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?

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)

step-by-step explanation:

no of data

divide by no of mode

pay attention in your classes and concentrate the best thing you can do to raise your test scores is to pay attention when you're supposed to be learning the material: in class!

take good notes.

do your homework.

use mnemonics and other tricks.

do practice tests.

note: the best i know.