Question

A random sample of 100 teletype operators indicates that their salaries fluctuate quite a bit. The sample Standard deviation of their daily salaries is $10.00. Construct a 90 percent confidence interval on the population standard deviation of the daily salaries.

Answer 1

Answer:

$\frac{(99)(10)^2}{123.225} \leq \sigma^2 \leq \frac{(99)(10)^2}{77.046}$

$80.341 \leq \sigma^2 \leq 128.495$

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

$8.963 \leq \sigma \leq 11.336$

Step-by-step explanation:

Data given and notation

s=10 represent the sample standard deviation

$\bar x$ represent the sample mean

n=100 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:

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

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

$df=n-1=100-1=99$

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(0.05,99)" "=CHISQ.INV(0.95,99)". so for this case the critical values are:

$\chi^2_{\alpha/2}=123.225$

$\chi^2_{1- \alpha/2}=77.046$

And replacing into the formula for the interval we got:

$\frac{(99)(10)^2}{123.225} \leq \sigma^2 \leq \frac{(99)(10)^2}{77.046}$

$80.341 \leq \sigma^2 \leq 128.495$

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

$8.963 \leq \sigma \leq 11.336$