Respuesta :
Answer:
[tex] 8.637 \leq \sigma^2 \leq 28.93[/tex]
[tex] 2.939 \leq \sigma \leq 5.379[/tex]
Step-by-step explanation:
Data given and notation
s represent the sample standard deviation
[tex]\bar x[/tex] 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:
[tex]\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}[/tex]
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:
[tex]df=n-1=23-1 =22[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], 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:
[tex]\chi^2_{\alpha/2}=36.78[/tex]
[tex]\chi^2_{1- \alpha/2}=10.98[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(22)(3.8)^2}{36.78} \leq \sigma^2 \leq \frac{(22)(3.8)^2}{10.98}[/tex]
[tex] 8.637 \leq \sigma^2 \leq 28.93[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 2.939 \leq \sigma \leq 5.379[/tex]
