SOLUTIONS y n. n 1 = 605, y 1 = 351. y1. p y n. n 2 = 195, y 2 = 41. y p H 0 : p 1 = p 2 vs. H 1 : p 1 p 2.

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1 STAT 400 UIUC Practice Problems # SOLUTIONS Stepaov Dalpiaz The followig are a umber of practice problems that may be helpful for completig the homework, ad will likely be very useful for studyig for exams The April 8, 994 issue of Time magazie reported the results of a telephoe poll of 800 adult Americas, 605 osmokers, who were asked the followig questio: Should the federal tax o cigarettes be raised by $.5 to pay for health care reform? Let p ad p equal the proportios of osmokers ad smokers, respectively, who say yes to this questio. y 35 osmokers ad y 4 smokers said yes. 605, y 35. y 35 ˆ 605 p , y 4. y 4 ˆ 95 y p ˆ y p 0.. a) With α 0.05, test H 0 : p p vs. H : p p. H 0 : p p vs. H : p p. Two tailed. The test statistic is z pˆ ( pˆ ) pˆ pˆ, where pˆ y pˆ pˆ. y

2 α The critical (rejectio) regio is z < z α/.96 or z > z α/.96. The observed value of z (test statistic) z ( ) is greater tha.96 (the test statistic does fall ito the rejectio regio), so Reject H 0. b) Fid a 95% cofidece iterval for p p. ( p ˆ pˆ ) ± z α pˆ ( pˆ ) pˆ ( pˆ ) 95% cofidece level α 0.05 α z α.96. ( ) ±.96 or [ 0.30, 0.44 ]. c) Fid a 95% cofidece iterval for p, the proportio of adult Americas who would say yes. p ˆ ± z pˆ ( pˆ ) α. 95% cofidece level α 0.05 α z α , [ 0.455, 0.55 ]

3 . A ew method of storig sap beas is believed to retai more ascorbic acid tha the old method. I a experimet, sap beas were harvested uder uiform coditios ad froze i 5 equal-size packages. Te of those packages were radomly selected ad stored accordig to the ew method, ad the other 5 packages were stored by the old method. Subsequetly, ascorbic acid determiatios (i mg/kg) were made, ad the followig summary statistics were calculated. New Method Old Method (sample) mea ascorbic acid (sample) stadard deviatio 0 45 a) Use Welch s T to costruct a 95% cofidece iterval for µ New µ Old. s s x y m s s x y m m degrees of freedom. t 0.05 ( 0 ).086, ( ) ± ± 7.6 or (.6, 5.6 ). b) Test H 0 : µ New µ Old vs. H : µ New > µ Old at a 5% level of sigificace. ( X Y ) δ 0 Test Statistic: T ( ) s s Critical Value: t ( 0 ).75. Reject H at α 0.05.

4 3. Assume that the distributios of X ad Y are N ( µ, σ ) ad N ( µ, σ ), respectively. Give the 4 observatios of X, ad the m 6 observatios of Y, 05, 30, 35, 50 6, 4, 46, 56, 66, 7 fid the p-value ( approximately ) for the test H 0 : µ µ vs. H : µ µ. Hit : Assume the populatio variaces are equal. x x x x x ( ) x x ( x x ) s x 350. y y y y y ( ) y y ( y y ) s y 80.

5 ( 4 ) 350 ( 6 ) 80 s s pooled 7.5. pooled 4 6 ( X Y ) δ ( 30 5) Test Statistic: T 0 s pooled m m d.f. t 0.05 ( 8 d.f. ).860. p-value ( tailed )

6 4. A radom sample of 9 adult elephats had the sample mea weight of,40 pouds ad the sample stadard deviatio of 450 pouds. A radom sample of 6 adult hippos had the sample mea weight of 5,700 pouds ad the sample stadard deviatio of 400 pouds. Assume that the two populatios are approximately ormally distributed, Costruct a 95% cofidece iterval for the differece betwee their overall average weights of adult elephats ad adult hippos. a) Assume that the overall stadard deviatios are equal. ( 9 ) 450 ( 6 ) 400 s 74,78.6 s pooled pooled ( X Y ) t ± tα spooled degrees of freedom m ( 3 ).069 (,40 5,700 ) ± ,540 ± ( 6,79.6, 6,900.4 ) 6 b) Do NOT assume that the overall stadard deviatios are equal. Use Welch s T. s s s s 5. 5 degrees of freedom t 0.05 ( 5 ).3 (,40 5,700 ) ±.3 6,540 ± ( 6,55.83, )

7 5. Six childre are tested for pulse rate before ad after watchig a violet movie with the followig results. Child Before After Usig the paired t test, test for differeces i the before ad after mea pulse rates. Use α 0.05 ad use a two-sided test. Before After Differece d Σ d s d d d OR s d ( d d ) H 0 : µ B µ A vs. H : µ B µ A. H 0 : µ D 0 vs. H : µ D 0. Test Statistic: T d 0 sd

8 5 degrees of freedom. α 0.05, α / 0.05, ± t α ±.57. Critical Values The test statistic is i the Rejectio Regio. Reject H 0 at α OR 5 degrees of freedom. t Right tail is less tha P-value ( two tails ) is less tha 0.0. ( p-value ) P-value < 0.05 α. Reject H 0 at α OR Cofidece iterval: 5 degrees of freedom. α 0.05, α / 0.05, s d ± t d α. t α.57. ± 0 ± 3. ( 6.8, 3. ) 95% cofidece iterval does NOT cover zero Reject H 0 at α 0.05.

