Respuesta :
Answer:
[tex]213.65-2.093\frac{49.028}{\sqrt{20}}=190.704[/tex]
[tex]213.65+2.093\frac{49.028}{\sqrt{20}}=236.596[/tex]
And the confidence interval would be given by:
[tex]190.704 \leq \mu \leq 236.596[/tex]
Step-by-step explanation:
Information given
136, 143, 153, 164, 166, 179, 184, 194, 199, 207, 216, 220, 228, 247, 252, 259, 275, 276, 282, 293
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= 213.65[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=49.028 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)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=20-1=19[/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 for this case is [tex]t_{\alpha/2}=2.093[/tex]
And replacing we got:
[tex]213.65-2.093\frac{49.028}{\sqrt{20}}=190.704[/tex]
[tex]213.65+2.093\frac{49.028}{\sqrt{20}}=236.596[/tex]
And the confidence interval would be given by:
[tex]190.704 \leq \mu \leq 236.596[/tex]
