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 ).

Respuesta :

Answer:

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

[tex]18-2.262\frac{2.75}{\sqrt{10}}=16.03[/tex]    

[tex]18+2.262\frac{2.75}{\sqrt{10}}=19.97[/tex]    

Step-by-step explanation:

Information given

[tex]\bar X=18[/tex] represent the sample mean

[tex]\mu[/tex] 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:

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

The degrees of freedom are given by:

[tex]df=n-1=10-1=9[/tex]

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

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

[tex]18-2.262\frac{2.75}{\sqrt{10}}=16.03[/tex]    

[tex]18+2.262\frac{2.75}{\sqrt{10}}=19.97[/tex]