Respuesta :
Point estimate of the population mean number of sit-ups is [tex]18.933$[/tex].
The margin of error is 5.615.
The 95 percent confidence interval of the population mean number of sit-ups is [tex][13.318,24.548]\end{aligned}$[/tex]
As per the given question 15 individuals were selected and the number of sit-ups that each could complete was recorded.
The results are {27, 12, 18, 5, 16, 25, 30, 13, 35, 10, 1, 13, 33, 24, 22}
(a):
The sample mean is,
[tex]$\overline{\mathrm{X}}=\frac{\sum \mathrm{x}}{\mathrm{n}}\\\\[/tex]
[tex]=\frac{27+12+18+5+16+25+30+13+35+10+1+13+33+24+22}{15} \\\\[/tex]
[tex]=18.9333$[/tex]
The point estimate of the population mean number of sit-ups is,
Point estimate [tex]$=\overline{\mathrm{X}}=18.933$[/tex]
(b):
The sample standard deviation is,
[tex]$\mathrm{s}=\sqrt{\frac{\sum X^2-\frac{\left(\sum X\right)^2}{n}}{n-1}}[/tex]
[tex]$\mathrm{s}=\sqrt{\frac{18.9333^2-\frac{\left(18.333\right)^2}{15}}{15-1}}=10.1381$[/tex]
The confidence level is, c=0.95
The significance level is [tex]$\alpha=1-c=1-0.95=0.05$[/tex]
The degrees of freedom is given by,
[tex]$=\mathrm{n}-1\\=15-1\\=14$[/tex]
The critical value is, [tex]$t_{\frac{\alpha}{2} , d f}=t_{\frac{0.05}{2,14}}=2.145$[/tex]
The Excel function is, TINV(0.05,14)
The margin of error is,
[tex]$\begin{aligned}\mathrm{E} & =\mathrm{t}_{\frac{\alpha}{2}, \mathrm{df}} \times \frac{\mathrm{s}}{\sqrt{\mathrm{n}}} \\& =2.145 \times \frac{10.1381}{\sqrt{15}} \\& =5.615\end{aligned}$[/tex]
(c):
The 95% confidence interval of the population mean number of sit-ups is,
[tex]\begin{aligned}95 \% \text { C. I. } & =\overline{\mathrm{X}} \pm \mathbf{E} \\& =18.933 \pm 5.615 \\& =[13.318,24.548]\end{aligned}$[/tex]
Therefore, [tex]$\overline{\mathrm{X}}=18.9333[/tex], [tex]$\begin{aligned}\mathrm{E} & =5.615\end{aligned}$[/tex] and 95% confidence interval of the population mean number of sit-ups is [tex][13.318,24.548]\end{aligned}$[/tex].
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