BUSI 275 Fall 2011 Midterm Exam Practise Questions

1. A quality control inspector has rated each batch produced today on a scale from A through E, where A represents the best quality and E is the worst. What is the level of measurement of this variable?
   Ordinal
2. Your customers’ average order size is $2601, with a standard deviation of $1275. Suppose 45 typical customers independently placed orders tomorrow. What are the chances that tomorrow’s average order per customer will be between $2450 and $2750?
   SE=190, z = -0.7945 and +0.7839, so area in between is 57%
3. Using the information in the preceding problem, produce a 95% confidence interval on average order per customer per day, assuming 20 customers per day.
   SE=285.10, z = +/- 1.96, so conf. int. is between $2042.22 and $3159.78.
4. The table below shows sales of some “light” foods:
   

   "Light" Food	Sales ($Millions)
   Entenmann's Fat Free baked goods	$125.5
   Healthy Request soup	$123.0
   Kraft Free processed cheese	$83.4
   Aunt Jemima Lite and Butter Lite pancake syrup	$58.0
   Fat Free Fig Newtons	$44.4
   Hellmann's Light mayonnaise	$38.0
   Louis Rich turkey bacon	$32.1
   Kraft Miracle Whip free	$30.3
   Ben & Jerry's frozen yogurt	$24.4
   Hostess Lights snack cakes	$19.3
   Perdue chicken/turkey franks	$3.8
   Milky Way II candy bar	$1.1

   My company is planning to launch a new brand of light food. Our goal is to reach at least the 20th percentile of current brands. What is our sales goal, in dollars?
   rank=2.4, round up to 3: 3rd entry is $19.3.
   Or, using PERCENTILE(): $20.32.
   The difference is due to the different way Excel computes percentiles.
5. Produce a relative frequency histogram for the variable "Airline" in the dataset AirlinePassengers.xls (linked).
6. Using the same dataset from the preceding problem, produce a joint frequency distribution for the two variables, "Male/Fem." and "Business or Pleasure". Are these two variables independent in this dataset? Why or why not?
   No: e.g., P(Business | Male) ≠ P(Business | Female)
7. You have followed up on people who received your catalog mailing. You found that 4% of these people ordered the hat, and 6% ordered the mittens. Of the people who ordered the hat, 55% also ordered the mittens. Of the people who did not order the hat, what percentage ordered the mittens?
   Given: P(H)=4%, P(M)=6%, P(M|H)=55%, the question is looking for P(M|noH).
   Then P(M and H) = (.04)(.55) = 2.2%.
   Since P(M and H) and P(M and noH) must add up to P(M), so P(M and noH) must equal 6% - 2.2% = 3.8%.
   We are looking for P(M|noH), which is P(M and noH) / P(noH).
   P(noH) must be 100% - P(H) = 96%, so
   P(M|noH) = 3.8% / 96% = 3.958333%.
8. Here are the satisfaction scores given by 12 randomly selected customers:
   

   89

   98

   96

   65

   99

   81

   76

   51

   82

   90

   96

   76

   Does the observed average differ significantly from the target score of 80?
   mean = 83.25, SD = 14.63, SE = 4.223.
   z = (80-83.25)/SE = -0.77, % area in left tail = 22.08%.
   No, does not significantly differ (technically, this question is a ch9 question if we consider an α, but the components are within ch7-8).
9. In a nationwide poll for a political candidate, we found that 309 out of 1105 registered voters claimed to be in favor of that candidate. If we report our findings with 95% confidence, what is our margin of error?
   SE = 1.3502%, z=1.96, so margin of error (the +/- in the confidence interval) is just (1.96)(1.3502%) = 2.65%
10. In order to pay your firm’s debts this year, you will need to be awarded at least 2 contracts. This is not usually a problem, since the yearly average is 5.1 contracts. What is the probability that you will not earn enough to pay your firm’s debts this year? (Hint: which distribution is appropriate?)
    POISSON(1, 5.1, 1) = 3.72%
11. Your client has a particularly complex database procedure it needs to run quickly. Your company can provide a new computer system to speed it up. You have run the procedure on your new system 14 times (independent runs) and obtained the following runtimes (in minutes):
    

    5

    8

    5

    6

    11

    9

    8

    10

    5

    11

    6

    5

    5

    10

    You would like to claim that your new system is very fast. In your advertising, you will say, "average processing time is as low as ___". Find the appropriate bound of the two-sided 95% confidence interval.
    t-score is TINV(.05,13) = 2.16.
    mean = 7.43, sd = 2.41, SE = 0.6438.
    Conf int is 6.038 to 8.819 (we want the lower bound).
12. You would like to estimate the average price of a stock to a precision of +/- 1%, with 95% confidence. Assume the volatility (coefficient of variation) of the stock is 12%. Assuming each day's price is independent and normally distributed, how many days should you track the stock in order to attain this precision in estimating its average price?
    SD = 0.12μ, so SE = 0.12μ/sqrt(n).
    Margin of error = 0.01μ.
    95% confidence implies a z-score of NORMSINV(0.975) = 1.96, so
    1.96 = 0.01μ / (0.12μ / sqrt(n)).
    Solving for n: n = ( (1.96)(0.12)/(0.01) )² = 554
13. (p.255 #41) The mean time between failures (MTBF) for a particular model of power supply unit (PSU) is 4,000 hours. Some PSUs may fail earlier; some later. What is the probability of a random PSU failing in less than 2,100 hours? (Hint: which probability distribution is appropriate?)