Respuesta :
Answer:
[tex]59-2.40\frac{10}{\sqrt{51}}=55.639[/tex]
[tex]59+ 2.40\frac{10}{\sqrt{51}}=62.361[/tex]
The 98% confidence interval would be given by (55.639;62.361)
Step-by-step explanation:
Information given
[tex]\bar X= 59[/tex] represent the sample mean
[tex] s= 10[/tex] represent the sample deviation
[tex] n= 51[/tex] represent the sample size
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
The degrees of freedom are given by:
[tex]df=n-1=51-1=50[/tex]
The Confidence is 0.98 or 98%, the significance would be [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex],and the critical value would be [tex]t_{\alpha/2}=2.40[/tex]
And replacing we got:
[tex]59-2.40\frac{10}{\sqrt{51}}=55.639[/tex]
[tex]59+ 2.40\frac{10}{\sqrt{51}}=62.361[/tex]
The 98% confidence interval would be given by (55.639;62.361)
