A lakefront resort is planning for its summer busy season. It wishes to estimate with 95% confidence the average number of nights each guest

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.

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Josephine 4 days 2021-10-10T21:39:31+00:00 1 Answer 0

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    2021-10-10T21:40:31+00:00

    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)    

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