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zheka24 [161]
3 years ago
5

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
Mathematics
1 answer:
SSSSS [86.1K]3 years ago
5 0

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

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