Question

lakefront resort is planning for its summer busy season. It wishes to estimate with 95% confidence the average number of nights each guest will stay in a consecutive visit. Using a sample of guests who stayed last year, the average number of nights per guest is calculated at 5 nights. The standard deviation of the sample is 1.5 nights. The size of the sample used is 120 guests and the resort desires a precision of plus or minus .5 nights. What is the standard error of the mean in the lakefront resort example? Within what range below can the resort expect with 95% confidence for the true population mean to fall?

Answer 1

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=120-1=119`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,119)".And we see that
`t_(\alpha/2)=1.98`

The standard error is given by:

`SE= t_(\alpha/2) (s)/(√(n))`

And replacing we got:

`SE= 1.98 (1.5)/(√(120))=0.271`

`5-1.98(1.5)/(√(120))=4.729`

`5+1.98(1.5)/(√(120))=5.271`

So on this case the 95% confidence interval would be given by (4.729;5.271)

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

`\bar X=5` represent the sample mean

`\mu` population mean (variable of interest)

s=1.5 represent the sample standard deviation

n=120 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)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=120-1=119`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,119)".And we see that
`t_(\alpha/2)=1.98`

The standard error is given by:

`SE= t_(\alpha/2) (s)/(√(n))`

And replacing we got:

`SE= 1.98 (1.5)/(√(120))=0.271`

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

`5-1.98(1.5)/(√(120))=4.729`

`5+1.98(1.5)/(√(120))=5.271`

So on this case the 95% confidence interval would be given by (4.729;5.271)