# You randomly select and weigh 27 samples of an allergy medication. The sample standard deviation is 1.22 milligrams. Assuming the weights are normally distributed, construct 90% confidence intervals for the population variance and standard deviation. Please interpret your result.

Question
Confidence intervals
You randomly select and weigh 27 samples of an allergy medication. The sample standard deviation is 1.22 milligrams. Assuming the weights are normally distributed, construct $$90\%$$ confidence intervals for the population variance and standard deviation. Please interpret your result.

2020-12-16
Step 1
From the given information, $$\displaystyle{S}={1.22}$$ milligrams and $$\displaystyle{n}={27}$$.
The level of confidence is 0.90.
$$\displaystyle{1}-\alpha={0.90}$$
$$\displaystyle\alpha={1}-{0.90}$$
$$\displaystyle\alpha={0.10}$$
Step 2
Then, a $$90\%$$ interval for the population variance is
$$\displaystyle{C}{I}={\left(\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{{1}-\frac{a}{{n}},{n}-{1}}}^{{2}}}}<\sigma^{2}<\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{\frac{a}{{n}},{n}-{1}}}^{{2}}}}\right)}$$</span>
$$\displaystyle={\left(\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{{1}-\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}<\sigma^{2}<\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}\right)}$$</span>
$$\displaystyle={\left(\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.95},{26}}}^{{2}}}}<\sigma^{2}<\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.5},{26}}}^{{2}}}}\right)}$$</span>
$$\displaystyle={\left(\frac{{{26}\times{1.22}^{2}}}{{38.855}}<\sigma^{2}<\frac{{{26}\times{1.22}^{2}}}{{15.379}}\right)}$$</span>
$$\displaystyle={\left({0.9952}<\sigma^{2}<{2.5163}\right)}$$</span>
Interpretation:
There is $$90\%$$ confidence that the population variance lies between 0.9952 and 2.5163.
Step 3
The $$90\%$$ confidence interval for the population standard deviation is
$$\displaystyle{C}{I}={\left(\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{{1}-\frac{a}{{n}},{n}-{1}}}^{{2}}}}<\sigma<\frac{{{\left({n}-{1}\right)}{S}^{2}}}{{{X}_{{\frac{a}{{n}},{n}-{1}}}^{{2}}}}\right)}$$</span>
$$\displaystyle={\left(\sqrt{{\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{X}_{{{1}-\frac{0.10}{{2}},{27}-{1}}}}}}<\sigma<\sqrt{{\frac{{{\left({27}-{1}\right)}{1.22}^{2}}}{{{X}_{{\frac{0.10}{{2}},{27}-{1}}}^{{2}}}}}}\right)}$$</span>
$$\displaystyle={\left(\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.95},{26}}}^{{2}}}}}}<\sigma<\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{{X}_{{{0.05},{26}}}^{{2}}}}}}\right)}$$</span>
$$\displaystyle{\left(\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{38.885}}}}<\sigma<\sqrt{{\frac{{{26}\times{1.22}^{2}}}{{15.379}}}}\right)}$$</span> [(Using excel function), (=CHISQ.INV(0.95,26)), (Using excel function), (=CHISQ.INV(0.05,26))]
$$\displaystyle={\left({0.9976}<\sigma<{1.5863}\right)}$$</span>
Interpretation:
There is $$90\%$$ confidence that the population standard deviation lies between 0.9976 and 1.5863.

### Relevant Questions

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
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The​ $$90\%$$ confidence interval is
The​ $$95\%$$ confidence interval is
Which interval is​ wider?
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