question 1 (5 points)
use the given degree of confidence and sample data to find a confidence interval for the population standard deviation $\sigma$. assume that the population has a normal distribution. round the confidence interval limits to one more decimal place than is used for the original set of data.
the daily intakes of milk (in ounces) for ten randomly selected people were:
23.3 28.4 10.5 16.4 26.4
18.1 20.4 17.3 27.4 13.2
find a 99% confidence interval for the population standard deviation $\sigma$
$0.88\text{ oz}

Question

question 1 (5 points)
use the given degree of confidence and sample data to find a confidence interval for the population standard deviation $\sigma$. assume that the population has a normal distribution. round the confidence interval limits to one more decimal place than is used for the original set of data.
the daily intakes of milk (in ounces) for ten randomly selected people were:
23.3 28.4 10.5 16.4 26.4
18.1 20.4 17.3 27.4 13.2
find a 99% confidence interval for the population standard deviation $\sigma$
$0.88\text{ oz} < \sigma < 3.38\text{ oz}$
$3.78\text{ oz} < \sigma < 13.95\text{ oz}$
$3.78\text{ oz} < \sigma < 12.53\text{ oz}$
$3.66\text{ oz} < \sigma < 12.53\text{ oz}$

Accepted Answer

# Explanation:
## Step1: Calculate Sample Variance
First, find the sample mean $\bar{x}$:
$$
\bar{x}=\frac{23.3 + 28.4+10.5 + 16.4+26.4+18.1+20.4+17.3+27.4+13.2}{10}=\frac{201.4}{10} = 20.14
$$
Then, calculate the sample variance $s^{2}$ using the formula $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$:
$$
\begin{align*}
\sum_{i=1}^{10}(x_{i}-\bar{x})^{2}&=(23.3 - 20.14)^{2}+(28.4 - 20.14)^{2}+(10.5 - 20.14)^{2}+(16.4 - 20.14)^{2}+(26.4 - 20.14)^{2}+(18.1 - 20.14)^{2}+(20.4 - 20.14)^{2}+(17.3 - 20.14)^{2}+(27.4 - 20.14)^{2}+(13.2 - 20.14)^{2}\
&=(3.16)^{2}+(8.26)^{2}+(- 9.64)^{2}+(-3.74)^{2}+(6.26)^{2}+(-2.04)^{2}+(0.26)^{2}+(-2.84)^{2}+(7.26)^{2}+(-6.94)^{2}\
&=9.9856+68.2276 + 92.9296+13.9876+39.1876+4.1616+0.0676+8.0656+52.7076+48.1636\
&=337.594
\end{align*}
$$
$$
s^{2}=\frac{337.594}{9}\approx37.5104
$$

## Step2: Find Chi - Square Critical Values
For a 99% confidence interval and $n = 10$ (so degrees of freedom $df=n - 1=9$):
- The lower critical value $\chi_{1-\alpha/2}^{2}=\chi_{0.005}^{2}$. From chi - square tables, $\chi_{0.005}^{2}=23.589$
- The upper critical value $\chi_{\alpha/2}^{2}=\chi_{0.995}^{2}$. From chi - square tables, $\chi_{0.995}^{2}=1.735$

## Step3: Calculate Confidence Interval for $\sigma$
The formula for the confidence interval for the population standard deviation $\sigma$ is:
$$
\sqrt{\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}}
$$
Substitute $n = 10$, $s^{2}\approx37.5104$, $\chi_{\alpha/2}^{2}=23.589$ and $\chi_{1-\alpha/2}^{2}=1.735$:
- Lower limit: $\sqrt{\frac{9\times37.5104}{23.589}}=\sqrt{\frac{337.5936}{23.589}}\approx\sqrt{14.31}\approx3.78$
- Upper limit: $\sqrt{\frac{9\times37.5104}{1.735}}=\sqrt{\frac{337.5936}{1.735}}\approx\sqrt{194.58}\approx13.95$

# Answer:
3.78 oz $<\sigma<$ 13.95 oz

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QUESTION IMAGE

Question

Explanation:

Step1: Calculate Sample Variance

Step2: Find Chi - Square Critical Values

Step3: Calculate Confidence Interval for \(\sigma\)

Answer: