BA 275Quizzes

Winter 2007

Quiz #4

Name (please print)
Section (circle one) / 12noon – 1:50pm / 2:00 – 3:50pm / 4:00 – 5:50pm

Question 1. Past experience indicates that the monthly long-distance telephone bill is normally distributed with a mean of $17.85 and a (population) standard deviation $3.87. After an advertising campaign aimed at increasing long-distance telephone usage, a random sample of 25 household bills was taken. You are concerned whether the campaign was successful, and would like to perform a test to find out.

  1. What are the null and the alternative hypotheses?

Null Hypothesis (H0) /  = $17.85
Alternative Hypothesis (Ha) /  > $17.85 (want to know if the usage has increased.)
  1. If the sample mean turns out to be $15, do you reject the null hypothesis? Why or why not? Assume  = 5%.

Circle one: / Reject H0 / Do not reject H0
The sample mean is $15 below $17.85. There is no evidence at all to support Ha. No further calculation is required. No evidence to reject H0.
  1. If the sample mean turns out to be $29.13, do you reject the null hypothesis? Why or why not? Assume  = 5%.

Circle one: / Reject H0 / Do not reject H0
The z score of the sample mean $29.13 = . It is 14.57 standard deviations above $17.85, very strong evidence in supporting Ha. If  = $17.85, the chance of observing a sample mean of $29.13 or more extreme is almost 0.0000 (p-value). Again, reject H0.
  1. Finally, the actual sample mean in your sample is $19.13. Do you reject the null hypothesis? Was the campaign successful? Assume  = 5%.

Circle one: / Reject H0 / Do not reject H0
Given  = 5%, the rejection region is: Reject H0 if z > 1.645. The z score of the sample mean $19.13 = → in the rejection region. If  = $17.85, the chance of observing a sample mean $19.13 or more extreme, P(>19.13 ) = P( Z > 1.653 ) = 0.0495. The p-value is smaller than , reject H0.
Was the campaign successful? (circle one) / Yes(by rejecting H0, we have evidence to say m > $17.85) / No

Question 2. The Survey of Study Habits and Attitudes (SSHA) is a psychological test that measures the motivation, attitude toward school, and study habits of students. The mean score for U.S. college students is about 115, and the standard deviation is about 30. A teacher suspects that older students (30 years or older) have better attitudes toward school and wishes to test H0:  = 115 vs. Ha:  > 115. To test the hypothesis, she decides to use a sample of n = 25, and to reject the null hypothesis H0 if the sample mean > 126.

You will receive 0 points if your answer does not appear inside the answer box. <

  1. Find the probability of a Type I error.

 = / Note that .
P( Type I error ) = P( reject H0 when H0 is true )
= P(  is 115 but a sample mean > 126 is observed )
= P( >126 when  = 115 ) = P( Z > 1.83 when  = 115 ) = 0.0336.
  1. Find the probability of a Type II error when  = 135.

 = / Note that .
P( Type II error ) = P( fail to reject H0 when Ha is true )
= P(  = 135 but a sample mean < 126 is observed )
= P( < 126 when  = 135 ) = P( Z < –1.5 when  = 135 ) = 0.0668.
  1. Find the power of the test when  = 135.

Power = / Note that .
P( reject H0 when Ha is true ) = 1 –  = 1 – 0.0668 = 0.9332.
  1. Suppose the sample mean turns out to be 131.2. Find the p-value of the test.

p-value / Note that .
P( observing to be 131.2 or more extreme when H0 is true )
= P(> 131.2 when  = 115 )
= P( Z > 2.7 when  = 115 ) = 0.0035.
  1. Construct a 95% confidence interval for the mean SSHA score for older students. Write your final answer in the following format: ( point estimate ) ± ( margin of error )

Answer / ( quick answer ) ± ( multiplier  )
( 131.2 ) ± ( 1.96  ), or
( 131.2 ) ± ( 2 )

Hsieh, P-H1