =18}$ 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, ${\displaystyle CI=\bar{x}\pm t_{\frac{\alpha }{2}}\left(\frac{s}{\sqrt{n}}\right)}$
${\displaystyle =18500\pm 2.1009\left(\frac{1750}{\sqrt{19}}\right)}$
${\displaystyle =18500\pm 843.464}$
${\displaystyle =(18500-843.464,18500+843.464)}$
${\displaystyle =(17656.536,19343.464)}$
${\displaystyle \approx (17657,19343)}$ 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.