Respuesta :
Answer:
[tex]\frac{(21)(5.437)^2}{41.402} \leq \sigma^2 \leq \frac{(21)(5.437)^2}{8.034}[/tex]
[tex] 14.996 \leq \sigma^2 \leq 77.278[/tex]
And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:
3.9<σ<8.8
Step-by-step explanation:
Data given:
65.2 71.9 72.8 73.1 73.1 73.5 75.5 75.7 75.8 76.1 76.2 76.2 77.0 77.9 78.1 79.6 79.7 79.9 80.1 82.2 83.7 93.8
The sample mean would be given by:
[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
We can calculate the sample deviation with this formula:
[tex]s = \frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}[/tex]
And we got:
s=5.437 represent the sample standard deviation
[tex]\bar x[/tex] represent the sample mean
n=22 the sample size
Confidence=99% or 0.99
The confidence interval for the population variance is given by:
[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 degrees of freedom given by:
[tex]df=n-1=22-1=21[/tex]
The Confidence is 0.99 or 99%, the value of significance is [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and the critical values are:
[tex]\chi^2_{\alpha/2}=41.402[/tex]
[tex]\chi^2_{1- \alpha/2}=8.034[/tex]
And the confidence interval would be:
[tex]\frac{(21)(5.437)^2}{41.402} \leq \sigma^2 \leq \frac{(21)(5.437)^2}{8.034}[/tex]
[tex] 14.996 \leq \sigma^2 \leq 77.278[/tex]
And the confidence interval for the deviation would be obtained taking the square root of the last result and we got:
3.9<σ<8.8
