BST 621 (Beasley) Homework 3 (Mid-Term) (200 points)
1. Suppose a researcher was interested in a smoking cessation treatment and administered a treatment of
nicotine patches to N = 20 randomly sampled patients. After 4 weeks, the results showed that 8 of these
people had quit smoking.
Note: You can compute your answers by hand or you can create a data set and use SAS Code of the form:
proc freq;table y / alpha=0.aa binomial(p=0.HH Level=2);run;
Where alpha = specifies (confidence = 1 -0.aa); for example alpha=0.aa is alpha= 0.10 for 90% confidence;
p = specifies the hypothesized value; for example is the hypothesized proportion is 0.62 then p=0.62.
1.a. What is the 95% CI for this sample proportion? (see Daniel, Section 6.5)
(3 points)
1.b. Assuming the population proportion of people who quit smoking after trying to quit for 4 weeks is π
= 0.25. What is the probability that a Proportional Difference this extreme or larger occurred by chance
assuming No Treatment effects? (see Daniel, Section 7.5)
P(p ≥ [π = 0.25]) =
(3 points)
1.c. How do the Standard Errors used for the one-sample Confidence Interval (1.a) and the one-sample Ztest (1.b) differ?
(2 points)
2. Frattola et al. (2000, Hypertension, 36, 622-628) found the Standard Deviation of 24-hour Diastolic
Blood Pressure (DSP) among Diabetics to be 12. The average 24-hour DSP was 76. Assume these values
are representative of the population parameters (µY = 76; σY = 12) and the shape the DSP distribution is
Normal. Now suppose N = 6 patients were given Lacidipine, which resulted in a mean of Y = 70.
2.a. What is the 95% symmetric CI for this DSP mean of Y = 70. (Daniel, Section 6.2)
(3 points)
2.b. Given the Frattola values as parameters, what is the probability of obtaining a mean DSP as extreme
as Y = 70 given that a population mean was µY = 76 (Daniel, Section 6.2)
P( Y ≠ 70 | [µY = 76; σY = 12]) =
(3 points)
2.c. How does the 95% CI in 2.a relate to the probability in 2.b?
(2 points)
3. Based on Frattola et al. (2000, Hypertension, 36, 622-628), suppose researchers randomly assigned nC
= 35 patients to a control condition given a placebo and nT = 35 patients to a treatment condition given a
new investigational drug. In the Microsoft Excel file (BST621-Assign3-BP.xls), the Control Group is
labeled zero (group=0) and the Treatment Group administered the new drug is labeled one (group=1). The
assumption is that population means of both variables for these groups are expected to be equal if there is
No Treatment effect of the new drug. The significance level was set at α = 0.05 for a two-tailed test.
(Daniel, Sections 6.4 and 7.3)
SPSS: Use Analyze-Compare Means-Independent Samples T-Test and
Use Analyze-Compare Means-One Way ANOVA
JMP: Change the group variable to be Nominal, then Use Analyze-Fit Y by X
Under the Oneway Analysis Banner select Means/Anova/Pooled t Means and Std Dev
SAS: Use PROC TTEST; CLASS group; VAR y; RUN; and
Use PROC GLM; CLASS group; MODEL y = group / solution; MEANS group / t hovtest;
LSMEANS group / adjust= t pdiff tdiff; RUN;
1
BST 621 (Beasley) Homework 3 (Mid-Term) (200 points)
3.a. Enter the following Results.
Control
DBP
Treatment
Meanj
SDj
Mean Diff
t
Treatment
Meanj
SDj
t
p-value
df
Control
Treatment
95% CI
Lower Bound
Upper Bound
Mean Diff
t
F
p-value
SE(MDiff)
nj
df
Control
nj
F
SE(MDiff)
Meanj
SDj
Meanj
SDj
95% CI (Mean Difference)
Lower Bound
Upper Bound
Mean Diff
nj
LDL
p-value
df
Control
HDL
F
SE(MDiff)
nj
SBP
(56 points total)
95% CI (Mean Difference)
Lower Bound
Upper Bound
Treatment
95% CI
Lower Bound
Upper Bound
Mean Diff
t
F
p-value
SE(MDiff)
df
3.b. Based on the 95% symmetric CIs, the output, or table of Percentiles of the t-distribution (Table E),
what was the critical value from the t-distribution?
