Answer :
Answer:
[tex]\frac{(4)(0.903)^2}{11.143} \leq \sigma^2 \leq \frac{(4)(0.903)^2}{0.484}[/tex]
[tex] 0.293 \leq \sigma^2 \leq 6.736[/tex]
And in order to obtain the confidence interval for the deviation we just take the square root and we got:
[tex] 0.541 \leq \sigma \leq 2.595[/tex]
Since the confidence interval cointains the 1 we don't have enough evidence to reject the hypothesis given by the claim
Step-by-step explanation:
Data provided
1.9, 2.4, 3.0, 3.5, and 4.2
We can calculate the sample mean and deviation from this data with these formulas:
[tex]\bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex] s=\frac{\sum_{i=1}^n (X_i-\bar X)^2}{n-1}[/tex]
And we got:
[tex]\bar X= 3[/tex]
s=0.903 represent the sample standard deviation
n=5 the sample size
Confidence=95% or 0.95
Confidence interval
We need to begin finding 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=5-1=4[/tex]
The Confidence level provided is 0.95 or 95%, the significance is then[tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical values for this case are:
[tex]\chi^2_{\alpha/2}=11.143[/tex]
[tex]\chi^2_{1- \alpha/2}=0.484[/tex]
And the confidence interval would be:
[tex]\frac{(4)(0.903)^2}{11.143} \leq \sigma^2 \leq \frac{(4)(0.903)^2}{0.484}[/tex]
[tex] 0.293 \leq \sigma^2 \leq 6.736[/tex]
And in order to obtain the confidence interval for the deviation we just take the square root and we got:
[tex] 0.541 \leq \sigma \leq 2.595[/tex]
Since the confidence interval cointains the 1 we don't have enough evidence to reject the hypothesis given by the claim