Question

An automatic filling machine is used to fill bottles with liquid detergent. A random sample of 18 bottles results in a sample standard deviation of fill volume LaTeX: s\:=s =0.038 oz. Assume that fill volume is approximately normal. Determine the upper bound of a two-sided 97.5% confidence interval on the population variance, LaTeX: \sigma^2σ 2. Round your answer to four decimal places.

Answer 1

Answer:

$\frac{(17)(0.038)^2}{32.64} \leq \sigma^2 \leq \frac{(17)(0.038)^2}{6.66}$

$0.0008 \leq \sigma^2 \leq 0.0037$

Step-by-step explanation:

Data given and notation

s=0.038 represent the sample standard deviation

$\bar x$ represent the sample mean

n=18 the sample size

Confidence=97.5% or 0.975

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=18-1=17$

Since the Confidence is 0.975 or 97.5%, the value of $\alpha=0.025$ and $\alpha/2 =0.0125$, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.0125,17)" "=CHISQ.INV(0.9875,17)". so for this case the critical values are:

$\chi^2_{\alpha/2}=32.64$

$\chi^2_{1- \alpha/2}=6.66$

And replacing into the formula for the interval we got:

$\frac{(17)(0.038)^2}{32.64} \leq \sigma^2 \leq \frac{(17)(0.038)^2}{6.66}$

$0.0008 \leq \sigma^2 \leq 0.0037$