Answer :
Answer:
a)[tex]\hat \sigma^2 =s^2 =30^2 = 900[/tex]
b) [tex] 567.277 \leq \sigma^2 \leq 1690.224[/tex]
Rounded to the nearest number would be:
[tex] 567 \leq \sigma^2 \leq 1690[/tex]
c) [tex] 23.818 \leq \sigma \leq 41.112[/tex]
And rounded :
[tex] 23.8 \leq \sigma \leq 41.1[/tex]
Step-by-step explanation:
Data given and notation
s=30 represent the sample standard deviation
[tex]\bar x=290[/tex] represent the sample mean
n=20 the sample size
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 .
Part a
The best point of estimate for the population variance is the sample variance, so on this case:
[tex]\hat \sigma^2 =s^2 =30^2 = 900[/tex]
Part b
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 next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:
[tex]df=n-1=20-1=19[/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(0.05,19)" "=CHISQ.INV(0.95,19)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=30.144[/tex]
[tex]\chi^2_{1- \alpha/2}=10.117[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(19)(30)^2}{30.144} \leq \sigma \leq \frac{(19)(30)^2}{10.117}[/tex]
[tex] 567.277 \leq \sigma^2 \leq 1690.224[/tex]
Rounded to the nearest number would be:
[tex] 567 \leq \sigma^2 \leq 1690[/tex]
Part c
In order to find the confidence interval for the deviation we just need to take the square root for the interval of the variance, and we got:
[tex] 23.818 \leq \sigma \leq 41.112[/tex]
And rounded :
[tex] 23.8 \leq \sigma \leq 41.1[/tex]