Be sure to use Excel Megastat to solve the following problem Smoking and short-t

Question

Be sure to use Excel Megastat to solve the following problem

Smoking and short-term illness case:

Besides the known long-term effects of smoking, some researchers believe that there may be a relationship between the average number of cigarettes smoked and the number of work days missed due to short illnesses. To help understand this, a sample of smokers was drawn. Each person was asked to report the average number of cigarettes smoked per day and the number of days absent from work due to colds last year (sick days).

Answer the following questions:

5. Using a significance level of = .05, is there sufficient evidence to conclude that the number of cigarettes smoked is useful in predicting the number of sick days from work?

6. Estimate with 95% confidence the average number of sick days for all individuals who smoke 30 cigarettes per day. Interpret the practical meaning of this interval estimate in the context of the problem.

7. Predict with 95% confidence the number of sick days of a single individual who smokes 30 cigarettes per day. Interpret the practical meaning of this interval estimate in the context of the problem.

8. Estimate the true population slope for this least squares regression line with 95% confidence. Interpret the practical meaning of this interval estimate in the context of the problem.

9. What are the variance and standard deviation of the random errors for this regression analysis?

10. Calculate and report the residual of the 10th observation in the data set.

cigarettes Days 44 18 41 18 17 18 35 15 33 14 44 15 39 18 35 10 43 16 43 14 41 16 47 19 41 13 30 12 15 18 39 18 36 11 41 13 37 10 43 21 32 13 17 13 34 15 32 11 19 10 44 26 30 13 42 25 34 5 36 8 42 15 47 15 55 19 27 15 28 15 45 22 29 10 37 13 52 13 34 16 48 23 40 15 29 16 34 10 35 19 57 23 48 11 41 10 40 13 36 13 26 7 29 15 52 16 28 16 35 11 42 15 54 19 56 19 37 9 41 17 32 16 63 20 31 15 36 15 31 12 25 15 40 25 55 16 37 21 28 15 56 21 24 16 61 19 40 16 50 14 52 10 45 14 42 19 47 23 22 0 26 12 38 13 46 13 27 9 45 21 43 15 17 11 45 16 40 15 41 17 37 15 30 10 37 14 31 9 49 10 44 15 16 2 25 4 30 5 20 6 36 11 39 14 49 13 37 19 20 11 48 12 28 5 51 16 36 21 42 6 47 9 17 11 22 16 42 21 32 11 31 7 49 16 30 11 31 11 38 18 44 14 41 17 40 20 32 17 48 15 29 17 40 18 45 18 40 15 35 18 37 17 41 16 25 16 35 19 44 15 58 24 27 13 42 22 26 9 48 15 30 11 36 13 26 8 27 6 39 14 17 17 44 21 41 19 26 14 30 14 36 13 31 15 55 17 45 17 27 17 25 9 42 15 18 6 49 15 60 19 40 18 29 15 32 12 35 16 45 15 42 11 52 13 52 16 8 16 36 15 33 12 38 17 38 11 38 14 36 16 44 11 36 14 35 12 30 13 39 21 37 7 26 22 51 18 42 10 44 19 41 15 32 16 56 19 47 22 39 11 18 10 33 12 49 14 16 7 38 9 27 8 26 15 27 8 25 15 14 11 40 9 38 9 38 13 62 18 47 14 54 9 37 6 40 17 56 17 38 16 53 22 52 19 21 11 34 16 23 5 44 13 36 17 45 14 23 10 26 12 54 20 37 19 56 11 38 14 59 18 39 16 38 21 42 20 44 17 32 17 45 13

Explanation / Answer

Answer:

Answer the following questions:

5. Using a significance level of = .05, is there sufficient evidence to conclude that the number of cigarettes smoked is useful in predicting the number of sick days from work?

Calculated F=56.13, P=0.000 which is < 0.05 level. Ho is rejected.

There is sufficient evidence to conclude that the number of cigarettes smoked is useful in predicting the number of sick days from work.

6. Estimate with 95% confidence the average number of sick days for all individuals who smoke 30 cigarettes per day. Interpret the practical meaning of this interval estimate in the context of the problem.

95% confidence interval =(12.336, 13.622).

We are 95% confidence that the predicted average number of sick days for all individuals who smoke 30 cigarettes per day fall in the interval (12.336, 13.622).

7. Predict with 95% confidence the number of sick days of a single individual who smokes 30 cigarettes per day. Interpret the practical meaning of this interval estimate in the context of the problem.

95% prediction interval=(5.078, 20.880).

We are 95% confidence that the number of sick days of a single individual who smokes 30 cigarettes per day falls in the interval (5.078, 20.880).

8. Estimate the true population slope for this least squares regression line with 95% confidence. Interpret the practical meaning of this interval estimate in the context of the problem.

95% confidence for slope= (0.1398, 0.2396).

We are 95% confident that true population slope for this least squares regression line falls in the interval (0.1398, 0.2396).

9. What are the variance and standard deviation of the random errors for this regression analysis?

Variance=15.9735

Standard deviation=3.997

10. Calculate and report the residual of the 10th observation in the data set.

Predicted value for 10th observation :

43

14

Predicted days =7.2865+0.1897*43 =15.4436

Residual =14-15.4436 = -1.4436

Regression Analysis

r²

0.197

n

231

r

0.444

k

1

Std. Error

3.997

Dep. Var.

Days

ANOVA table

Source

SS

df

MS

F

p-value

Regression

896.6419

1  

896.6419

56.13

1.46E-12

Residual

3,657.9295

229  

15.9735

Total

4,554.5714

230  

Regression output

confidence interval

variables

coefficients

std. error

   t (df=229)

p-value

95% lower

95% upper

Intercept

7.2865

0.9889

7.369

3.11E-12

5.3381

9.2350

cigarettes

0.1897

0.0253

7.492

1.46E-12

0.1398

0.2396

Predicted values for: Days

95% Confidence Interval

95% Prediction Interval

cigarettes

Predicted

lower

upper

lower

upper

Leverage

30

12.979

12.336

13.622

5.078

20.880

0.007

43

14

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.