A researcher obtained a sample of size 16 from a normal population with unknown mean and...

Question:

A researcher obtained a sample of size 16 from a normal population with unknown mean and variance. A graduate student constructed a 95% confidence interval for the population mean and for the population variance. The next day, the student could only find the result for the confidence interval for the population mean which was (16.89, 18.15). Is this information enough to recalculate the 95% confidence interval for the population variance? If so, obtain this interval.

Confidence Interval

The confidence interval is the interval estimation for estimating the value of population parameters, and it is provided with the lower and upper limit for the value of the population parameters at the specified level of confidence level.

Answer and Explanation:

Given information

Sample size: 16

Confidence level: 95%

Confidence interval for the population mean:(16.89, 18.15)


The value of sample variance is calculated by using the margin of error formula as follows.

{eq}\begin{align*} M.E &= {t_{\alpha /2}}*\dfrac{s}{{\sqrt n }}\\ \left( {18.15 - 16.89} \right) &= 2.13145*\dfrac{s}{{\sqrt {16} }}\\ s &= 2.3646 \end{align*} {/eq}

Therefore, the value of sample standard deviation is 2.3646.

The confidence interval for the population variance is calculated as follows.

{eq}\begin{align*} P\left( {\dfrac{{\left( {n - 1} \right){s^2}}}{{\chi _{1 - \alpha /2}^2}} < {\sigma ^2} < \dfrac{{\left( {n - 1} \right){s^2}}}{{\chi _{\alpha /2}^2}}} \right) &= 0.95\\ P\left( {\dfrac{{\left( {16 - 1} \right)2.3646}}{{27.4884}} < {\sigma ^2} < \dfrac{{\left( {16 - 1} \right)2.3646}}{{6.26214}}} \right) &= 0.95\\ P\left( {1.2903 < {\sigma ^2} < 5.6640} \right) &= 0.95 \end{align*}{/eq}


Learn more about this topic:

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Finding Confidence Intervals with the Normal Distribution

from Statistics 101: Principles of Statistics

Chapter 9 / Lesson 3
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