Answer to Question #255196 in Statistics and Probability for Thafeng

Question

a)   A random sample of employee files is drawn revealing an average of 2.8 overtime hours worked per week with a standard deviation of 0.7. The sample size is 500.

(i)   What is the standard error of the mean based on this information?                        (2)

(ii) What is the 98% confidence interval for the number of overtime hours worked?                                                                                                                                           (5)

How would your result be affected if the sample size had been 100?                   

Answer 1

(i)

$$SE=\dfrac{s}{\sqrt{n}}=\dfrac{0.7}{\sqrt{500}}\approx0.031305$$

(ii) The critical value for $\alpha = 0.02$ and $df = n-1 = 500-1=499$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.333844.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(2.8-2.333844\times\dfrac{0.7}{\sqrt{500}}, 2.8+2.333844\times\dfrac{0.7}{\sqrt{500}})$$

$$=(2.727, 2.873)$$

Therefore, based on the data provided, the 98% confidence interval for the population mean is $2.727 < \mu < 2.873,$ which indicates that we are 98% confident that the true population mean $\mu$  is contained by the interval $(2.727, 2.873)$

(iii) The critical value for $\alpha = 0.02$ and $df = n-1 = 100-1=99$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.364606.$ The corresponding confidence interval is computed as shown below:

$$CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(2.8-2.364606\times\dfrac{0.7}{\sqrt{100}}, 2.8+2.364606\times\dfrac{0.7}{\sqrt{100}})$$

$$=(2.6345, 2.9655)$$

Therefore, based on the data provided, the 98% confidence interval for the population mean is $2.6345 < \mu < 2.9655,$ which indicates that we are 98% confident that the true population mean $\mu$  is contained by the interval $(2.6345, 2.9655)$

Decreasing the sample size causes the error bound to increase, making the confidence interval wider.