6. Three different distributions There are 572 fulil service restaurants in verm

Question

6. Three different distributions There are 572 fulil service restaurants in vermont The mean number of seats per restaurant is 65,.7. Source: Data based on the 2002 Economic Census from U.S. Census Bureau. suppose that the true population mean = 65.7 and standard deviation = 20.5 are unknown to the Vermont tourism board. They select a simple random sample of 55 full-service restaurants located within the state to estimate . The mean number of seats per restaurant in the sample is M-69.9, with a sample standard deviation of s 19.1. The standard deviation of the distribution of sample means (that is, the standard error ) is 2.76 . (Note: Although and are unknown to the Vermont tourism board, they are known to you for the purposes of calculating these answers.) The standard or typical average difference between the mean number of seats in the 572 full-service restaurants in Vermont (-65.7) and one randomly selected full-service restaurant in Vermont is 20.5 The standard or typical average difference between the mean number of seats in the sample of 55 restaurants (M = 69.9) and one randomly selected restaurant in that sample is The standard or typical average difference between the mean number of seats in the 572 full-service restaurants in Vermont ( = 65.7) and the sample mean of any sample of size 55 is The z-score that locates the mean number of seats in the Vermont tourism board's sample (M = 69.9) in the distribution of sample means is

Explanation / Answer

Solution:

6)

a) Sigma(X) = sigma/sqrt(n) = 20.5/sqrt(55) = 2.7642
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b) The population standard deviation,

sigma = 20.5.
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c) The sample standard deviation.

s = 19.1
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d) The standard error in part a,

sigma(X) = 2.7642

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e) z = (M - u)*sqrt(n)/sigma = (69.9-65.7)*sqrt(55)/20.5 = 1.5194

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f) Using a table the right tailed area for the z score in e is

P(z > 1.5194) =1 P ( Z < 1.5194 )
= 1 0.9357
= 0.0643

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