Question

A random sample of 9 wheels of cheese yielded the following weights in pounds has a sample mean of 20.90 and a sample variance of 3.45. Assume the weights of wheels of cheese have a normal distribution. Find a 90% confidence interval for the population variance.

Answer 1

Answer:

$2.002 \leq \sigma^2 \leq 11.365$

Step-by-step explanation:

1) Data given and notation

s represent the sample standard deviation

$s^2$ represent the sample variance

n=9 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 .

2) Calculating the confidence interval

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

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

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

$s=sqrt{\frac{\sum_{i=1}^9 (x_i -\bar x)^2}{n-1}} The sample variance given was [tex]s^2=3.45$

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

$df=n-1=9-1=8$

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(0.05,8)" "=CHISQ.INV(0.95,8)". so for this case the critical values are:

$\chi^2_{\alpha/2}=15.507$

$\chi^2_{1- \alpha/2}=2.732$

And replacing into the formula for the interval we got:

$\frac{(9)(3.45)}{15.507} \leq \sigma \frac{(9)(3.45)}{2.732}$

$2.002 \leq \sigma^2 \leq 11.365$