Question

a) In order to determine the average price of hotel rooms in Atlanta, a sample of 61 hotels were selected. It was determined that the average price of the rooms in the sample was $111.9. The population standard deviation is known to be $19. We would like to test whether or not the average room price is significantly different from $110.

Compute the test statistic.

b) In order to determine the average price of hotel rooms in Atlanta, a sample of 61 hotels were selected. It was determined that the test statistic (z) was $-1.01. We would like to test whether or not the average room price is significantly different from $110. Population standard deviation is known to us.

Compute the p-value.

c) In order to determine the average price of hotel rooms in Atlanta. Using a 0.1 level of significance, we would like to test whether or not the average room price is significantly different from $110. The population standard deviation is known to be $16. A sample of 64 hotels was selected. The test statistic (z) is calculated and it is -1.8.

We conclude that the average price of hotel rooms in Atlanta is NOT significantly different from $110. (Enter 1 if the conclusion is correct. Enter 0 if the conclusion is wrong.)

Answer 1

The test statistic formula is given as below:

Z = (Xbar - µ)/[σ/sqrt(n)]

From given data, we have

µ = 110

Xbar = 111.9

σ = 19

n = 61

Z = (111.9 - 110)/(19/sqrt(61))

Z = 0.7810

Test statistic = 0.7810

Ansb:

Given . Z=-1.01

P-value=P(|Z|>1.01)

=2*P(Z>1.01)

=2*0.1563

=0.3125

#P-value=0.3125

Ansc:

H0: = 110

Ha: 110

Given . Z=-1.8

P-value=P(|Z|>1.8)

=2*P(Z>1.8)

=2*0.0359

=0.0719

We have sufficient evidence to support the claim if p-value > (level of significance)

Here, p-value = 0.0719, = 0.1

Since p-value < 0.1 , Reject H0.

We have sufficient evidence to conclude that he average price of hotel rooms in Atlanta is significantly different from $110

So, conclusion in question is wrong.

Answer : 0