Question

The following sample of weights was taken from 8 muffins off the assembly line. Construct the 90% confidence interval for the population variance for all muffins that come off the assembly line. Round your answers to two decimal places. 12.9,12.5,13.0,13.1,12.3,12.9,12.2,13.2

Answer 1

Answer:

$\frac{(7)(0.378)^2}{14.067} \leq \sigma^2 \leq \frac{(7)(0.378)^2}{2.167}$

$0.0711 \leq \sigma^2 \leq 0.462$

And rounded we got:

$0.07 \leq \sigma^2 \leq 0.46$

Step-by-step explanation:

Data given and notation

s represent the sample standard deviation

$\bar x$ represent the sample mean

n=8 the sample size

Confidence=90% or 0.90

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:

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

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}} And in order to find the sample mean we just need to use this formula: [tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$

The sample mean obtained on this case is $\bar x= 12.7625$ and the deviation s=0.378

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

$df=n-1=8-1=7$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,7)" "=CHISQ.INV(0.95,7)". so for this case the critical values are:

$\chi^2_{\alpha/2}=14.067$

$\chi^2_{1- \alpha/2}=2.167$

And replacing into the formula for the interval we got:

$\frac{(7)(0.378)^2}{14.067} \leq \sigma^2 \leq \frac{(7)(0.378)^2}{2.167}$

$0.0711 \leq \sigma^2 \leq 0.462$

And rounded we got:

$0.07 \leq \sigma^2 \leq 0.46$