Mathematics, 27.09.2019 02:30 naomicervero
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.
Answers: 2
![\left[0.219, \hspace{0.1cm} 0.807\right]\\\\](/tpl/images/0266/6775/fb06a.png)
![\left [15.112, \hspace{0.3cm}15.688\right]](/tpl/images/0266/6775/5f77e.png)
for the population variance will be calculated.
![\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 ]\\\\](/tpl/images/0266/6775/01fae.png)
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]\\\\](/tpl/images/0266/6775/a3302.png)
![\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 ]\\\\](/tpl/images/0266/6775/2042b.png)
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] \\\\](/tpl/images/0266/6775/b11b8.png)
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An article in the journal of sports science (1987, vol. 5, pp. 261-271) presents the results of an i...
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