IBA) Tutorial 6 Full solutions

Business Statistics (BK/IBA) Tutorial 6 – Full solutions Testing contingency table Q1 (based on Doane & Seward, 4/E, 15.2) Teenagers make up a large ...
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Business Statistics (BK/IBA) Tutorial 6 – Full solutions Testing contingency table Q1

(based on Doane & Seward, 4/E, 15.2) Teenagers make up a large percentage of the market for clothing. Below are data on running shoe ownership in four world regions (excluding China). a. At  = .01, does this sample show that running shoe ownership depends on world region? b. Which combinations contribute most to the dependence? c. Can you estimate the proportion of Europeans who possess running shoes? d. Can you estimate the proportion of Europeans among the running shoe owners? (See J. Paul Peter and Jerry C. Olson, Consumer Behavior and Marketing Strategy, 9th ed. [McGraw-Hill, 2004], p. 64.)

A1

a. Age Group and Region are dependent; b. Adults in Europe and Adults in Latin America; c. No; d. No. a. Five step-procedure: (i) : Age Group and World Region are independent; : Age Group and World Region are dependent;  = 0.01

Sol

(ii) Sample statistic: = ∑,



   

; reject for “too large” values

(iii) Under , ~  ; provided all expected cell frequencies ≥ 5 (iv) Calculations: table with observed and expected frequencies, and each cell’s contribution to  ! :



 ! = 19.31; $% = ;. = 11.34 As we see, all expected frequencies are ≥ 5, so no problem with the -test condition.

(v) Reject  because  ! ≥ $% , and conclude that Age Group and Region are dependent. b. Adults in Europe and Adults in Latin America contribute the most (7.24 and 4.77 of 19.31). c. No. The cross table is only for persons that do possess running shoes.

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d. No. You can see that 100 running shoe owners from 4 continents were sampled, but this of  course doesn’t mean that of the running shoes is found in Europe. *

Q2

(based on Doane & Seward, 4/E, 15.5) Marketing researchers prepared an advance notification card announcing an upcoming mail survey and describing the purpose of their research. Half the target customers received the prenotification, followed by the survey. The other half received only the survey. The survey return rates are shown below. Research question: At  = .025, is return rate independent of prenotification? a. Use the approach with a contingency table. b. Use the approach with a two-sample proportion test. (Data are from Paul R. Murphy, Douglas R. Dalenberg, and James M. Daley, “Improving Survey Responses with Postcards,” Industrial Marketing Management 19, no. 4 [November 1990], pp. 349–355.)

A2

a. Return Rate and Pre-notification Status are dependent; b. persons with a pre-notification returned the questionnaire significantly more often. a. Five step-procedure: (i) : Return Rate and Pre-notification Status are independent; : Return Rate and Prenotification Status are dependent;  = 0.25

Sol

(ii) Sample statistic: = ∑,



   

; reject for “too large” values

(iii) Under , ~  ; provided all expected cell frequencies ≥ 5 (iv) Calculations: table with observed and expected frequencies, and each cell’s contribution to  ! :



 ! = 5.42; $% = ;. + = 5.024 As we see, all expected frequencies are ≥ 5, so no problem with the -test condition.

(v) Just reject  because  ! ≥ $% , and conclude that Return Rate and Pre-notification Status are dependent.

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b. You must choose the two sub-populations and then define the “success”. For instance, group 1 received a pre-notification, and group 2 did not receive a pre-notification; “success” indicates if the survey was returned. The five-step procedure then gives: (i)  : - = - ;  : -_1 ≠ - ;  = 0.025 (ii) Sample statistic 0 − 0 ; reject for “too small” and “too large” values. 2  (iii) Under , ~;50,16; assumptions *

(iv) 0 = 0 = 0̅ = =*9= = ?>; @ ! = .A = 2.328; @$% = @. +/ = 2.24 (v) Just reject  because @ ! ≥ @$% , and conclude that Return Rate for persons with and without a Pre-notification are not the same. Post-hoc managerial conclusion: - > - : persons with a pre-notification returned the questionnaire significantly more often. Power and design Q1

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Q2

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(Doane & Seward, 4/E, 8.55) Inspection of a random sample of 19 American Airlines MD-80 aircraft showed that 15 needed repairs to fix a wiring problem that might compromise safety. a. How large a sample would be needed to estimate the true proportion of jets with the wiring problem, with 90 percent confidence and an error of ±6 percent? b. Would the airline actually conduct further sampling, or just inspect all the planes? Explain. (Source: www.comcvast.net/news, accessed March 26, 2008.) a. < = 125. H5H6

a. G = @3

I

⇒ < = @

H5H6 K

= 1.645

.A=×.  .>

= 124.7, so take 125.

b. Conduct inspection of a sample of 125. A fleet size is much larger. Note: always round upward. + Note: the value of - is based on the small-sample estimate of . But as soon as the true - is = smaller than 0.79, we would need a higher sample size in order to reach the desired precision. It would be safer to use for - a value closer to 0.5. For instance, - = 0.5 yields < = 188. (Doane & Seward, 4/E, 9.1) If you repeated a hypothesis test 1,000 times (in other words, 1,000 different samples from the same population) how many times would you expect to commit a Type I error, assuming the null hypothesis were true, if a.  = .05; b.  = .01; c.  = .001? a. 50; b. 10; c. 1. a. 0.05 × 1000 = 50 times b. 0.01 × 1000 = 10 times c. 0.001 × 1000 = 1 time (Doane & Seward, 4/E, 9.2) Define Type I and Type II error for each scenario, and discuss the cost(s) of each type of error. a. A 25-year-old ER patient in Minneapolis complains of chest pain. Heart attacks in 25 year olds are rare, and beds are scarce in the hospital. The null hypothesis is that there is no heart attack (perhaps muscle pain due to shoveling snow).

