a quality control manager at a potato chip company uses the standard deviation in a production line as a measure of consistency. a bag of potato chips is intended to have a mean of 10 ounces, and the manager would like the standard deviation to be as low as possible. he gathers a random sample of 18 containers. estimate the population standard deviation of the amount of chips at a 90% level of confidence. (data checksum: 180.25) a) find the sample standard deviation: round to 3 decimal places. b) find the lower and upper $\chi^{2}$ critical values at 90% confidence: round to 3 decimal places. lower: upper: c) we are 90% confident that the standard deviation of the amount of chips is between and. round to 2 decimal places.

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Accepted Answer

# Explanation:
## Step1: Calculate sample mean
The sample mean $\bar{x}=\frac{\sum x}{n}$. Given $\sum x = 180.25$ and $n = 18$, so $\bar{x}=\frac{180.25}{18}\approx10.014$.
## Step2: Calculate sample variance $s^{2}$
The formula for sample variance is $s^{2}=\frac{\sum(x - \bar{x})^{2}}{n - 1}$.
$$
\begin{align*}
\sum(x-\bar{x})^{2}&=(9.96 - 10.014)^{2}+(10.19 - 10.014)^{2}+(9.95 - 10.014)^{2}+(9.9 - 10.014)^{2}+(10.04 - 10.014)^{2}+(9.9 - 10.014)^{2}+(10 - 10.014)^{2}+(10.15 - 10.014)^{2}+(10.09 - 10.014)^{2}+(9.85 - 10.014)^{2}+(10.03 - 10.014)^{2}+(9.97 - 10.014)^{2}+(10.07 - 10.014)^{2}+(10.04 - 10.014)^{2}+(10.21 - 10.014)^{2}+(10.08 - 10.014)^{2}+(10.11 - 10.014)^{2}+(9.71 - 10.014)^{2}\
&=(- 0.054)^{2}+(0.176)^{2}+(-0.064)^{2}+(-0.114)^{2}+(0.026)^{2}+(-0.114)^{2}+(-0.014)^{2}+(0.136)^{2}+(0.076)^{2}+(-0.164)^{2}+(0.016)^{2}+(-0.044)^{2}+(0.056)^{2}+(0.026)^{2}+(0.196)^{2}+(0.066)^{2}+(0.096)^{2}+(-0.304)^{2}\
&=0.002916 + 0.030976+0.004096 + 0.012996+0.000676+0.012996+0.000196+0.018496+0.005776+0.026896+0.000256+0.001936+0.003136+0.000676+0.038416+0.004356+0.009216+0.092416\
&=0.286
\end{align*}
$$
Then $s^{2}=\frac{0.286}{18 - 1}\approx0.01682$.
## Step3: Calculate sample standard deviation $s$
$s=\sqrt{s^{2}}=\sqrt{0.01682}\approx0.1297\approx0.130$.
## Step4: Find $\chi^{2}$ critical values
The degree of freedom $df=n - 1=18-1 = 17$. For a $90\%$ confidence interval, $\alpha=1 - 0.90 = 0.10$, and $\frac{\alpha}{2}=0.05$.
From the $\chi^{2}$ - distribution table, $\chi_{1-\frac{\alpha}{2},df}^{2}=\chi_{0.95,17}^{2}=8.672$, $\chi_{\frac{\alpha}{2},df}^{2}=\chi_{0.05,17}^{2}=27.587$.
## Step5: Calculate confidence interval for population standard deviation
The formula for the confidence interval for the population standard deviation $\sigma$ is $\sqrt{\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2},df}^{2}}}\leq\sigma\leq\sqrt{\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2},df}^{2}}}$.
Substitute $n = 17$, $s^{2}=0.01682$, $\chi_{0.05,17}^{2}=27.587$, $\chi_{0.95,17}^{2}=8.672$ into the formula.
$$
\begin{align*}
\sqrt{\frac{17\times0.01682}{27.587}}&\leq\sigma\leq\sqrt{\frac{17\times0.01682}{8.672}}\
\sqrt{\frac{0.28594}{27.587}}&\leq\sigma\leq\sqrt{\frac{0.28594}{8.672}}\
\sqrt{0.01036}&\leq\sigma\leq\sqrt{0.03297}\
0.102&\leq\sigma\leq0.182
\end{align*}
$$

# Answer:
a) $0.130$
b) Lower: $8.672$, Upper: $27.587$
c) $0.10$ and $0.18$

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate sample mean

Step2: Calculate sample variance $s^{2}$

Step3: Calculate sample standard deviation $s$

Step4: Find $\chi^{2}$ critical values

Step5: Calculate confidence interval for population standard deviation

Answer: