# 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

Question
Confidence intervals
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

2021-02-25
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.

### Relevant Questions

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.

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