THe following data set contains data on Blood Alcohol Content (BAC) for a set of

Question

THe following data set contains data on Blood Alcohol Content (BAC) for a set of men who drank a particular number of beers over a one hour period. We wish to predict BAC from the number of beers consumed using simple regression. The commands you need are in Minitab under Stat>Regression. To do a prediction, click on the "Options" box in the regression command. Type the value of the X variable into the box "Prediction Intervals for new observations."

Beers BAC

5 0.1

2 0.03

9 0.19

8 0.12

3 0.04

7 0.095

3 0.07

5 0.06

3 0.02

5 0.05

4 0.07

6 0.1

5 0.085

7 0.09

1 0.01

4 0.05

Create a 90% confidence interval for the slope coefficient.

A.) The degrees of freedom for the t-statistic are

B.)The correct t value (the value from the t-distribution that is necessary for calculating the confidence interval) is (three decimals)

C.) I am 90% confident that the population slope coefficient, 1 is in the interval (round your values to the nearest integer): [ , ]

Explanation / Answer

Let independent variable x = beers and

dependent variable y = BAC.

Here we have to do regression in MINITAB.

Steps in MINITAB :

Stat --> Regression --> Regression --> Response : y --> Predictors : x --> Options : Prediction interval for new observations : x --> Confidence level : 90% --> ok --> Results : select second option --> ok --> ok

Output is,

Regression Analysis: y versus x

The regression equation is
y = - 0.0127 + 0.0180 x

Predictor Coef SE Coef T P
Constant -0.01270 0.01264 -1.00 0.332
x 0.017964 0.002402 7.48 0.000

S = 0.0204410 R-Sq = 80.0% R-Sq(adj) = 78.6%

Analysis of Variance

Source DF SS MS F P
Regression 1 0.023375 0.023375 55.94 0.000
Residual Error 14 0.005850 0.000418
Total 15 0.029225

Predicted Values for New Observations

New
Obs Fit SE Fit 90% CI 90% PI
1 0.07712 0.00513 ( 0.06808, 0.08615) ( 0.04000, 0.11424)
2 0.02323 0.00847 ( 0.00831, 0.03815) (-0.01574, 0.06220)
3 0.14897 0.01128 ( 0.12910, 0.16884) ( 0.10785, 0.19009)
4 0.13101 0.00920 ( 0.11480, 0.14722) ( 0.09152, 0.17049)
5 0.04119 0.00671 ( 0.02937, 0.05301) ( 0.00330, 0.07909)
6 0.11305 0.00733 ( 0.10014, 0.12595) ( 0.07480, 0.15129)
7 0.04119 0.00671 ( 0.02937, 0.05301) ( 0.00330, 0.07909)
8 0.07712 0.00513 ( 0.06808, 0.08615) ( 0.04000, 0.11424)
9 0.04119 0.00671 ( 0.02937, 0.05301) ( 0.00330, 0.07909)
10 0.07712 0.00513 ( 0.06808, 0.08615) ( 0.04000, 0.11424)
11 0.05915 0.00547 ( 0.04952, 0.06879) ( 0.02188, 0.09642)
12 0.09508 0.00585 ( 0.08477, 0.10539) ( 0.05763, 0.13253)
13 0.07712 0.00513 ( 0.06808, 0.08615) ( 0.04000, 0.11424)
14 0.11305 0.00733 ( 0.10014, 0.12595) ( 0.07480, 0.15129)
15 0.00526 0.01049 (-0.01321, 0.02373) (-0.03520, 0.04573)
16 0.05915 0.00547 ( 0.04952, 0.06879) ( 0.02188, 0.09642)

Values of Predictors for New Observations

New
Obs x
1 5.00
2 2.00
3 9.00
4 8.00
5 3.00
6 7.00
7 3.00
8 5.00
9 3.00
10 5.00
11 4.00
12 6.00
13 5.00
14 7.00
15 1.00
16 4.00

The degrees of freedom for the t-statistic are 14.

The correct t value is 7.480.

90% confidence interval for population slope coefficient is,

b - E < 1 < b + E

where b is slope for x.

E is the margin of error.

E = tc * Sb1

tc is critical value for t-distribution.

Sb1 is the standard error for x.

Sb1 = 0.002402

tc we can find by using EXCEL.

syntax is,

=TINV(probability, deg_freedom)

where, probability = 1 - c

c is confidence level = 90% = 0.90

probability = 1 - 0.90 = 0.10

deg_freedom = 14

tc = 1.7613

E = 1.7613 * 0.002402 = 0.0042

lower limit = 0.0180 - 0.0042 = 0.0138

upper limit = 0.0180 + 0.0042 = 0.0222

I am 90% confident that the population slope coefficient, 1 is in the interval [0.0138, 0.0222].

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