Question

How much money does the average professional hockey fan spend on food at a single hockey​ game? That question was posed to 10 randomly selected hockey fans. The sampled results show that sample mean and standard deviation were $ 18.00 and $ 2.75​, respectively. Use this information to create a 90​% confidence interval for the mean. Express the answer in the form x overbar plus or minus t Subscript alpha divided by 2 Baseline (s divided by StartRoot n EndRoot ).

Answer 1

Answer:

Now we have everything in order to replace into formula (1):

$18-2.262\frac{2.75}{\sqrt{10}}=16.03$    

$18+2.262\frac{2.75}{\sqrt{10}}=19.97$    

Step-by-step explanation:

Information given

$\bar X=18$ represent the sample mean

$\mu$ population mean (variable of interest)

s=2.75 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 given by:

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

The Confidence is 0.90 or 90%, the significance is $\alpha=0.1$ and $\alpha/2 =0.05$, and the critical vaue would be $t_{\alpha/2}=2.262$

Now we have everything in order to replace into formula (1):

$18-2.262\frac{2.75}{\sqrt{10}}=16.03$    

$18+2.262\frac{2.75}{\sqrt{10}}=19.97$