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

`2.002 \leq \sigma^2 \leq 11.365`

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:

`((n-1)s^2)/(\chi^2_(\alpha/2)) \leq \sigma ((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{(\sum_(i=1)^9 (x_i -\bar x)^2)/(n-1)} </p><p>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:

`((9)(3.45))/(15.507) \leq \sigma ((9)(3.45))/(2.732)`

`2.002 \leq \sigma^2 \leq 11.365`