Question

The following data was collected from a local high school and represents the number of points the women's basketball team scored in their first 10 games: 62, 55, 49, 71, 39, 66, 56, 57, 44, 48. If we wanted to calculate a two-sided 95% confidence interval for the population mean, what would the correct critical value be

Answer 1

Answer:

The degrees of freedom are:

$df=n-1=10-1=9$

The confidence level is 0.95 or 95%, the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$, the critical value for this case would be $t_{\alpha/2}=2.262$

And replacing we got:

$54.7-2.262\frac{9.956}{\sqrt{10}}=47.58$    

$54.7+2.262\frac{9.956}{\sqrt{10}}=61.82$    

Step-by-step explanation:

Information given

In order to calculate the mean and the sample deviation we can use the following formulas:  

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)  

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)  

$\bar X= 54.7$ represent the sample mean

$\mu$ 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

The degrees of freedom are:

$df=n-1=10-1=9$

The confidence level is 0.95 or 95%, the significance would be $\alpha=0.05$ and $\alpha/2 =0.025$, the critical value for this case would be $t_{\alpha/2}=2.262$

And replacing we got:

$54.7-2.262\frac{9.956}{\sqrt{10}}=47.58$    

$54.7+2.262\frac{9.956}{\sqrt{10}}=61.82$