Please help me!!! I missed class the other day and do not udnerstand this for th
ID: 3259114 • Letter: P
Question
Please help me!!! I missed class the other day and do not udnerstand this for the life of me..
Please use Excel Megastat to solve the problem.
Smoking and short-term illness case: (Data at the bottom of the Q's)
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).
Use the data file from Blackboard à Lab assignment #4 folder à Lab assignment #4 data. Run the analysis in excel Megastat and answer the following questions. Note that most answers do not need manual calculations and may be reported directly from the Megastat output (there is plenty of space after each question, answers should not be longer than the allocated space).
As understanding of the values and building the skill of reporting data/information, you are asked to interpret the meaning of numerous values.
Answer the following questions:
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.
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.
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.
What are the variance and standard deviation of the random errors for this regression analysis?
Calculate and report the residual of the 10th observation in the data set.
Data-
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 13Explanation / Answer
Answer:
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
Answer the following questions:
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% CI=(12.336, 13.622). we are 95% confident that true average number of sick days for all individuals who smoke 30 cigarettes per day falls in the interval (12.336, 13.622).
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% PI=(5.078, 20.880). we are 95% confident that true number of sick days for single individual who smoke 30 cigarettes per day falls in the interval (5.078, 20.880).
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.
Slope =0.1897
95% CI for slope = (0.1398, 0.2396).
We are 95% confident that for who smoke one more cigarette increases the number of sick days falls in the interval (0.1398, 0.2396).
What are the variance and standard deviation of the random errors for this regression analysis?
Variance=15.9735
standard deviation =3.997
Calculate and report the residual of the 10th observation in the data set.
10th observation (43,14)
Predicted value y =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
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