 # The number of points TEDU basketball team scores against Stanford basketball team is normally distributed. Construct a 95\% two-sided confidence interval for the mean of the number of points scored by TEDU. UkusakazaL 2021-08-06 Answered
The number of points TEDU basketball team scores against Stanford basketball team is normally distributed. Suppose that over the last 10 games between the two teams TEDU scored the points given below:
a) Construct a $95\mathrm{%}$ two-sided confidence interval for the mean of the number of points scored by TEDU.
b) The coach of the TEDU team finds out that the variance of the number of points scored is 25. Given this new information, construct a $95\mathrm{%}$ two-sided confidence interval for the mean of the number of points scored by TEDU.
c) Compare the confidence intervals found in part (a) and (b). Explain the reason behind the difference.
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Part (a)
The table containing the mean, variance is shown below.
$\begin{array}{|ccc|}\hline X& X-\overline{X}& \left(X-\overline{X}{\right)}^{2}\\ 74& 9& 81\\ 59& -6& 36\\ 62& -3& 9\\ 61& -4& 16\\ 70& 5& 25\\ 62& -3& 9\\ 66& 1& 1\\ 62& -3& 9\\ 75& 10& 100\\ 59& -6& 36\\ Total& & 322\\ \hline\end{array}$
The mean is 65.
Use the equation $V\left(X\right)=\frac{\sum {\left(X-\stackrel{―}{X}\right)}^{2}}{n-1}$ to find the variance.
$V\left(X\right)=\frac{322}{9}$
$=35.78$
Use the equation $S.D=\sqrt{V\left(X\right)}$ to find the standard deviation.
$S.D=\sqrt{35.78}$
$=5.98$
The t value corresponding to $95\mathrm{%}$ confidence interval is 2.26.
Use the equation to construct the confidence interval.
$C.I=65±2.26\frac{{5.98}^{2}}{10}$
$=65±4.28$
Therefore, the upper and lower limits are $65+4.28$ and $65-4.28$
Part (b)
Given that, the variance of the distribution is 25.
$V\left(X\right)=25$
$S.D=\sqrt{25}$
$=5$
Use the equation to construct the interval.
$C.I=65±2.26×\frac{{5}^{2}}{10}$
$C.I=65±5.65$
The upper and lower limits of the confidence interval is $65+5.65$ and $65-5.65.$
Part (c)
The standard deviation is defined as the degree of variance of value from mean value.
In part (b) the standard deviation is higher compared to part (a).
Therefore, the deviation is higher. So, the confidence interval got widened.