Question

In: Advanced Math

QUESTION 1 In order to determine the average price of hotel rooms in Atlanta, a sample...

QUESTION 1

  1. In order to determine the average price of hotel rooms in Atlanta, a sample of 38 hotels were selected. It was determined that the average price of the rooms in the sample was $109.3. The population standard deviation is known to be $18. We would like to test whether or not the average room price is significantly different from $110.

    Compute the test statistic.

QUESTION 2

  1. In order to determine the average price of hotel rooms in Atlanta, a sample of 39 hotels were selected. It was determined that the test statistic (z) was $-1.99. 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.

QUESTION 3

  1. 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 -1.38.

    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.)

QUESTION 4

  1. 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 p-value associated with the test statistic (z) is calculated and it is 0.03.

    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.)

Solutions

Expert Solution

QUESTION 1:

SE = /

= 18/

= 2.9200

Test Statistic is given by:

Z = (109.3 - 110)/2.9200

= - 0.2397

So,

Test Statistic is: - 0.2397

QUESTION 2:

Z score = - 1.99

Two Tail Test

By Technology, Cumulative Area Under Standard Normal Curve = 0.0233

So,

P - Value = 0.0233 X 2 = 0.0466

QUESTION 3:

Z calculated = - 1.38

= 0.10

From Table, critical values of Z = 1.645

Since calculated value of Z = - 1.38 is greater than critical value of Z = - 1.645, the difference is not significant. Fail to reject null hypothesis.

So,

correct option:

1

QUESTION 4:

p - value = 0.03

= 0.10

Since p - value is less than , the difference is significant. Reject null hypothesis.

So,

Correct option:

0


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