Respuesta :
Answer:
The 99% confidence interval is given by:
3.9<σ<8.8
The distribution needs to be approximately normal.
Step-by-step explanation:
1) Data given and notation
Data: 65.1 71.9 72.7 73.1 73.4 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.3 79.6 79.7 79.9 80.1 82.2 83.6 93.8
We can calculate the sample standard deviation with this formula:
[tex]s=\frac{\sum_{i=1}^n (x_i -\bar x)^2}{n-1}[/tex]
s=5.438 represent the sample standard deviation
[tex]\bar x[/tex] represent the sample mean
n=22 the sample size
Confidence=99% or 0.99
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=22-1=21[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical values.
The excel commands would be: "=CHISQ.INV(0.005,21)" "=CHISQ.INV(0.995,21)". so for this case the critical values are:
[tex]\chi^2_{\alpha/2}=41.401[/tex]
[tex]\chi^2_{1- \alpha/2}=8.034[/tex]
And replacing into the formula for the interval we got:
[tex]\frac{(21)(5.438)^2}{41.401} \leq \sigma \frac{(21)(5.438)^2}{8.034}[/tex]
[tex] 15.0 \leq \sigma^2 \leq 77.3[/tex]
Now we just take square root on both sides of the interval and we got:
[tex] 3.9 \leq \sigma \leq 8.8[/tex]
So the 99% confidence interval is given by:
3.9<σ<8.8
For the conditions in order to calculate this interval we assume this:
The distribution needs to be approximately normal.
