Formulas:

Each problem is worth 11 points.

1. A real estate company appraised the market value of 20 homes in a prestigious neighborhood. The sample mean was $960,000 with standard deviation $63,000. Find a 95% confidence interval for the market value of all homes in the neighborhood.

   Use t.

   

   

   

   We are 95% certain that the mean market value for all homes is between $930,315 and $989,485.

2. Find the p-value for the hypothesis test H_o: m > 12, H_a: m < 12, if s = 5, = 10, and n = 49.

   H_o: m = 12
   H_a: m < 12
   Assume H_o: m = 12

   Test Statistic:
   

   Rejection Region:
   

   p = 0.5 - 0.4974 = 0.0026

3. In a survey of 250 voters prior to an election, 44% indicated that they would vote for the incumbent candidate. Can we conclude with 95% confidence that the incumbent will lose the majority vote?

   Let p be the portion of voters who vote for the incombent.
   H_o: p = 0.50
   H_a: p < 0.50
   Assume H_o: p = 0.50
   

   Check for normality:
   n p = (250)(0.50) = 125 > 5
   n q = (250)(0.50) = 125 > 5
   

   Rejection Region:
   

   Test Statistic:
   

   Reject H_o. We are 95% certian that the incumbent will lose the majority of the vote.

4. A specialist in hypertension claims that regular aerobic exercise can reduce high blood pressure more effectively than medication can. To test the claim, 50 high blood pressure patients undergo an exercise regimen while another 60 patients take standard medication. The exercise patients experience an average 14.3 percent reduction in their blood pressure with standard deviation 1.6 percent, and the medicated patients experience an average 13.2 percent reduction in their blood pressure with standard deviation 1.8 percent. Can we conclude with 95% confidence that the specialist is correct?

   Two population, use z independent samples.

   Let m₁ be the mean reduction using exercise and m₂ be the mean reduction using medication

   H_o: m₁ - m₂ = 0
   H_a: m₁ - m₂ > 0
   Assume H_o: m₁ - m₂ = 0
   

   Rejection Region:
   

   Test Statistic:
   
   
   
   = 2.78

   Reject H_o. We are 95% certain that exercise is more effective.

5. A production manager would like to estimate the mean time required for workers to complete a task on an assembly line. How large a sample is required to estimate the mean time to within 5 seconds with 90% confidence. Assume s is 80 seconds.

   Find n so that z_a/2 s/ < 5

   1.645 80/ < 5
   > (1.645 80)/5
   n > [(1.645 80)/5]² = 692.7

   Round up to n = 693.

6. A company wants to locate a new facility where the average commuting time for its employees is under 30 minutes. For one proposed site, a random sample of 20 employees have a mean commuting time of 26 minutes with standard deviation 11 minutes. Is this an acceptable location? Let a = 0.05.

   H_o: m = 30
   H_a: m < 30
   Assume H_o: m = 30
   

   Rejection Region:
   

   Test Statistic:
   

   Draw no conclusion.

7. A test is conducted to determine if there is a difference between the size of tips received by waiters and waitresses. A randomly selected wiater and waitress from each of 6 restaurants earned the following tips. Attempt to show that there is a difference between the tips received by waiters and waitresses. Let a = 0.05.

                                       Restaurant
                1          2          3          4          5          6          
   Waiter     12.3%      10.8%      14.2%      19.3%      13.7%      20.6%      
   Waitress   13.1        8.2       12.4       18.1       14.7       18.4       
   
   difference -0.8        2.6        1.8        1.2       -1.0        2.2
   
   

   Matched pairs.

   H_o: m₁ - m₂ = 0 (m = 0)
   H_a: m₁ - m₂ 0 (m = 0)
   Assume H_o: m₁ - m₂ = 0 (m = 0)
   

   Rejection Region:
   

   Test Statistic:
   
   
   

   Draw no conclusion.

8. A bank wishes to determine the average time that its customers have to wait in line. A random sample of 100 customers wait an average of 7.2 minutes with standard deviation 2 minutes. Estimate with 90% confidence, the mean waiting time for all customers.

   

   u = + z_a/2 s/ = 7.2 + 1.645 x 2/10 = 7.2 + 0.33

   We are 90% certain that the average amount owed for all delinquent accounts is between 6.87 and 7.53.

9. Social scientists are interested in measuring the variability of incomes among different countries, because greater variability may indicate social inequities. A random sample of 25 individuals from one country had a mean income of $20,000 with standard deviation $8000. A random sample of 25 individuals from one country had a mean income of $20,000 with standard deviation $8000. A random sample of 30 individuals from a second country had a mean income of $30,000 with standard deviation $7000. Is there sufficient evidence to conclude, with 95% confidence, that one country's income is more variable than the other?

   H_o: s₁ = s₂
   H_a: s₁ s₂
   Assume H_o: s₁ = s₂
   Rejection Region:
   
   

   Test Statistic:
   
   

   Draw no conclusion.