Question

Find the 95% confidence intervalfor the variance and standard deviation for the time ittakes a customer to place a telephone order with a largecatalogue company if a sample of 23 telephone ordershas a standard deviation of 3.8 minutes. Assume thevariable is normally distributed. Do you think that thetimes are relatively consistent?

Answer 1

Answer:

$8.637 \leq \sigma^2 \leq 28.93$

$2.939 \leq \sigma \leq 5.379$

Step-by-step explanation:

Data given and notation

s represent the sample standard deviation

$\bar x$ represent the sample mean

n=23 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 .

Calculating the confidence interval

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

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

The sample standard deviation for this case was s = 3.8

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

$df=n-1=23-1 =22$

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 tabele to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.025,22)" "=CHISQ.INV(0.975,22)". so for this case the critical values are:

$\chi^2_{\alpha/2}=36.78$

$\chi^2_{1- \alpha/2}=10.98$

And replacing into the formula for the interval we got:

$\frac{(22)(3.8)^2}{36.78} \leq \sigma^2 \leq \frac{(22)(3.8)^2}{10.98}$

$8.637 \leq \sigma^2 \leq 28.93$

Now we just take square root on both sides of the interval and we got:

$2.939 \leq \sigma \leq 5.379$