9 6. Crosses of mice will produce either gray, brow, or albio offsprig. Medel s model predicts that the probability of a gray offsprig is 9 / 6 ; the probability of a brow offsprig is 3 / 6 ; ad the probability of a albio offsprig is 4 / 6. a) A experimet to assess the validity of Medel s theory produces the followig data: 35 gray offsprig; 0 brow offsprig; ad 5 albio offsprig. Test H 0 : p 9 / 6, p 3 / 6, p 3 4 / 6 vs H : ot H 0 i) at α 0.05, ii) at α 0.0. gray brow albio O / / / 6 0 ( O ) ( 35 45) ( 0 5) ( 5 0) χ cells ( O ) k 3 d.f. Rejectio Regio: Reject H 0 if χ χ α i) χ χ < χ α Do NOT Reject H 0 at α ii) χ χ > χ α Reject H 0 at α 0.0.

10 b) Suppose the experimet i part (a) is repeated, but with twice as may observatios. Suppose also that we happeed to get the same proportios, amely, 70 gray offsprig; 40 brow offsprig; ad 50 albio offsprig. Repeat part (a) i this case, usig α 0.0. black brow albio O / / / 6 40 ( O ) ( 70 90) ( 40 30) ( 50 40) χ cells ( O ) k 3 d.f. Rejectio Regio: Reject H 0 if χ χ α χ χ > Reject H 0 at α 0.0. χ α 9.0.

11 7. Suppose we toss a 6-sided die 0 times ad cout how may times each outcome ( through 6) occurs. We obtai the followig results: Outcome Observed frequecy We wat to use the chi-square goodess-of-fit test to test the hypothesis that the die is fair (balaced) usig a 5% level of sigificace. a) State the ull hypothesis. H 0 : p / 6, p / 6, p 3 / 6, p 4 / 6, p 5 / 6, p 6 / 6 vs H : ot H 0 b) Calculate the values of the chi-square test statistic. O ( O ) ( 4 0) ( 4 0) ( 8 0) ( 7 0) ( 0) ( 5 0) χ cells ( O ) c) Fid the critical value χ α. Rejectio Regio: Reject H 0 if χ χ α k 6 5 d.f. χ d) Test the hypothesis that the die is fair (balaced) usig a 5% level of sigificace. 0.7 χ >X χ α.07. Do NOT Reject H 0 at α 0.05.

12 8. A article i Busiess Week reports profits ad losses of firms by idustry. A radom sample of 00 firms is selected, ad for each firm i the sample, we record whether the compay made moey or lost moey, ad whether or ot the firm is a service compay. The data are summarized i the table below. Use a 0% level of sigificace to test whether the two evets the compay made profit this year ad the compay is i the service idustry are idepedet. Idustry Type Service Noservice Profit 3 38 Loss 8 O ( O ) Q ( ) ( ) degree of freedom. χ 0. 0 ( ).706. Q >.706 χ α. Reject H 0.

13 9. A group of 50 childre (5 boys ad 5 girls) were asked to idetify their favorite color. We obtai the followig data: Favorite Color Sex Red Gree Blue Pik Purple Boys Girls A toy maufacturer wats to kow if the color prefereces of boys ad girls differ. Perform χ test of homogeeity usig a % level of sigificace. H 0 : I all 5 respose categories ( Red, Gree, Blue, Pik, Purple ), the probabilities are equal for these populatios (Boys, Girls). H A : Not H 0. Favorite Color Sex Red Gree Blue Pik Purple Total O ( O ) Boys Girls Total Q ( ) ( 5 ) 4 degrees of freedom. χ 0. 0 ( 4 ) 3.8. Q > 3.8 χ α. Reject H 0.

14 0. A breakfast cereal maufacturer wats to kow whether idividual prefereces for types of breakfast cereal sweeteer are associated with the age of the buyer. I a survey, sweeteer prefereces were matched to the cosumers age group. From a radom sample of 500 resposes, the results were as follows: Age (Years) Cereal Over 60 Sugar Sweeteed Fruit Sweeteed Natural Test whether sweeteer prefereces ad age are idepedet at a % level of sigificace. Age (Years) Cereal Over 60 Total Sugar Sweeteed Fruit Sweeteed Natural Total O ( O ) Q ( 3 ) ( 4 ) 6 degrees of freedom. Q < 6.8 χ 0. 0 ( 6 ) 6.8. χ α ( k ). Do NOT Reject H 0.