Question

Any medical item used in the care of hospital patients is called a factor. For example, factors can be intravenous tubing, intravenous fluid, needles, shave kits, bedpans, diapers, dressings, medications, and even code carts. The coronary care unit at Bayonet Point Hospital (St. Petersburg, Florida) investigated the relationship between the number of factors per patient, x, and the patient’s length of stay (in days), y. The data for a random sample of 50 coronary care patients are saved in FACTORS file.

a. Construct a scatter plot of the data.

b. Find the least squares line for the data and plot it on your scatter plot.

c. Define ?̂ 1 in the context of the problem.

d. Test the hypothesis that the number of factors per patient (x) contributes no information for the prediction of the patient’s length of stay (y) when a linear model is used (use ? = 0.05). Draw the appropriate conclusions.

e. Find a 95% confidence interval for ?1. Interpret your results.

f. Find the coefficient of determination for the linear model you constructed in part b. Interpret your result.

g. Find a 95% prediction interval for the length of stay of a coronary care patient who is administered a total of ? = 231 factors. h. Find a 95% confidence interval for the mean length of stay of a coronary care patient who is administered a total of ? = 230 factors.

Below is the table used for FACTORS.TXT doc

LOS FACTORS

9 231

7 323

8 113

5 208

4 162

4 117

6 159

9 169

6 55

3 77

4 103

6 147

6 230

3 78

9 525

7 121

5 248

8 233

4 260

7 224

12 472

8 220

6 383

9 301

7 262

11 354

7 142

9 286

10 341

5 201

11 158

6 243

6 156

7 184

4 115

6 202

5 206

6 360

3 84

9 331

7 302

2 60

2 110

5 131

4 364

7 180

6 134

15 401

4 155

8 338

Answer 1

(a)

Following is the scatter plot of data:

(b)

Following is the output of regression analysis generated by excel:

(c)

It shows for each unit increase in factors, LOS is increased by 0.0148 units.

(d)

Hypotheses are:

The p-value of t test is: 0.0000

Since p-value is less than 0.05 so we reject the null hypothesis.

(e)

The 95% confidence interval for slope is

(0.0092, 0.0203)

It shows that we are 95% confident that true slope lies in the above interval.

f.

The coefficient of determination is: 0.374

It is that 37.4% of variation in dependent variable LOS is accounted for the model.

g. Find a 95% prediction interval for the length of stay of a coronary care patient who is administered a total of ? = 231 factors.

The required prediction interval is (2.448, 10.981).

h. Find a 95% confidence interval for the mean length of stay of a coronary care patient who is administered a total of ? = 230 factors.

The required confidence interval is (6.099, 7.300).