Stat 250.2 Fall 1999

Exam #2 Sample Questions

The following questions are meant to show you the types of questions I have asked on past midterm exams covering the same material.  The sample questions are not meant to be all inclusive, but rather to give you an idea of the format of my typical exam questions.  The coverage of topics may vary from semester-to-semester as well, so just because there is not a sample question pertaining to a particular topic does not mean it won't be addressed on the exam.

If I were taking this exam, and wanted to do well on it, I would:



Questions 1 and 2 pertain to the following situation.  The following is Minitab output for a confidence interval for µ, the average head circumference (in cm) of all Penn State students:
 
Confidence Intervals
Variable     N      Mean    StDev  SE Mean       95.0 % CI
head        63    56.468    2.298    0.290  (  55.889,  57.047)
__F__ 1.  Using the above confidence interval, which, if any, of the following null hypotheses would we reject?
   (i) H0: µ = 54
   (ii) H0: µ = 56
   (iii) H0: µ = 58
 
(A) None  (B) Only (i)  (C) Only (ii) (D) Only (iii)
(E) (i) and (ii) (F) (i) and (iii) (G) (ii) and (iii) (H) All
__F__ 2. Which, if any, of the following statements are true?
(i) Everything else remaining the same, a 95% confidence interval based on 100 students would be shorter than the 95% interval given above.
(ii) Everything else remaining the same, a 99% confidence interval would be shorter than the 95% interval given above.
(iii) Everything else remaining the same, if the head circumferences of the sampled students varied more, a 95% confidence interval would be longer than the 95% interval given above.
 
(A) None  (B) Only (i)  (C) Only (ii) (D) Only (iii)
(E) (i) and (ii) (F) (i) and (iii) (G) (ii) and (iii) (H) All


 __C__ 3. Which, if any, of the following statements correctly define(s) "p-value"?
(i) The p-value is the probability that the null hypothesis is true.
(ii) The p-value is the probability that we would observe a sample as extreme as we did under the assumption that the null hypothesis is true.
(iii) The p-value is the probability that the alternative hypothesis is true.
 
(A) None  (B) Only (i)  (C) Only (ii) (D) Only (iii)
(E) (i) and (ii) (F) (i) and (iii) (G) (ii) and (iii) (H) All

4.  The number of years a person lives after being diagnosed with a certain disease is normally distributed with a mean of 4 years and a standard deviation of 1 year.   Suppose a random sample of 16 people are selected from this diseased population.  What is the probability that the sample will live an average of less than 3.5 years?

P(X-bar < 3.5) = P[Z < (3.5 -4)/(1/ sqrt(16)] = P(Z < - 2) = 0.023


Questions 5 to 10 pertain to the following situation.  The average serum cholesterol level in a certain group of patients is 240 milligrams.  A new medication is designed to lower the cholesterol level in this population after one month.  A sample of 60 people in the population used the medication for 30 days, after which their average cholesterol level was 232 milligrams with a standard deviation of 18 milligrams.   The manufacturer of the new medication would like to see if the average cholesterol level in the population has been reduced.

5. Using statistical notation, specify the appropriate null and alternative hypotheses.

     H0: µ = 240
     HA: µ < 240

6.  Calculate the p-value for the hypothesis test.

P(X-bar < 232) = P[Z < (232 -240)/(18/ sqrt(60)] = P(Z <  -3.44) = 0.000...
 

7. (4 pts.).  Make a decision, "reject" or "do not reject," at the a = 0.05 level.  Explain your reasoning.

Reject H0, since p = 0.000 is small and less than 0.05.
 

8. (2 pts.) Based on the decision you made in question #7, what type of error might you have made?

   Type I
 

9. (4 pts.)  Calculate a 95% confidence interval for µ, the average serum cholesterol level in the population after one month.
 

   232 ± 1.96(18/ sqrt(60))
   232 ± 4.6
   227.4 < µ < 236.6

10. (2 pts.)  Is your confidence interval in question #9 consistent with your decision in question #7?  Explain.

Yes, we'd still reject H0, since µ=240 is not in interval.
 


 __A__ 11.  A researcher was interested in evaluating whether Vitamin E delayed the admission of Alzheimer's patients to nursing homes.  The researcher recruited 90 patients who were taking Vitamin E as a dietary supplement and 80 patients who were not taking Vitamin E.  The researcher then recorded the number of days it took for each patient to be admitted to a nursing home.  What type of study did the researchers conduct?
 
 
(A) an observational study (B) an experiment

Question 12 pertains to the following situation.  Can we conclude that, on average, lymphocytes and tumor cells differ in size?  The following Minitab output tests the appropriate hypothesis using data on the cell diameters of 40 lymphocytes and 50 tumor cells.
 

Two Sample T-Test and Confidence Interval
Two sample T for Lympho vs Tumor
         N      Mean     StDev   SE Mean
Lympho  40      6.95      1.60      0.25
Tumor   50     17.92      2.97      0.42

95% CI for mu Lympho - mu Tumor:
T-Test mu Lympho = mu Tumor (vs not =): T= -21.05  P=0.0000  DF=  88
Both use Pooled StDev = 2.46
 

__C__ 12. Which, if any, of the following could possibly describe the confidence interval for µLympho- µTumor that is missing from the output?

