Respuesta :
Answer:
[tex]71.35-2.093\frac{22.48}{\sqrt{20}}=60.83[/tex]
[tex]71.35+2.093\frac{22.48}{\sqrt{20}}=81.87[/tex]
Step-by-step explanation:
Information given
90 34 41 106 84 53 55 48 41 75 49 97 92 73 74 80 94 102 56 83
In order to calculate the mean and the sample deviation we can use the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)
[tex]\bar X=71.35[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=22.48 represent the sample standard deviation
n=20 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=20-1=19[/tex]
Since the Confidence is 0.95 or 95%, the significance is [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and the critical value would be [tex]t_{\alpha/2}=2.093[/tex]
And replacing we got:
[tex]71.35-2.093\frac{22.48}{\sqrt{20}}=60.83[/tex]
[tex]71.35+2.093\frac{22.48}{\sqrt{20}}=81.87[/tex]
