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

defazajx

defazajx

Answered question

2021-02-24

If it was discovered 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, then create a 95% confidence interval for the true average monthly salary earned by all employees of People Plus Pty. Define your response.

Answer & Explanation

Ezra Herbert

Ezra Herbert

Skilled2021-02-25Added 99 answers

Step 1 The sample size is 19, mean is R18500, standard deviation is R1750. The degrees of freedom is, df=n1
=191
=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, CI=x±tα2(sn)
=18500±2.1009(175019)
=18500±843.464
=(18500843.464,18500+843.464)
=(17656.536,19343.464)
(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.

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