Question

The quality control manager of a chemical company randomly sampled twenty 100-pound bags of fertilizer to estimate the variance in the pounds of the impurities. The sample variance was found to be 6.62. Find a 95% confidence interval for the population variance in the pounds of impurities. State any assumption you need to make to be able to answer this problem.

Answer 1

Answer:

The 95% confidence interval for the variance in the pounds of impurities would be $3.829 \leq \sigma^2 \leq 14.121$.

Step-by-step explanation:

1) Data given and notation

$s^2 =6.62$ represent the sample variance

s=2.573 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=95% or 0.95

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:

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

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

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

$df=n-1=20-1=19$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,19)" "=CHISQ.INV(0.975,19)". so for this case the critical values are:

$\chi^2_{\alpha/2}=32.852$

$\chi^2_{1- \alpha/2}=8.907$

And replacing into the formula for the interval we got:

$\frac{(19)(6.62)}{32.852} \leq \sigma^2 \frac{(19)(6.62)}{8.907}$

$3.829 \leq \sigma^2 \leq 14.121$

So the 95% confidence interval for the variance in the pounds of impurities would be $3.829 \leq \sigma^2 \leq 14.121$.