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A beachfront hotel 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 single 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 beachfront hotel desires a precision of plus or minus 0.5 nights.

Required:
a. What is the standard error of the mean in the lakefront resort example?
b. Within what range below can the resort expect with 95% confidence for the true population mean to fall?

Respuesta :

fichoh

Answer:

0.1369

(4.732, 5.268)

Step-by-step explanation:

Given that:

Mean, m = 5

Standard deviation, s = 1.5

Error margin, E = 0.5

Sample size, n = 120

Zcritical at 95% = 1.96

Standard Error of the mean (S. E) :

S. E = s /sqrt(n)

S. E = 1.5 / sqrt(120

S.E = 1.5 / 10.954451

S. E = 0.1369306

S. E = 0.1369

Confidence interval (C. I)

Mean ± (Zcritical * S.E)

5 ± (1.96 * 0.1369)

Lower boundary = 5 - 0.268324 = 4.731676

Upper boundary = 5 + 0.268324 = 5.268324

(4.732, 5.268)

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