Answer:
The degrees of freedom are:
[tex]df=n-1=10-1=9[/tex]
The confidence level is 0.95 or 95%, the significance would be [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], the critical value for this case would be [tex]t_{\alpha/2}=2.262[/tex]
And replacing we got:
[tex]54.7-2.262\frac{9.956}{\sqrt{10}}=47.58[/tex]
[tex]54.7+2.262\frac{9.956}{\sqrt{10}}=61.82[/tex]
Step-by-step explanation:
Information given
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= 54.7[/tex] represent the sample mean
[tex]\mu[/tex] population mean
s=9.956 represent the sample standard deviation
n=10 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:
[tex]df=n-1=10-1=9[/tex]
The confidence level is 0.95 or 95%, the significance would be [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], the critical value for this case would be [tex]t_{\alpha/2}=2.262[/tex]
And replacing we got:
[tex]54.7-2.262\frac{9.956}{\sqrt{10}}=47.58[/tex]
[tex]54.7+2.262\frac{9.956}{\sqrt{10}}=61.82[/tex]
