Use the one-standard-deviation chi-square interval procedure to obtain the specified confidence interval for thepopulation standard deviation Ϭ. Assume that the population has a normal distribution.
-The mean replacement time for a random sample of 20 washing machines is $10.8$ years and the standard deviation is $2.7$ years. Construct a $99 \%$ confidence interval for the standard deviation, $\sigma$, of the replacement times of all washing machines of this type.
A)$1.8$ to $5.1 \mathrm { yr }$
B) $1.9$ to $5.7 \mathrm { yr }$
C) $2.0$ to $4.3 \mathrm { yr }$
D) $1.9$ to $4.5 \mathrm { yr }$

Question

Use the one-standard-deviation chi-square interval procedure to obtain the specified confidence interval for thepopulation standard deviation Ϭ. Assume that the population has a normal distribution.
-The mean replacement time for a random sample of 20 washing machines is $10.8$ years and the standard deviation is $2.7$ years. Construct a $99 \%$ confidence interval for the standard deviation, $\sigma$, of the replacement times of all washing machines of this type.  
A)$1.8$ to $5.1 \mathrm { yr }$
B) $1.9$ to $5.7 \mathrm { yr }$
C) $2.0$ to $4.3 \mathrm { yr }$
D) $1.9$ to $4.5 \mathrm { yr }$

Quizplus · Accepted Answer

The correct answer is D.The one-standard-deviation chi-square interval procedure for the population standard deviation is given by:$$\frac{(n-1)s^2}{\chi^2_{\alpha/2,n-1}} \le \sigma^2 \le \frac{(n-1)s^2}{\chi^2_{1-\alpha/2,n-1}}$$where $s$ is the sample standard deviation, $n$ is the sample size, and $\chi^2_{\alpha/2,n-1}$ and $\chi^2_{1-\alpha/2,n-1}$ are the chi-square critical values with $(n-1)$ degrees of freedom and areas $\alpha/2$ and $1-\alpha/2$, respectively.For a 99% confidence interval, $\alpha = 0.01$ and $\alpha/2 = 0.005$. Using a chi-square table with $(n-1) = 19$ degrees of freedom, we find that $\chi^2_{0.005,19} = 30.144$ and $\chi^2_{0.995,19} = 9.236$.Plugging in the values, we get:$$\frac{(20-1)(2.7)^2}{30.144} \le \sigma^2 \le \frac{(20-1)(2.7)^2}{9.236}$$$$\Rightarrow 1.9 \le \sigma^2 \le 4.5$$$$\Rightarrow 1.9 \le \sigma \le 4.5$$Therefore, the 99% confidence interval for the standard deviation is $1.9$ to $4.5$ years.

Use the One-Standard-Deviation Chi-Square Interval Procedure to Obtain the Specified 10.810.810.8

Use the One-Standard-Deviation Chi-Square Interval Procedure to Obtain the Specified $10.8$