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Q4

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b. Approaching O’Hare for landing, a British Air flight from London has been in a holding pattern for 45 minutes due to bad weather. Landing is expected within 15 minutes. The flight crew could declare an emergency and land immediately, but an FAA investigation would be launched. The null hypothesis is that there is enough fuel to stay aloft for 15 more minutes. c. You are trying to finish a lengthy statistics report and print it for your evening class. Your color printer is very low on ink, and you just have time to get to Staples for a new cartridge. But it is snowing and you need every minute to finish the report. The null hypothesis is that you have enough ink. a. Null hypothesis: The man is not having a heart attack. Type I error: I admit him when he does not have a heart attack. Type II error: I fail to admit him, he does have a heart attack. Better to make a Type I than Type II error. b. Type I error: I reject the null and let them land even though they could have stayed up for 15 minutes (or more). Type II: Don’t let the plane land and the plane runs out of fuel. It is more costly to make a Type II error. c. Type I error: I reject the null and rush out to Staples, get caught in the snow and fail to finish the report (when if I had stayed I would have finished it). Type II error: I run out of ink and can’t finish the report. Better to stay and try to finish the report, in fact better to print out some of it than none of it. Better to make a Type I error. (Doane & Seward, 4/E, 9.4) A hotel installs smoke detectors with adjustable sensitivity in all public guest rooms. a. State the null and alternative hypotheses. b. Define Type I and II error. What are the consequences of each? c. Which is more to be feared, and by whom? d. If the hotel decides to reduce M risk, what would be the consequences? Who would be affected? a. Null hypothesis: There is no fire. Alternative hypothesis: There is an fire. b. Type I error: The hotel is evacuated while there is no fire. Guests will complain, ask for a refund, not return the next year, write bad reviews on internet. Type II error: The hotel is not evacuated while there is a fire. Guests may be injured or die, the hotel’s management may receive huge claims and loose permits. c. Type II error is the worst, for managers and for guests. d. The consequence would be that the probability of a type I error might increase. More fire alarms, for instance, reduce M, but increase . We have a sample (size < = 100) from a normally distributed population with unknown N and O = 12. We want to test the hypothesis  : N ≤ 100 against  : N > 100. It is decided to reject the null hypothesis when Q̅ ≥ 102. a. Find the probability of a type I error. b. Find the power of this test for N = 101, N = 102 and N = 105. c. Give an interpretation of the type I error and give an interpretation of the power for N = 105. a. 0.0475; b. 0.2033, 0.5000 and 0.9938 a. RST 5reject  6 = R[\ 5reject  6 = R[\ 5]^ ≥ 1026 = = R 7_ ≥

  :  /√

≈ R5_ ≥ 1.676 = 0.0475

= R 7_ ≥

  :  /√

≈ R5_ ≥ 0.836 = 0.2033

= R 7_ ≥

 

:  /√

≈ R5_ ≥ 0.006 = 0.5000

b. Rbcde5N = 1016 = R[\ 5reject  6 = R[\ 5]^ ≥ 1026 = Rbcde5N = 1026 = R[\ 5reject  6 = R[\ 5]^ ≥ 1026 =

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Rbcde5N = 1056 = R[\+ 5reject  6 = R[\+ 5]^ ≥ 1026 =

= R 7_ ≥

 +  /√

: ≈ R5_ ≥ −2.506 = 0.9938

c. If the true N equals 100, there is a probability of 4.75% that we will “reject”  anyway if we run the experiment described above. If the true N equals 105, there is a probability of more than 99% that we will “detect” that  is false if we run the experiment described above. Q6

A6 Sol

We run < = 16 times an experiment. For each experiment the result is either “success” or “failure”. Let ] be the number of successes in these 16 experiments. The experiments are identical and independent from one another. Denote the probability of a success by -. We want to   test the hypothesis  : - = against  : - ≠ .



a. Find the rejection region (large and small values) such that the probability of a type I error is at most 5%. How large is the probability of type I error? b. Find the power of this test for - = 0.6, - = 0.7 and - = 0.8. a. Reject for Q ≤ 3 and for Q ≥ 13; b. 0.0660, 0.2459, 0.5981 a. Reject for Q ≤ 3 and for Q ≥ 13. Note that RST 5] ≤ 36 = RH\.+ 5] ≤ 36 = 0.0002 + 0.0018 + 0.0085 = 0.0105, but that  RH\.+ 5] ≤ 46 > × 5% = 2.5%.

b. Rbcde5- = 0.66 = RH\.> 5reject  6 = = RH\.> 5X ≤ 36 + RH\.> 5X ≥ 136 = 0.0009 + 0.0651 = 0.0660 Rbcde5- = 0.76 = RH\.A 5reject  6 = = RH\.A 5X ≤ 36 + RH\.A 5X ≥ 136 = 0.0000 + 0.2459 = 0.2459 Rbcde5- = 0.86 = RH\.? 5reject  6 = = RH\.? 5X ≤ 36 + RH\.? 5X ≥ 136 = 0.0000 + 0.5981 = 0.5981

Old exams & Remaining questions As “mixed exercises”, we propose to study one old exam at the tutorial session. To synchronize the activities, please use the exam of 26 May 2014. In addition, or instead, the tutorial may be devoted to exercises of the previous sessions that still need attention.

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