Question

An Article in the Journal of Sports Science (1987, Vol. 5, pp. 261-271) presents the results of an investigation of the hemoglobin level of Canadian Olympic ice hockey players. A random sample of 20 players is selected and the hemoglobin level is measured. The resulting sample mean and standard deviation(g/dl) are and Use this information to calculate a 95% two-sided confidence interval on the mean hemoglobin level and a 95% two-sided confidence interval on the variance. Assume the data are normally distributed. Round your answers to 2 decimal places.

Answer 1

Answer:

The 95% confidence interval for the population variance is $\left[0.219, \hspace{0.1cm} 0.807\right]$

The 95% confidence interval for the population mean is $\left [15.112, \hspace{0.3cm}15.688\right]$

Step-by-step explanation:

To solve this problem, a confidence interval of $(1-\alpha) \times 100%$ for the population variance will be calculated.

$Sample variance: S^2=(0.6152)^2 \\Sample size n=20 \\Confidence level (1-\alpha)\times100\%=95\% \\ \alpha: \alpha=0.05 \\ \chi^2 values (for a 95\% confidence and n-1 degree of freedom)\\ \chi^2_{\left (1-\frac{\alpha}{2};n-1\right )}=\chi^2_{(0.975;19)}=8.907\\ \chi^2_{\left (\frac{\alpha}{2};n-1\right )}=\chi^2_{(0.025;19)}=32.852$

Then, the $(1-\alpha) \times 100%$ confidence interval for the population variance is given by:

$\left [\frac{(n-1)S^2}{\chi^2_{\left (\frac{\alpha}{2};n-1\right )}}, \hspace{0.3cm}\frac{(n-1)S^2}{\chi^2_{\left (1-\frac{\alpha}{2};n-1\right )}} \right ]$Thus, the 95% confidence interval for the population variance is:$\\\\\left [\frac{(19-1)(0.6152)^2}{32.852}, \hspace{0.1cm}\frac{(19-1)(0.6152)^2}{8.907} \right ]=\left[0.219, \hspace{0.1cm} 0.807\right]$

On other hand,

A confidence interval of $(1-\alpha) \times 100%$ for the population mean will be calculated

$Sample mean: \bar X=15.40 \\Sample variance: S^2=(0.6152)^2 \\Sample size n=20 \\Confidence level (1-\alpha)\times100\%=95\% \\ \alpha: \alpha=0.05 \\T values (for a 95\% confidence and n-1 degree of freedom) T_{(\alpha/2;n-1)}=T_{(0.025;19)}=2.093\\\\ Then, the (1-\alpha) \times 100 \% confidence interval for the population mean is given by:$

\$\left[ \bar X - T_{(\alpha/2;n-1}\sqrt{\frac{\S^2}{n}}, \hspace{0.3cm}\bar X + T_{(\alpha/2;n-1}\sqrt{\frac{\S^2}{n}} \right ]$Thus, the 95\% confidence interval for the population mean is:$\\\\\left [15.40 - 2.093\sqrt{\frac{(0.6152)^2}{19}}, \hspace{0.3cm}15.40 + 2.093\sqrt{\frac{(0.6152)^2}{19}} \right ]=\left [15.112, \hspace{0.3cm}15.688\right]$