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

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

`18-2.262(2.75)/(√(10))=16.03`

`18+2.262(2.75)/(√(10))=19.97`

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)(s)/(√(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(2.75)/(√(10))=16.03`

`18+2.262(2.75)/(√(10))=19.97`