(2 points)
HDL tCV =
DBP tCV =
3.c. For the DBP, SBP, HDL, and LDL interpret each 95% CI.
(8 points)
3.d. For all analyses, how do t-statistic and F-statistic relate?
(2 points)
3.e. For all analyses, how do the 95% CI’s relate to the p-value?
(2 points)
3.f. In symbolic notation, what was the null hypothesis for the t-tests in the previous analyses?
(The null hypothesis was basically the same for all variables.)
(2 points)
2
BST 621 (Beasley) Homework 3 (Mid-Term) (200 points)
4. Based on these data, conduct a Power analysis for the DBP and SBP results. (Daniel, Section 7.9)
JMP: Use Analyze-Fit Y by X (Under the Oneway Analysis Banner select the Power . . . option
In the To: row under the Number column insert a “large number” in the By: row of
In the By: row under the Number column insert a 1.
Select the Solve for Power and Solve for Least Significant Number boxes and click Done
SAS: proc power; onewayanova groupmeans = X.XXX | Y.YYY stddev = S.SSS
alpha = 0.05 ntotal = (Value or .) power = 0.8 (Value or . ) ; run;
4.a. What was the “Observed Power” for the DBP results at a two-tailed α = 0.05.?
(3 points)
4.b. Holding these data (Means and SDs) constant, what would a future Total Sample Size (N) need to be
for the DBP results to be statistically significant at a two-tailed α = 0.05.?
(3 points)
4.c. Holding these data (Means and SDs) constant, what would a future Total Sample Size (N) need to be
for the DBP analysis to have 80% Power (1 – β = 0.80) at a two-tailed α = 0.05.?
(3 points)
4.d. Given YOUR DECISION on whether or not to Reject the Null Hypothesis for DBP, what is the
Probability of a Type I Error?
(2 points)
4.e. Given YOUR DECISION on whether or not to Reject the Null Hypothesis for DBP, what is the
Probability of a Type II Error?
(2 points)
5.a. What was the “Observed Power” for the SBP results at a two-tailed α = 0.05.?
(3 points)
5.b. Holding these data (Means and SDs) constant, what would a future Total Sample Size (N) need to be
for the SBP results to be statistically significant at α = 0.05.?
(3 points)
5.c. Holding these data (Means and SDs) constant, what would a future Total Sample Size (N) need to be
for the SBP analysis to have 80% Power (1 – β = 0.80) at α = 0.05.?
(3 points)
5.d. Given YOUR DECISION on whether or not to Reject the Null Hypothesis for SBP, what is the
Probability of a Type I Error?
(2 points)
5.e. Given YOUR DECISION on whether or not to Reject the Null Hypothesis for SBP, what is the
Probability of a Type II Error?
(2 points)
3
BST 621 (Beasley) Homework 3 (Mid-Term) (200 points)
6. How would missing data affect the interpretation of these results?
(3 points)
7. Write a brief summary interpretation of these results.
(6 points)
8. Suppose a test statistic has a Type I error rate of α = 0.05. That is, 5% of the time the test will reject the
null hypothesis when the null hypothesis is actually true. Now suppose the K=4 tests conducted in the
previous analyses were independent.
8.a.. List all possible combinations for r = 0 to 4 Type I errors and using the binomial distribution
function calculate the probability of each occurrence.
(5 points)
P[r]
KCr
r=0
r=1
r=2
r=3
r=4
8.b. What is the probability of at least one Type I error, P(r ≥ 1) = _____________.
