BST 621 Homework 3: Confidence Intervals and Hypothesis Testing | Assignments Data Analysis & Statistical Methods

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]) =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 Z-

test (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 (Y = 76; Y = 12) and the shape the DSP distribution is

Normal. Now suppose N = 6 patients were given Lacidipine, which resulted in a mean of

= 70.

2.a. What is the 95% symmetric CI for this DSP mean of

= 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

= 70 given that a population mean was Y = 76 (Daniel, Section 6.2)

≠ 70 | [π = 0.25]) =Y = 76; Y = 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  = 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;

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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.. 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]) =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 Z- test (1.b) differ? (2 points)
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 ( Y = 76;  Y = 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  Y = 76 (Daniel, Section 6.2) P( Y^ ≠ 70 | [π = 0.25]) = Y = 76;  Y = 12]) = (3 points) 2.c. How does the 95% CI in 2.a relate to the probability in 2.b? (2 points)
Based on Frattola et al. (2000, Hypertension , 36, 622-628), suppose researchers randomly assigned n C = 35 patients to a control condition given a placebo and n T = 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  = 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;

3.a. Enter the following Results. (56 points total) 95% CI (Mean Difference) Control Treatment Mean Diff Lower Bound Upper Bound DBP Mean j SD j t F p-value SE(MDiff) n j df 95% CI (Mean Difference) Control Treatment Mean Diff Lower Bound Upper Bound SBP Mean j SD j t F p-value SE(MDiff) n j df 95% CI Control Treatment Mean Diff Lower Bound Upper Bound HDL Mean j SD j t F p-value SE(MDiff) n j df 95% CI Control Treatment Mean Diff Lower Bound Upper Bound LDL Mean j SD j t F p-value SE(MDiff) n j 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) DBP t CV = HDL t CV = 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)

How would missing data affect the interpretation of these results? (3 points)
Write a brief summary interpretation of these results. (6 points)
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)

K C r P[π = 0.25]) =r]

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) =

Suppose a researcher was interested in a smoking cessation treatment. Suppose the researcher randomly assigned n C = 80 patients to a control condition given a placebo (group = 0) and n T = 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  = 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.) (2 points) 9.b. Enter the following Results. (21 points total) 95% CI Control Treatment p Diff Lower Bound Upper Bound QUIT p j Z ^2 p-value n j SE(pDiff) 95% CI Control Treatment p Diff Lower Bound Upper Bound ITT p j Z ^2 p-value n j SE(pDiff) Note : For the pooled variance Z -statistic it can be hand-computed as: 1 1 1 2 (1 ) (1 ) p p Z p p p p n n      ; where =^ 1 1 2 2 1 2 p n p n p n n    (Daniel, Section 7.6) 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 ^2 -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)

BST 621 Homework 3: Confidence Intervals and Hypothesis Testing, Assignments of Data Analysis & Statistical Methods

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K C r P[π = 0.25]) =r]