(i)  ( -6.95, 17.92)
(ii)  ( -12.01,  -9.33)
(iii) ( -10.97, 10.97)
 
(A) None  (B) Only (i)  (C) Only (ii) (D) Only (iii)
(E) (i) and (ii) (F) (i) and (iii) (G) (ii) and (iii) (H) All


Questions 13 and 14 pertain to the following situation.  Researchers wished to know if they could conclude that premature babies begin walking at a later age than babies carried to full term.  They recorded the ages, in months, of two samples of such babies, and generated the following three sets of Minitab output:
 
Two Sample T-Test and Confidence Interval
Two sample T for FullTerm vs Premature
           N      Mean     StDev   SE Mean
FullTerm  12     10.38      1.40      0.40
Prematur  12     12.21      1.45      0.42

95% CI for mu FullTerm - mu Prematur: ( -3.04,  -0.63)
T-Test mu FullTerm = mu Prematur (vs not =): T= -3.16  P=0.0046 DF=22
Both use Pooled StDev = 1.42

Two Sample T-Test and Confidence Interval
Two sample T for FullTerm vs Premature
           N      Mean     StDev   SE Mean
FullTerm  12     10.38      1.40      0.40
Prematur  12     12.21      1.45      0.42

95% CI for mu FullTerm - mu Prematur: ( -3.04,  -0.63)
T-Test mu FullTerm = mu Prematur (vs <): T= -3.16  P=0.0023  DF=22
Both use Pooled StDev = 1.42

Two Sample T-Test and Confidence Interval
Two sample T for FullTerm vs Premature
           N      Mean     StDev   SE Mean
FullTerm  12     10.38      1.40      0.40
Prematur  12     12.21      1.45      0.42

95% CI for mu FullTerm - mu Prematur: ( -3.04,  -0.63)
T-Test mu FullTerm = mu Prematur (vs >): T= -3.16  P=1.0  DF=  22
Both use Pooled StDev = 1.42
__B__ 13. What is the p-value for the researchers' desired hypothesis?
 
 
(A) 0.0046 (B) 0.0023 (C) 1.0  (D) -3.16
 

__E__ 14. Which, if any, of the following would be valid conclusions that the researchers could draw from these three sets of output?  There is sufficient evidence to conclude that...

(i)  premature babies walk, on average, at a later age than full term babies
(ii)  premature babies walk, on average, at a different age than full term babies
(iii) premature babies walk, on average, at an earlier age than full term babies
 
(A) None  (B) Only (i)  (C) Only (ii) (D) Only (iii)
(E) (i) and (ii) (F) (i) and (iii) (G) (ii) and (iii) (H) All

Questions 15 and 16 pertain to the following situation.  Does sensory deprivation have an effect on a person's alpha-wave frequency?  Twenty volunteer subjects were randomly divided into two groups.  Subjects in group A were subjected to a 10-day period of sensory deprivation, while subjects in group B served as controls.  After the 10-day period, the alpha-wave frequency component of subjects' electroencephalograms were measured, and the following Minitab output generated:
 
Two Sample T-Test and Confidence Interval
Two sample T for GrpB vs GrpA
       N      Mean     StDev   SE Mean
GrpB  10    11.080     0.459      0.15
GrpA  10    10.280     0.598      0.19

95% CI for mu GrpB - mu GrpA: ( 0.30,  1.30)
T-Test mu GrpB = mu GrpA (vs not =): T= 3.36  P=0.0035  DF=  18
Both use Pooled StDev = 0.533

T-Test of the Mean
Test of mu = 11.000 vs mu not = 11.000

Variable     N      Mean    StDev   SE Mean        T          P
GrpA        10    10.280    0.598     0.189    -3.81     0.0042
GrpB        10    11.080    0.459     0.145     0.55       0.59
 

__B__ 15. What type of study did the researchers conduct?
 

(A) an observational study (B) an experiment
 

__A__ 16. Which of the following is the correct decision?

(A) P-value is 0.0035.  Conclude that sensory deprivation does have an effect on a person's alpha-wave frequency.
(B) P-value is 0.0035.  Conclude that sensory deprivation does not have an effect on a person's alpha-wave frequency.
(C) P-value is 0.0042.  Conclude that sensory deprivation does have an effect on a person's alpha-wave frequency.
(D) P-value is 0.0042.  Conclude that sensory deprivation does not have an effect on a person's alpha-wave frequency.
(E) P-value is 0.59.  Conclude that sensory deprivation does have an effect on a person's alpha-wave frequency.
(F)  P-value is 0.59.  Conclude that sensory deprivation does not have an effect on a person's alpha-wave frequency.


Questions 17 to 18 pertain to the following set-up.  The number of years a person lives after being diagnosed with a certain disease is normally distributed with a mean of 4 years and a standard deviation of 1 year.

17. What is the probability that a randomly selected individual with the disease will live more than 5.5 years?

  P(X > 5.5) = P(Z > (5.5- 4)/1) = P(Z > 1.50) = 1 - 0.9332 = 0.0668
 
 

18. What is the probability that a randomly selected individual will live between 3.5 and 5.2 years?

   P(3.5 < X < 5.2) = P[(3.5 - 4)/1 < Z < (5.2 - 4)/1]  = P( - 0.50 < Z < 1.20)
Then, P(Z < -0.50) =  0.3085 and P(Z < 1.2) = 0.8849
So P( - 0.5 < Z < 1.2) = 0.8849 - 0.3085 = 0.5764



19. The weight of an airline passenger's suitcase is normally distributed with a mean of 45 pounds and a standard deviation of 2 pounds.  If 15% of the suitcases are overweight, what is the maximum weight allowed by the airline?
   P(Z > 1.04) = 0.15, so X = 45 + 1.04(2) = 47.08