Question

A 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 for 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 means to fall? Show the calculation; otherwise, the answer will not be accepted.

Answer 1

`SE=(1.5)/(√(120))=0.137`

The 95% confidence interval would be given by (4.729;5.271)

Explanation:

1) 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".

The standard error of a statistic is "the standard deviation of its sampling distribution or an estimate of that standard deviation"

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` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

We use the t distirbution for this case since we don't know the population standard deviation
`\sigma`.

Where the standard error is given by:
`SE=(s)/(√(n))`

And the margin of error would be given by:
`ME=t_(\alpha/2)(s)/(√(n))`

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. And we see that
`t_(\alpha/2)=1.98`

The standard error would be given by:

`SE=(1.5)/(√(120))=0.137`

Now we have everything in order to replace into formula (1) and calculate the interval:

`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)