Respuesta :
Answer:
Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical value would be [tex]t_{\alpha/2}=1.669[/tex]
And replacing we got
[tex]17.5-1.669\frac{4.2}{\sqrt{65}}=16.63[/tex]
[tex]17.5+1.669\frac{4.2}{\sqrt{65}}=18.37[/tex]
For the 95% confidence the critical value is [tex]t_{\alpha/2}=1.998[/tex]
[tex]17.5-1.998\frac{4.2}{\sqrt{65}}=16.46[/tex]
[tex]17.5+1.998\frac{4.2}{\sqrt{65}}=18.54[/tex]
Step-by-step explanation:
Information given
[tex]\bar X¿ 17.5[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s¿4.2 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 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and the critical value would be [tex]t_{\alpha/2}=1.669[/tex]
And replacing we got
[tex]17.5-1.669\frac{4.2}{\sqrt{65}}=16.63[/tex]
[tex]17.5+1.669\frac{4.2}{\sqrt{65}}=18.37[/tex]
For the 95% confidence the critical value is [tex]t_{\alpha/2}=1.998[/tex]
[tex]17.5-1.998\frac{4.2}{\sqrt{65}}=16.46[/tex]
[tex]17.5+1.998\frac{4.2}{\sqrt{65}}=18.54[/tex]
