Answer:
[tex] 5.1 - 2.015 \frac{19.1}{\sqrt{45}}= -0.637[/tex]
[tex] 5.1 + 2.015 \frac{19.1}{\sqrt{45}}= 10.837[/tex]
And for this case since the confidence interval contains the value 0 we don't have significant evidence that we have a net change in the levels
Step-by-step explanation:
For this case we have the following info given:
[tex] n=45[/tex] represent the sample size
[tex] \bar X= 5.1[/tex] represent the sample mean
[tex] s= 19.1[/tex] represent the sample deviation
We can calculate the confidence interval for the mean with the following formula:
[tex] \bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
The confidence level is 0.90 then the significance level would be [tex]\alpha=0.1[/tex] and the degrees of freedom are given by:
[tex] df= n-1 = 45-1=44[/tex]
And the critical value for this case would be:
[tex] t_{\alpha/2}=2.015[/tex]
And replacing we got:
[tex] 5.1 - 2.015 \frac{19.1}{\sqrt{45}}= -0.637[/tex]
[tex] 5.1 + 2.015 \frac{19.1}{\sqrt{45}}= 10.837[/tex]
And for this case since the confidence interval contains the value 0 we don't have significant evidence that we have a net change or efectiveness in the levels
