AMS572.01 Midterm Exam Fall, 2003

Instructions: This is a close book exam. Anyone who cheats in the exam shall receive a grade of F. Please provide complete solutions for full credit. Good luck!

1. In a study of hypnotic suggestion, 10 male volunteers were randomly allocated to an experimental group and a control group. Each subject participated in a two-phase experimental session. In the first phase, respiration was measured while the subject was awake and at rest. In the second phase, the subject was told to imagine that he was performing muscular work, and respiration was measured again. For subjects in the experimental group, hypnosis was induced between the first and second phases; thus, the suggestion to imagine muscular work was “hypnotic suggestion” for experimental subjects and “waking suggestion” for control subjects. The accompanying table shows the measurements of total ventilation (liters of air per minute per square meter of body area) for all 10 subjects.

Experimental Group Control Group Subject Rest Work Subject Rest Work 1 6 6 6 6 5 2 7 9 7 5 5 3 5 8 8 5 5 4 7 12 9 6 6 5 6 7 10 5 4

(a) Use suitable tests to investigate (Use α =.05 for each test. Please report the p-value for each test and state the assumption(s) of the test.) (i) the response of the experimental group to suggestion;

Assume that the difference d =work - rest is normal.

d=2.2, sd = 1.92 and n = 5

The hypotheses are H0 :md 0 v.s Ha:m d > 0 . The test statistic is d - 0 2.2 t0 = = = 2.56. sd / n 1.95/ 5

Since t4,0.05 = 2.132 and t0> t 4,0.05 = 2.132 , we reject H0 at a = 0.05.

t4,0.05=2.132 < 2.56 < t 4,0.025 = 2.776

� 0.025-p < value 0.05.

(ii) the response of the control group to suggestion;

Paired t-test. d =work - rest . Assume that the difference is normal. d= -0.4, sd = 0.55 and n = 5

The hypotheses are H0 :md = 0 v.s Ha: d  0 . The test statistic is d -0 - 0.4 t0 = = = -1.63. sd / n 0.55/ 5

Since t4,0.05 = 2.132 and t0 1.63  t 4,0.05  2.132 , we cannot reject H0 at a = 0.05.

t4,0.11.533  1.63  t 4,0.05  2.132 0.9 p  value  0.95.

(iii) The differences between the responses of the experimental and control groups:

We need to check whether the two population variances are equal by the F-test. 2 2 d1 22, s 1  3.7, n 1  5 and d20.4, s 2  0.3, n 2  5. 2 2 2 2 H0:s 1= s 2 v.s Ha :s1 s 2 The test statistic is 2 2 F0= s 1/ s 2 = 3.7 / 0.3 = 12.3 > F 4,4,0.05 = 6.39.

Therefore, we reject H0 at the significance level a = 0.1 and we can't use pooled variance t-test.

Using unequal-variance t test for two independent samples, 2 2 w1 s 1/ n 1  0.74 , w2 s 2/ n 2  0.06 and 2 (w1+ w 2 ) 0.64 df=2 2 = = 4.64. w1/( n 1- 1) + w 2 /( n 2 - 1) 0.1378 Therefore, df 5 . H : m= m H : m m The test statistic of hypotheses 0 d1 d 2 v.s a d1 d 2 .

d- d -2.2 - 0.4 t=1 2 = = - 2.9. 0 2 2 37 / 5 0.3/ 5 s1/ n 1+ s 2 / n 2 +

Since |t0 |> t 5,0.025 , we reject H0 . p-value = 2P(T > 2.9) and 0.025 < P(T > 2.9) < 0.05 Therefore 0.05 < p-value < 0.10.

(b) Please write up the entire SAS program necessary to answer questions raised in (a). Please include the data step as well as tests for testing for various assumptions.

data hypnosis; input group rest work @@; diff=work-rest; datalines; 1 6 6 1 7 9 1 5 8 1 7 12 1 6 7 2 6 5 2 5 5 2 5 5 2 6 6 2 5 4 ; run; proc univariate data=hypnosis normal; class group; var diff; run;

proc ttest data=hypnosis; class group; var diff; run;

proc npar1way data=hypnosis; class group; var diff; run;

2. Thanksgiving was coming up and Harvey's Turkey Farm was doing a land-office business. Harvey sold 100 gobblers to Nedicks for their famous Turkey-dogs. Nedicks found that 90 of Harvey's turkeys were in reality peacocks.

(a) Estimate the proportion of peacocks at Harvey's Turkey Farm and find a 95% confidence interval for the true proportion of turkeys that Harvey owns.

Let p be the proportion of peacocks and q be the proportion of turkeys. )90 ) ) p=100 =0.9, q = 1 - p = 0.1 and n =100 . 95% C.I on q is ) ) ) ) q(1- q ) 0.1 0.9 1-p = q� z � 0.1 1.96 [0.0412,0.1588]. a / 2 n 100

(b) How large a random sample should we select from Harvey's Farm to guarantee the length of the 95% confidence interval to be no more than 0.06? (Note: please first derive the general formula for sample size calculation based on the length of the CI for inference on one population proportion, large sample situation. Please give the formula for the two cases: (i) we have an estimate of the proportion and (ii) we do not have an estimate of the proportion to be estimated. (iii) Finally, please plug in the numerical values and obtain the sample size for this particular problem.)

L=0.06, E = L / 2 = 0.03. ) ) q(1- q ) (i) E= z a / 2 n ) ) 2 (z )2 q(1- q )骣 1.96 n =a / 2 = 琪鬃0.1 0.9∐ 385 E2 n 桫0.03 ) (ii) if q is unknown, 2 ) ) ) 1  1 1 q(1 q )   q     2  4 4

2 2 4(z )2 1 骣 z 骣1.96 n =a/ 2 = 琪 a / 2 . \n = 琪 1068. L2 4 桫 L 桫0.06

3. A random sample of Democrats and a random sample of Republicans were polled on an issue. Of 200 Republicans, 90 would vote yes on the issue; of 100 democrats, 58 would vote yes. Can we say that more Democrats than Republicans favor the issue at the 1% level of significance?

) 58 Democrats: p1=100 =0.58, n 1 = 100, x 1 = 58, n 1 - x 1 = 42. ) 90 Republicans: p2=200 =0.45, n 2 = 200, x 2 = 90, n 2 - x 2 = 110.

) x1+ x 2 58+ 90 Hypotheses are H: p= p v.s H: p> p . p = = . 0 1 2 a 1 2 n+ n 100 + 200 ) ) 1 2 p1- p 2 0.58- 0.45 z0 =) ) = ∐2.12 . p(1- p )(1/ n1 + 1/ n 2 ) 0.49� 0.51(1/100 1/ 200) p- value = P( z > z0 ) = 0.017 > 0.01.

We cannot reject H0 at a = 0.01 . Therefore, Democrats favor the issue as same as Republicans at the 1% significance level.