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

Answer:

$SE=\frac{1.5}{\sqrt{120}}=0.137$

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

Step-by-step 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:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (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: $SE=\frac{s}{\sqrt{n}}$

And the margin of error would be given by: $ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

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:

$SE=\frac{1.5}{\sqrt{120}}=0.137$

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

$5-1.98\frac{1.5}{\sqrt{120}}=4.729$    

$5+1.98\frac{1.5}{\sqrt{120}}=5.271$

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