Question

A random sample of 25 round trip flights between Philadelphia and Dallas has an average airfare of 393.50 with a sample Standard Deviation of $50.30. The critical value for an 80% confidence interval around this sample mean

Answer 1

Answer:

$t_{\alpha/2}=1.318$

And the confidence interval would be given by:

$393.5-1.318\frac{50.30}{\sqrt{25}}=380.24$    

$393.5+1.318\frac{50.30}{\sqrt{25}}=406.76$

Step-by-step explanation:

Information given

$\bar X=393.50$ represent the sample mean

$\mu$ population mean

s=50.30 represent the sample standard deviation

n=25 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 given by:

$df=n-1=25-1=24$

The Confidence level is 0.80 or 80%, the value of significance $\alpha=0.2$ and $\alpha/2 =0.1$, the critical value for this case would be:

$t_{\alpha/2}=1.318$

And the confidence interval would be given by:

$393.5-1.318\frac{50.30}{\sqrt{25}}=380.24$    

$393.5+1.318\frac{50.30}{\sqrt{25}}=406.76$