Respuesta :
Answer:
[tex] 0.609 \leq \sigma \leq 0.772[/tex]
And the best conclusion would be:
D. This conclusion is safe because 1.40 °F is outside the confidence interval.
Step-by-step explanation:
1) Data given and notation
s=0.68 represent the sample standard deviation
[tex]\bar x =98.90[/tex] represent the sample mean
n=98 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 .
2) 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=98-1=97[/tex]
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.05,97)" "=CHISQ.INV(0.95,97)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=120.990[/tex]
[tex]\chi^2_{1- \alpha/2}=75.282[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(97)(0.68)^2}{120.990} \leq \sigma \leq \frac{(97)(0.68)^2}{75.282}[/tex]
[tex] 0.371 \leq \sigma^2 \leq 0.596[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 0.609 \leq \sigma \leq 0.772[/tex]
And the best conclusion would be:
D. This conclusion is safe because 1.40 °F is outside the confidence interval.
