Question

The problem and the data: A pharmaceutical company has developed a new drug designed to reduce a smoker's reliance on tobacco. Since certain dosages of the drug may reduce one's pulse rate to dangerously low levels, the product-testing division of the pharmaceutical company wants to model the relationship between decrease in pulse rate, y (beats per minute), and dosage, (cubic centermeters). Different dosages of the drug were administered to eight randomly selected patients, and 30 minutes later the decrease in each patient's pulse rate was recorded. The results are given in the following table. Patient Dosage (2) Decrease in Pulse Rate (y) 2.0 12 4.0 20 1.5 6 1.0 3 3.0 16 3.5 20 2.5 13 3.0 18 2 3 4 5 6 7 8 1. Plot the data set. Does the plot (visually) suggest a linear relationship between dosage () and decrease in pulse rate (y)? (10 pts.)

all part of one big question. please answer parts 5-10 and show work

Answer 1

1. To construct the scatter plot consider the following steps in MS Excel: 1. Enter the given data in the MS Excel Spreadsheet. 2. Select the entered data and go to CHARTS >> SCATTER. The obtained scatter plot is as follows: Scatter Plot 25 20 .. 15 Decrease in Pulse Rate 10 5 . 0 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 Dosage From the graph it can be seen that as Dosage increases, the decrease in pulse rate is also increasing. Thus, the scatter plot suggests that there is a linear relationship between Dosage and Decrease in Pulse Rate. 2.

Consider the following table: Patient x? y2 Dosage (x) 2 4 1.5 Decrease in Pulse Rate (y) 12 20 6 4 2 16 2.25 3 3 1 4 5 6 7 8 Total xy 24 80 9 3 48 70 32.5 54 320.5 3 3.5 2.5 3 20.5 16 20 13 18 108 144 400 36 9 256 400 169 324 1738 9 12.25 6.25 9 59.75 From the table it can be assumed that n=8, Xx=20.5, y=108, Zr? =59.5, Xy? =1738, X xy=320.5. Now, (2x)-(20.5) = 420.25 (2 y)=(108) 11664 X=- n [20.5] - 2.5625 8

v=- 11 Σ» [108] 1 -- 8 13.5 2 (Σ Ν SS, x2. XX n (420.25) =59.5- 8 16.9688 SS=Σ». (11664) n =1738 8 280 (Σ) Σν) SS, =Συν- (20.5) (108) =320.5-. 8 n =143.75

SSE n-2 SS vy SS- SS2 YY n-2 (6.9688)(280)–(43.75) 8-2 = 2.49

3. The simple linear regression model is given as follows: y = bo + byx Here be is intercept and b, is slope. n2 n2 7 The formula for computing the slope is given as follows: x;y; - 2* | Xy & X; Xx n>xy; i=1 i=1 n2 i=1 8(320.5)-(20.5)(108) 8(59.75)-(20.5) 6.0606 Now, the intercept can be computed as follows: 60 = J – b, x =13.5-(6.0606)(2.5625) -2.0303 Thus, the simple linear regression model is ĝ=-2.0303+ 6.0606.x.

To plot the regression line on the scatter plot in Excel, choose the chart menu and click on Trendline. Scatter Plot 25 20 15 Decrease in Pulse Rate 10 5 0 0 0.5 1 1.5 3 3.5 4 4.5 2 2.5 Dosage 4. For a 1 unit increase in dosage, on an average there will be 2.0303 units decrease in the pulse rate. 5. The decrease in pulse rate when the dosage is 4 is given as follows: =-2.0303 +6.0606 x =-2.0303 +6.0606(4) 22.2121

The decrease in pulse rate when the dosage is 5 is given as follows: y=-2.0303+6.0606x =-2.0303 +6.0606(5) 28.2727 = 6. The correlation coefficient can be computed as follows: n(Xxy)-(2x)(y) 1 X 8(320.5)-(20.5)(108) V8(59.5)-(20.5) 18(1738)–(108) 350 353.3836 0.9904 7. The coefficient of determination is the square of the coefficient of correlation. Therefore, Coefficient of Determination = (Correlation Coefficient)? =(0.9904) 2 = 0.9809 8.

To test the linear relationship between x and y, the following hypotheses will be tested: H:b=0, H :b, +0 The level of significance is 0.05.

t= The test statistic is given as follows: b-b2 se/VSE 6.0606 2.49/ 16.9688 = 6.4253 The degrees of freedom are given as follows: df =n-2 =8-2 = 6 The critical value of t distribution at 0.05 level of significance with 6 degrees of freedom is 2.4469 (from t distribution tables). The decision rule is to reject the null hypothesis if the critical value of the t distribution is less than the value of test statistic. For the given problem, the critical value of the t distribution is less than the value of test statistic, therefore, reject the null hypothesis and conclude that there is linear relationship between x and y. 9. The formula for a two-sided 95% prediction interval for an individual decrease in pulse rate at dosage level xo =4 is given as follows: in+1 (3-7) CI a/2. (n-2) Sxx = +b, xo Et + n

The critical value of t distribution at 0.05 level of significance with 6 degrees of freedom is 2.4469 (from t distribution tables). The confidence interval can be computed as follows: 8+1 4-2.5625 CI=[-2.0303+ 6.0606(4)]+2.44 + 8 6.6988 = 22.2121+2.8321 (19.38, 25.0442 10. Let X be the decrease in the pulse rate at a dosage level xo = 4. The mean and standard deviation are given as follows: u=-2.0303+6.0606(4) = 22.2121 O=Se = 2.49 o The probability that the individual will have a dangerous decrease in pulse rate is given as follows: X-u 25 - 22.2121 P[X < 25)=P 2.49 =P[Z <1.12] = NORM.S.DIST (1.12, TRUE) (use the function in MS Excel) =0.8686