Respuesta :
Answer:
[tex] 51.278 -2.896 \frac{22.979}{\sqrt{9}}= 29.096[/tex]
[tex] 51.278 +2.896 \frac{22.979}{\sqrt{9}}= 73.460[/tex]
And the interval would be:
[tex] (29.10 \leq \mu \leq 73.46)[/tex]
Step-by-step explanation:
For this problem we have the following dataset given:
44 32.8 59.2 31.4 12.7 68.5 84.7 72.5 55.7
We can find the mean and sample deviation with the following formulas:
[tex] \bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex] s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}[/tex]
And replacing we got:
[tex]\bar X= 51.278[/tex]
[tex] s= 22.979[/tex]
The confidence interval for the mean is given by:
[tex] \bar X \pm t_{\alpha/2} \frac{s}{\sqrt{n}}[/tex]
The degrees of freedom are:
[tex] df=n-1= 9-1=8[/tex]
The confidence would be 0.98 and the significance [tex]\alpha=0.02[/tex] then the critical value would be:
[tex] t_{\alpha/2}= 2.896[/tex]
Ad replacing we got:
[tex] 51.278 -2.896 \frac{22.979}{\sqrt{9}}= 29.096[/tex]
[tex] 51.278 +2.896 \frac{22.979}{\sqrt{9}}= 73.460[/tex]
And the interval would be:
[tex] (29.10 \leq \mu \leq 73.46)[/tex]
