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.. 7.-8 8.-8 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 0.58. 95, y 4. y 4 ˆ 95 y 35 4 39 p ˆ 0.49. 605 95 800 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
α 0.05. The critical (rejectio) regio is z < z α/.96 or z > z α/.96. The observed value of z (test statistic) 0.58 0. z 8.988 0.49 0.5 ( ) 605 95 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 0.05. 0.580.4 605 0.0.79 95 z α.96. ( 0.58 0. ) ±.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 0.05. z α.96. 0.490.5 0.490.5 0.49.96, 0.49.96 [ 0.455, 0.55 ]. 800 800
. 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 435 40 (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 0 0 0 0 0 45 5 5 45 5 0.69867 0 degrees of freedom. t 0.05 ( 0 ).086, ( 435 40 ) ±.086 0 45 0 5 5 ± 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 ( 435 40 ) s s 0 0 0 45 5.890. Critical Value: t ( 0 ).75. Reject H 0.05 0 at α 0.05.
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 50 30. x x 4 05 30 35 50 x ( ) x x 65 0 5 400 0 050 5 0 5 0 ( x x ) 050 3 s x 350. y y 906 5. y y 6 6 4 46 56 66 7 y ( ) y y 65 00 5 5 5 400 0 400 5 0 5 5 5 0 ( y y ) 400 5 s y 80.
( 4 ) 350 ( 6 ) 80 s 306.5. s pooled 7.5. pooled 4 6 ( X Y ) δ ( 30 5) Test Statistic: T 0 s pooled m 7.5 0 4 6.859. m 4 6 8 d.f. t 0.05 ( 8 d.f. ).860. p-value ( tailed ) 0.05 0.0.
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 48.07 pooled ( X Y ) t 0.05 9 6 ± tα spooled 9 6 3 degrees of freedom m ( 3 ).069 (,40 5,700 ) ±.069 48.07 9 6,540 ± 360.4 ( 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 450 400 9 6 450 400 9 9 6 6 t 0.05 ( 5 ).3 (,40 5,700 ) ±.3 6,540 ± 384.7 ( 6,55.83, 694.7 ) 450 400 9 6
5. Six childre are tested for pulse rate before ad after watchig a violet movie with the followig results. Child Before After 0 96 08 3 89 94 4 04 5 90 0 6 85 98 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 0 96 89 04 90 85 After 08 94 0 98 Differece 0 5 8 3 d 0 5 8 3 6 60 0. 6 Σ d 00 44 5 64 44 69 646. s d d d 646 6 0 6 46 9.. 5 OR s d ( d d ) 0 4 5 4 4 9 5 46 9.. 5 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 0 0 9. 6 8.076.
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 0.05. OR 5 degrees of freedom. t 0.005 4.03. Right tail is less tha 0.005. P-value ( two tails ) is less tha 0.0. ( p-value 0.00047 ) P-value < 0.05 α. Reject H 0 at α 0.05. OR Cofidece iterval: 5 degrees of freedom. α 0.05, α / 0.05, 0.57 9. 6 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.
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 35 0 5 80 9 / 6 45 80 3 / 6 5 80 4 / 6 0 ( O ) ( 35 45) ( 0 5) ( 5 0) 45 5..6667.5 0 χ cells ( O )..6667.5 5.389. k 3 d.f. Rejectio Regio: Reject H 0 if χ χ α i) χ 0.05 5.99. 5.389 χ < χ α 5.99. Do NOT Reject H 0 at α 0.05. ii) χ 0.0 4.605. 5.389 χ > χ α 4.605. Reject H 0 at α 0.0.
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 70 40 50 60 9 / 6 90 60 3 / 6 30 60 4 / 6 40 ( O ) ( 70 90) ( 40 30) ( 50 40) 90 30 4.4444 3.3333.5 40 χ cells ( O ) 4.4444 3.3333.50 0.778. k 3 d.f. Rejectio Regio: Reject H 0 if χ χ α χ 0.0 9.0. 0.778 χ > Reject H 0 at α 0.0. χ α 9.0.
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 3 4 5 6 Observed frequecy 4 4 8 7 5 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 4 4 8 7 5 0 0 0 0 0 0 ( O ) ( 4 0) ( 4 0) ( 8 0) ( 7 0) ( 0) ( 5 0) 0 0 0 0 0 0.80 0.80 3.0 0.45 3.0.5 χ cells ( O ).80 0.80 3.0 0.45 3.0.5 0.7. c) Fid the critical value χ α. Rejectio Regio: Reject H 0 if χ χ α k 6 5 d.f. χ 0. 05.07. 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.
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 ) 3 38 70 8 4 0.5749 0.38095 8 30 8.333333 0.888889 40 60 00 Q 3.74603. ( ) ( ) degree of freedom. χ 0. 0 ( ).706. Q 3.74603 >.706 χ α. Reject H 0.
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 35 5 35 0 0 Girls 5 0 5 30 5 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 35 5 35 0 0 5 30.5 30 0.5 0.83333 0.7778 0.83333 5.00000 0.7778 Girls 5 0 5 30 5 5 30.5 30 0.5 0.83333 0.7778 0.83333 5.00000 0.7778 Total 60 45 60 40 45 50 Q 4.4444. ( ) ( 5 ) 4 degrees of freedom. χ 0. 0 ( 4 ) 3.8. Q 4.4444 > 3.8 χ α. Reject H 0.
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 5 5 6 40 4 60 Over 60 Sugar Sweeteed 50 60 55 35 Fruit Sweeteed 5 35 55 35 Natural 5 55 40 30 Test whether sweeteer prefereces ad age are idepedet at a % level of sigificace. Age (Years) Cereal 5 5 6 40 4 60 Over 60 Total Sugar Sweeteed 50 60 55 35 00 40 60 60 40.5 0 0.4667 0.65 Fruit Sweeteed 5 35 55 35 50 30 45 45 30 0.83333.. 0.83333 Natural 5 55 40 30 50 30 45 45 30 0.83333. 0.55556 0 Total 00 50 50 00 500 O ( O ) Q 3.6389. ( 3 ) ( 4 ) 6 degrees of freedom. Q 3.6389 < 6.8 χ 0. 0 ( 6 ) 6.8. χ α ( k ). Do NOT Reject H 0.