Practice Exam Problems. Certain dosages of a new drug developed to reduce a smok
ID: 3326259 • Letter: P
Question
Practice Exam Problems. Certain dosages of a new drug developed to reduce a smoker's reliance on nicotine may reduce one's pulse rate to low levels. To investigate the drug's effect on pulse rate, different dosages of the drug were administered to 8 patients, and 30 minutes later the decrease in each patient's pulse rate was recorded. 1. Patient 2 3456 7 15 2.0 2.0 2.5 3.0 3.0 4.0 4.0 8 14 15 16 18 20 2123 Decrease in Pulse The sample summaries below should be useful: | -22.0 I -135.0 2x2-66 s2y2-2435.0 2w-400.01 x 22.0 y= 135.0 | r-66.5 Describe a regression model that may explain the relationship between the drug's dosage level and the decrease in a patient's pulse rate. Explain the meaning of each symbol used in your model, in the context of the problem. What factors other than the drug's dosage level may affect the decrease in a patient's pulse? a. Calculate the sample correlation coefficient between the drug's dosage level and the decrease in a patient's pulse rate b. obtain a 95% confidence interval for the mean decrease in a patient's pulse rate given the drug's dosage level is 3.5, what is the meaning of 95% confidence? c, d. Suppose that each of the measurements in y (decrease in pulse) were actually 1 less than those recorded in the table above. For example, a measurement of 8 should actually be 7. Show how the answers of part (b) and part (c) above are affected. 2 A research lah has develoned a new nrocess in making a snecialty rone, The lab would like to show that a roneExplanation / Answer
Question a.
Here the linear regression model would be the best model to describe the model.
y^ = a + bx
a is the intercept and b is the slope here
x = Dosage
y = Decrease in Pulse
Regression Analysis: y versus x
The regression equation is
y = 3.64 + 4.54 x
Predictor Coef SE Coef T P
Constant 3.635 2.502 1.45 0.196
x 4.5417 0.8679 5.23 0.002
S = 2.12582 R-Sq = 82.0% R-Sq(adj) = 79.0%
y^ = 3.635 + 4.5417x
The value of the intercept is 3.635 this is the decrease in the pulse rate when the amount of the dosage is 0. The value of the slope is 4.5417. We can say that if there is one unit increment in the dosage there the pulse rate is going to be decreased by 4.5417 units.
Question b.
The value of the correlation is = sqrt (0.82) = 0.906
The value of the sample correlation is 0.906
Question c.
Analysis of Variance
Source DF SS MS F P
Regression 1 123.76 123.76 27.39 0.002
Residual Error 6 27.11 4.52
Total 7 150.87
Predicted Values for New Observations
New
Obs Fit SE Fit 95% CI 95% PI
1 19.531 0.994 (17.098, 21.964) (13.789, 25.274)
The 95% confidence interval is 17.098 and 21.964. This value tells us that when the dosage level is 3.5 then the population mean of the decrease pulse rate will lie in between 17.098 and 21.964.
Question d.
Regression Analysis: y versus x
The regression equation is
y = 2.64 + 4.54 x
Predictor Coef SE Coef T P
Constant 2.635 2.502 1.05 0.333
x 4.5417 0.8679 5.23 0.002
S = 2.12582 R-Sq = 82.0% R-Sq(adj) = 79.0%
Part b.
The value of the sample correlation will remain the same as 0.906 after changing the value of y’s.
Part c.
Analysis of Variance
Source DF SS MS F P
Regression 1 123.76 123.76 27.39 0.002
Residual Error 6 27.11 4.52
Total 7 150.87
Predicted Values for New Observations
New
Obs Fit SE Fit 95% CI 95% PI
1 18.531 0.994 (16.098, 20.964) (12.789, 24.274)
Values of Predictors for New Observations
New
Obs x
1 3.50
We got the confidence interval as (16.098 and 20.964). The width of the confidence interval remain the same just the lower limit and upper limit values got changed.
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