Answer:
[tex]\frac{(17)(0.038)^2}{32.64} \leq \sigma^2 \leq \frac{(17)(0.038)^2}{6.66}[/tex]
[tex] 0.0008 \leq \sigma^2 \leq 0.0037[/tex]
Step-by-step explanation:
Data given and notation
s=0.038 represent the sample standard deviation
[tex]\bar x[/tex] represent the sample mean
n=18 the sample size
Confidence=97.5% or 0.975
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:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=18-1=17[/tex]
Since the Confidence is 0.975 or 97.5%, the value of [tex]\alpha=0.025[/tex] and [tex]\alpha/2 =0.0125[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.0125,17)" "=CHISQ.INV(0.9875,17)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=32.64[/tex]
[tex]\chi^2_{1- \alpha/2}=6.66[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(17)(0.038)^2}{32.64} \leq \sigma^2 \leq \frac{(17)(0.038)^2}{6.66}[/tex]
[tex] 0.0008 \leq \sigma^2 \leq 0.0037[/tex]