AMS572



AMS572.01 Final Exam Fall, 2007

Name: __________________________ ID: ____________ Signature: _____________

Instruction: 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. The exam goes from 5:00-7:30pm. Good luck!

1 (for all students in class). To determine whether glaucoma affects the corneal thickness, measurements were made in 8 people affected by glaucoma in one eye but not in the other. The corneal thickness (in microns) were as follows:

|Person |1 |

|Old Method |13 15 13 12 14 |

(a). Please construct a 95 % percent confidence interval for the mean difference between the compressive strengths by two methods.

(b). At the significance level .05, can you conclude that the new method increases the compressive strength?

(c). What assumptions do you need for the inference in parts (a) and (b)?

(d). Please write a SAS program to check for assumptions in (c) and perform the test in (b).

Solution: Inference on two population means. Two small and independent samples.

New method: [pic]

Old method: [pic]

Under the normality assumption, we test if the two population variances are equal [pic] vs [pic].

Test statistic is

[pic], [pic] and [pic].

Since F0 is between 0.16 and 6.39, we cannot reject H0 . Therefore it is reasonable to assume that[pic].

a) 95% C. I. for difference is

[pic]

where [pic]

Therefore 95% C.I. is [-0.46, 2.86].

b) Using t-test with hypotheses [pic] v.s. [pic],

[pic]

[pic] Thus, we cannot reject H0. There isn't enough evidence to conclude that the new method increases the strength.

c) (1) Both populations are normally distributed

(2) [pic]

d) The SAS program is as follows.

Data blocks;

Input method $ strength @@;

Datalines;

new 15 new 14 new 13 new 15 new 16

old 13 old 15 old 13 old 12 old 14

;

Run;

Proc univariate data = blocks normal;

Class method;

Var strength;

Run;

Proc ttest data = blocks;

Class method;

Var strength;

Run;

3b (for AMS PhD students ONLY). Please derive the likelihood ratio test for H0: μ = μ0 versus Ha: μ ≠ μ0, when the population is normal and population variance σ2 is known.

Solution:

For a 2-sided test of H0: μ = μ0 versus Ha: μ ≠ μ0, when the population is normal and population variance σ2 is known, we have:

[pic] and [pic]

The likelihoods are:

[pic]

There is no free parameter in [pic], thus [pic].

[pic]

There is only one free parameter μ in [pic]. Now we shall find the value of μ that maximizes the log likelihood [pic].

By solving [pic], we have [pic]

It is easy to verify that [pic] indeed maximizes the loglikelihood, and thus the likelihood function.

Therefore the likelihood ratio is:

[pic]

Therefore, the likelihood ratio test that will reject H0 when [pic]is equivalent to the z-test that will reject H0 when [pic], where c can be determined by the significance level α as [pic].

4 Fifteen subjects were randomly assigned to three treatment groups X, Y and Z (with 5 subjects per treatment). Each of the three groups has received a different method of speed-reading instruction. A reading test is given, and the number of words per minute is recorded for each subject. The following data are collected:

|X |Y |Z |

|700 |480 |500 |

|850 |460 |550 |

|820 |500 |480 |

|640 |570 |600 |

|920 |580 |610 |

a) Are these treatments equally effective? Test at α = 0.05.

b) If these treatments are not equally good, please use Tukey’s procedure with α = 0.05 to make pairwise comparisons.

c) Please write a SAS program to answer the above two questions.

Solution : This is one-way ANOVA with 3 samples and 5 observations per sample.

(a) The ANOVA Procedure

Dependent Variable: WORDS

Sum of

Source DF Squares Mean Square F Value Pr > F

Model 2 215613.3333 107806.6667 16.78 0.0003

Error 12 77080.0000 6423.3333

Corrected Total 14 292693.3333

The p-value of the ANOVA F-test is 0.0003, less than the significance level α = 0.05. Thus we conclude that the three reading methods are not equally good.

(b) The ANOVA Procedure

Tukey's Studentized Range (HSD) Test for WORDS

NOTE: This test controls the Type I experimentwise error rate, but it generally has a higher Type

II error rate than REGWQ.

Alpha 0.05

Error Degrees of Freedom 12

Error Mean Square 6423.333

Critical Value of Studentized Range 3.77278

Minimum Significant Difference 135.22

Means with the same letter are not significantly different.

Tukey Grouping Mean N GROUP

A 786.00 5 X

B 548.00 5 Z

B

B 518.00 5 Y

Thus we found that method X is superior to methods Y and Z, while methods Y and Z are not significantly different at the family-wise error rate of 0.05 using Tukey’s Studentized Range test.

(c) The SAS code is as follows:

DATA READING;

INPUT GROUP $ WORDS @@;

DATALINES;

X 700 X 850 X 820 X 640 X 920 Y 480 Y 460 Y 500

Y 570 Y 580 Z 500 Z 550 Z 480 Z 600 Z 610

;

RUN ;

PROC ANOVA DATA=READING ;

TITLE ‘Analysis of Reading Data’ ;

CLASS GROUP;

MODEL WORDS = GROUP;

MEANS GROUP / TUKEY;

RUN;

5. 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.

Solution: Let [pic]be the proportion of peacocks and [pic]be the proportion of turkeys.

[pic] and [pic].

95% C.I on [pic]is

[pic].

(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.)

Solution: [pic]

i) [pic]

[pic]

ii) if [pic] is unknown,

[pic]

[pic][pic]. [pic]

6. (*Extra credit problem for everyone in class) A study in which infants were fed baked beans showed that such infants tended to gain more weight than a control group. The z-value calculated was 1.96, which is equivalent to a p-value of 0.05 (2-sided). In a second independent study, the z-value calculated was 1.64 corresponding to a p-value of 0.10 (2-sided). What conclusion can be reached if we combine the results of these 2 studies in the following scenarios? (For each scenario, please derive the test statistic utilizing all the information available and report the p-value for the newly constructed test. Please show the entire derivation and calculation for full credit.)

a) We do not know the sample size for each study.

b) We do not know the exact sample size for each study; however, we do know that n1=k*n2, where n1 is the sample size for study 1, n2 is the sample size for study 2, and k is a known constant (for example k = 2).

Solution:

First, we examine the case when the two population variances are equal and known.

[pic]

I leave the other possible scenarios for you to ponder over and explore for this extra credit problem.

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