(3 points)
8.c. Algebraically reduce the formula for the binomial distribution function and Write the general formula
for the probability of at least one failure,
(4 points)
P(r ≥ 1) =
4
BST 621 (Beasley) Homework 3 (Mid-Term) (200 points)
9. Suppose a researcher was interested in a smoking cessation treatment. Suppose the researcher randomly
assigned nC = 80 patients to a control condition given a placebo (group = 0) and nT = 80 patients to a
treatment condition given nicotine patches (group =1) . The assumption is that population proportions of
quitting for these groups are expected to be equal if there is No Treatment effect for the nicotine patches.
In the Microsoft Excel file (BST621-Assign3-smoke.xls), participants who quit smoking were given a
value of quit = 1. Those who did not quit were given a value of quit = 0. Several (33) participants did not
return or could not be contacted to report their smoking status. An Intent-to-Treat (ITT) analysis was
conducted, where these missing cases were assumed to be still smoking. The significance level was set at
α = 0.05 for two-tailed tests.
SPSS: Use Analyze-Descriptive Statistics-Crosstabs and under the Statistics options select the
Chi-Square and Phi and Cramer’s V boxes
JMP: Change the group, quit, and itt variables to Nominal the Use Analyze-Fit Y by X
SAS: proc freq ; tables group*quit / chisq measures riskdiff; run;
9.a. In symbolic notation, what was the null hypothesis for the previous analysis?
(The null hypothesis was the same for both variables.)
9.b. Enter the following Results.
Control
QUIT
(21 points total)
Treatment
p Diff
95% CI
Lower Bound
Upper Bound
pj
Z
nj
χ2
p-value
SE(pDiff)
Control
ITT
(2 points)
Treatment
p Diff
95% CI
Lower Bound
Upper Bound
pj
Z
nj
χ2
p-value
SE(pDiff)
Note: For the pooled variance Z-statistic it can be hand-computed as:
Z=
p1 − p1
pn +p n
; where = p = 1 1 2 2 (Daniel, Section 7.6)
n1 + n2
p (1 − p ) p (1 − p )
+
n1
n2
9.c. For both QUIT and ITT, interpret each 95% CI.
(4 points)
9.d. For both QUIT and ITT, how do z-statistic and χ2-statistic relate?
(3 points)
9.e. For both the QUIT and ITT, how do the 95% CI’s relate to the p-value?
(3 points)
5
BST 621 (Beasley) Homework 3 (Mid-Term) (200 points)
10. Based on these data, conduct a Power analysis for the QUIT results.
SAS: proc power; twosamplefreq test=pchi groupproportions = (.15 .25)
alpha = 0.05 ntotal = (Value or .) power = 0.8 (Value or . ) ; run;
10.a. What was the “Observed Power” for the QUIT results at a two-tailed α = 0.05.?
(3 points)
10.b. Holding these data (proportions) constant, what would a future Total Sample Size (N) need to be for
the QUIT results to be statistically significant at a two-tailed α = 0.05.?
(3 points)
10.c. Holding these data (proportions) constant, what would a future Total Sample Size (N) need to be for
theQUIT analysis to have 80% Power (1 – β = 0.80) at a two-tailed α = 0.05.?
(3 points)
10.d. Holding these data (proportions) constant, what would a future Total Sample Size (N) need to be for
the QUIT results to be statistically significant at a two-tailed α = 0.01.?
(3 points)
10.e. Holding these data (proportions) constant, what would a future Total Sample Size (N) need to be for
the QUIT analysis to have 80% Power (1 – β = 0.80) at a two-tailed α = 0.01.?
(3 points)
10.f. Given YOUR DECISION on whether or not to Reject the Null Hypothesis for QUIT, what is the
Probability of a Type I Error?
(2 points)
10.g. Given YOUR DECISION on whether or not to Reject the Null Hypothesis for QUIT, what is the
Probability of a Type II Error?
(2 points)
11. How would missing data affect the interpretation of these results?
(3 points)
12. Write a brief summary and interpretation of these results.
(6 points)
6