The following are scores on IQ tests of a random sample of 18 students at a large eastern university.

130, 122, 119, 142, 136, 127, 120, 152, 141, 132, 127, 118, 150, 141, 133, 137, 129, 142

a) Construct a 95 percent confidence interval estimate of the average IQ score of all students at the university.

b) Construct a 95 percent lower confidence interval estimate.

c) Construct a 95 percent upper confidence interval estimate.

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(a)From the given information,Let the sample size n=18The objective of the study is to construct a 95 \% confidence interval estimate of the average IQ score of all students at the university.The 95 \% confidence level for the average IQ score of all students at the university is,\bar{x}-t_{\alpha / 2}\left(\frac{s}{\sqrt{n}}\right)<\mu<\bar{x}+t_{\alpha / 2}\left(\frac{s}{\sqrt{n}}\right)Here,\bar{x} be the sample mean.s be sample standard deviation.The sample mean for the given data is,\begin{aligned}\bar{x} & =\frac{\sum x}{n} \\& =\frac{130+122+119+\cdots+142}{18} \\& =\frac{2398}{18} \\& =133.22\end{aligned}The sample standard deviation for the given data is,\begin{aligned}s & =\sqrt{\frac{1}{n-1} \sum(x-\bar{x})^{2}} \\& =\sqrt{\frac{1}{(18-1)}\left[(130-133.22)^{2}+(122-133.22)^{2}+\cdots+(142-133.22)^{2}\right]} \\& =10.2128\end{aligned}The degrees of freedom is,\begin{aligned}d f & =n-1 \\& =18-1 \\& =17\end{aligned}Since \sigma is unknown and s must replace it, then the t distribution is used for the confidence interval. Hence, at 95 \% confidence level with 17 degrees of freedom t_{\alpha / 2}=2.11.Substitute the values is the above equation, that is,\begin{aligned}\bar{x}-t_{\alpha / 2}\left(\frac{s}{\sqrt{n}}\right) & <\mu<\bar{x}+t_{\alpha / 2}\left(\frac{s}{\sqrt{n}}\right) \\133.22-2.11\left(\frac{10.2128}{\sqrt{18}}\right) & <\mu<133.22+2.11\left(\frac{10.2128}{\sqrt{18}}\right) \\133.22-5.0791 & <\mu<133.22+5.0791 \\128.14 & <\mu<138.30\end{aligned}Hence, the 95 \% confidence interval estimate of the average IQ score of all students at the university is in between 128.14 to 138.30 .(b)The objective is to construct the 95 \% lower confidence interval estimate.That is,\begin{aligned}\bar{x}-t_{\alpha}\left(\frac{s}{\sqrt{n}}\right) & =133.22-1.74\left(\frac{10.2128}{\sqrt{18}}\right) \quad \quad\left[\text { From t-table, } t_{\alpha}=t_{0.05}=1.74\right] \\& =133.22-4.1885 \\& =129.03\end{aligned}Hence, the 95 \% lower confidence interval estimate is, 129.03 .(c)The objective is to construct the 95 \% upper confidence interval estimate.That is,\begin{aligned}\bar{x}+t_{\alpha}\left(\frac{s}{\sqrt{n}}\right) & =133.22+1.74\left(\frac{10.2128}{\sqrt{18}}\right) \quad\left[\text { From t-table, } t_{\alpha}=t_{0.05}=1.74\right] \\& =133.22+4.1885 \\& =137.41\end{aligned}Hence, the 95 \% upper confidence interval estimate is, 137.41 . ...