Respuesta :
Answer:
[tex]75-2.58\frac{13.3}{\sqrt{33}}=69.027[/tex]
[tex]75+2.58\frac{13.3}{\sqrt{33}}=80.793[/tex]
And the 95% confidence interval would be between (69.027;80.793)
Step-by-step explanation:
Information given
[tex]\bar X=75[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
[tex]\sigma=13.3[/tex] represent the population standard deviation
n=33 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=33-1=32[/tex]
The Confidence level is 0.99 or 99%, the significance would be [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], the critical value for this case would be [tex]z_{\alpha/2}=2.58[/tex]
And replacing we got:
[tex]75-2.58\frac{13.3}{\sqrt{33}}=69.027[/tex]
[tex]75+2.58\frac{13.3}{\sqrt{33}}=80.793[/tex]
And the 95% confidence interval would be between (69.027;80.793)
