Respuesta :
Answer:
95% confidence
[tex]19.5-1.998\frac{5.2}{\sqrt{65}}=18.211[/tex]
[tex]19.5+1.998\frac{5.2}{\sqrt{65}}=20.789[/tex]
For the 90% confidence interval the critical value would be [tex]t_{\alpha/2}=1.669[/tex] and replacing we got:
[tex]19.5-1.669\frac{5.2}{\sqrt{65}}=18.424[/tex]
[tex]19.5+1.669\frac{5.2}{\sqrt{65}}=20.576[/tex
Step-by-step explanation:
Information given
[tex]\bar X=19.5[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]s=5.2[/tex] represent the sample standard deviation
n=65 represent the sample size
Confidence interval
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=65-1=64[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value would be [tex]t_{\alpha/2}=1.998[/tex]
Now we have everything in order to replace into formula (1):
[tex]19.5-1.998\frac{5.2}{\sqrt{65}}=18.211[/tex]
[tex]19.5+1.998\frac{5.2}{\sqrt{65}}=20.789[/tex]
For the 90% confidence interval the critical value would be [tex]t_{\alpha/2}=1.669[/tex] and replacing we got:
[tex]19.5-1.669\frac{5.2}{\sqrt{65}}=18.424[/tex]
[tex]19.5+1.669\frac{5.2}{\sqrt{65}}=20.576[/tex]
