Respuesta :
Answer:
[tex]17.5-1.796\frac{3.61}{\sqrt{12}}=15.63[/tex]
[tex]17.5+1.796\frac{3.61}{\sqrt{12}}=19.37[/tex]
Step-by-step explanation:
Data given
20 13 21 18 19 22 19 15 12 12 18 21
We can calculate the sample mean and 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 we got:
[tex]\bar X = 17.5[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=3.61 represent the sample standard deviation
n=12 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=12-1=11[/tex]
Since the Confidence is 0.90 or 90%, the significance is [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], the critical value would be given by [tex]t_{\alpha/2}=[/tex]
Now we have everything in order to replace into formula (1):
[tex]17.5-1.796\frac{3.61}{\sqrt{12}}=15.63[/tex]
[tex]17.5+1.796\frac{3.61}{\sqrt{12}}=19.37[/tex]
