Question

Construct a confidence interval that will capture the true population mean with 95% confidence. Population standard deviation is unknown.
Data: 18
19,19,19,19,19,19,20,20,20,20,20,20,20,20,21,21,21,21,21,21,21, 22,22,22,22,23,23,23,23,23,23,23,23,23,23,23,23,23,24,24,24,24,24,24,24,24,24,24,24,24,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,25,26,26,26,26,26,26,26,26,,26,26,26,27,27,27,27,27,28,28,28,28,28,28,28,28,28,28,29,29,29,30,30,30,31

Answer 1

A confidence interval that will capture the true population mean with 95% confidence. The value is ( 23.6459    , 24.8293 ) .

What is population mean?

A population in statistics is a group of comparable objects or occurrences that are relevant to a particular topic or experiment. A statistical population can be a collection of real things or a hypothetical, possibly limitless collection of objects derived from experience.

Data: 18

for given data

$\begin{aligned}\bar{X} & =\frac{\sum X i}{n} \\& =2448 / 101 \\& =24.2376\end{aligned}$

as population variance is unknown so we used sample variance

$\begin{aligned}S^2 & =\frac{1}{n-1}\left(\sum X i^2-n \times \bar{X}^2\right) \\S^2 & =\frac{1}{100}\left(60232-101 \times 24.2376^2\right) \\& =8.9830\end{aligned}$

Now

$\frac{\bar{X}-\mu}{S / \sqrt{n}} \sim t_{n-1}$

hence 95 % of population mean is

As n=101 which is large so critical value is 1.98397

$\begin{aligned}& =\left(\bar{X}-\sqrt{\frac{S}{n}} \times t_{(n-1, \alpha / 2)}, \bar{X}+\sqrt{\frac{S}{n}} \times t_{(n-1, \alpha / 2)}\right) \\& =\left(24.2376-\sqrt{\frac{8.9830}{101}} \times 1.98397,24.2376+\sqrt{\frac{8.9830}{101}} \times 1.98397\right) \\& =(23.6459,24.8293)\end{aligned}$

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