Question #255196

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?                   


1
Expert's answer
2021-10-25T14:44:15-0400

(i)

SE=sn=0.75000.031305SE=\dfrac{s}{\sqrt{n}}=\dfrac{0.7}{\sqrt{500}}\approx0.031305

(ii) The critical value for α=0.02\alpha = 0.02 and df=n1=5001=499df = n-1 = 500-1=499 degrees of freedom is tc=z1α/2;n1=2.333844.t_c = z_{1-\alpha/2; n-1} = 2.333844. The corresponding confidence interval is computed as shown below:


CI=(Xˉtc×sn,Xˉ+tc×sn)CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})

=(2.82.333844×0.7500,2.8+2.333844×0.7500)=(2.8-2.333844\times\dfrac{0.7}{\sqrt{500}}, 2.8+2.333844\times\dfrac{0.7}{\sqrt{500}})

=(2.727,2.873)=(2.727, 2.873)

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


(iii) The critical value for α=0.02\alpha = 0.02 and df=n1=1001=99df = n-1 = 100-1=99 degrees of freedom is tc=z1α/2;n1=2.364606.t_c = z_{1-\alpha/2; n-1} = 2.364606. The corresponding confidence interval is computed as shown below:


CI=(Xˉtc×sn,Xˉ+tc×sn)CI=(\bar{X}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{X}+t_c\times\dfrac{s}{\sqrt{n}})

=(2.82.364606×0.7100,2.8+2.364606×0.7100)=(2.8-2.364606\times\dfrac{0.7}{\sqrt{100}}, 2.8+2.364606\times\dfrac{0.7}{\sqrt{100}})

=(2.6345,2.9655)=(2.6345, 2.9655)

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


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



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