Math 533 week 6 more help

MATH 533 WEEK 6 - MORE
HELP
B. Heard
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MATH 533 WEEK 6 - MORE HELP
Example 1 – The results for the output of a certain component are
tracked consistently to make sure that that the component performs
as advertised. Suppose the component’s manufacturer wants to use
simple linear regression to predict the output in volts (y) from the
setting on one of it’s machines (x). Find a 90% confidence interval for
the true slope of the line and interpret what the results mean.
Data on following page.
Also, remember the general form of the equation is
y = β0 + β1x + ε

Batch
Output/Volt
s
Setting
1 3.3 100
2 3.6 108
3 4.1 138
4 4 90
5 3.9 104
6 4.1 95
7 3.9 113
8 3.7 149
9 3.7 119
10 4 92
11 3.5 290
12 3.7 136
13 4 184
14 3.6 139
15 3.4 164
16 3.4 263
17 3.9 151
18 3.6 144
19 3.8 108
20 3.4 143
21 4 113
22 3.9 100
23 3.9 125
24 4 121
25 3.6 137
Remember the output in volts is (y) and the setting is (x).

First copy and paste data into Minitab…

Now go to Stat >> Regression >> Regression (remembering which
one is x and which one is y)

I get my results in the session window…

What I will need now are the following two things…
The slope (β1) and the standard error (sβ1)… I have them both as seen below

Now I will need my value for tα/2, Go to Graph >> Probability Distribution Plot,
click View Probability, then Ok

Set distribution to t and your degrees of freedom to your sample size
minus 2. (In this case 25-2 = 23

Click the Shaded Area Tab, Probability Radial Button, Both Tails
button and then enter 0.10 for probability (1 – your confidence)

After clicking OK, you will see that your t value is 1.714
0.4
0.3
0.2
0.1
0.0
X
Density
-1.714
0.05
1.714
0.05
0
Distribution Plot
T, df=23

Now I can easily get my confidence interval by using…
(β1) +/- (tα/2)(sβ1) (My Betas should have rooftops on them)
Let’s look at the results…

(β1) +/- (tα/2)(sβ1)
(-0.00223) +/- (1.714)(0.0009321) = (-0.0038, -0.0006) Rounded to 4 Decimals
0.4
0.3
0.2
0.1
0.0
X
Density
-1.714
0.05
1.714
0.05
0
Distribution Plot
T, df=23

Interpret results
The y-intercept would have no practical interpretation because a
machine setting of 0 is outside the range of the sample data.
However, the slope does have meaning, because for each additional
unit in setting, the output in volts is estimated to change by the value
of the slope.

Example 2 – A study was done on the results of students quiz grades
based on the number of hours studied. 8 Different students were
examined. Based on the given data find the least squares line
relating score (y) to hours studied (x).
Student Score
Hours
Studied
1 88.5 30
2 88.3 27.6
3 86.3 25.3
4 76.9 23.4
5 87.2 28.9
6 90.4 34.8
7 75.6 21.8
8 85.7 26.1

Copy Data into Minitab, then Go to Stat >> Regression >>
Regression

Regression Equation is y = 53.627+1.147x ( to three decimals)
Note I used the
coefficients circled
because they gave me
the accuracy I needed.

Interpretations
The y intercept has no meaning since 0 is not in the observed range
of study hours.
However, the slope does have meaning. For each hour increase in
study hours, the score is estimated to increase by the slope.
Remember that β0 is your y-intercept and β0 is your slope. My β’s should have
“rooftops” on them – You know a “^”…

Predict the score of a student who studies 30 hours.
Just plug and chug.
y = 53.627+1.147x
Plug a 30 in for x
y = 53.627+1.147(30)
y = 88.037 (three decimals)
Always pay attention to the required accuracy (number of decimals)

Example 3 – A study was done on ranking the total driving
performance of some really bad golfers like me. The method required
knowing the golfer’s average driving distance and driving accuracy
(shots in the fairway). In the study, they constructed a straight line
model relating driving accuracy (y) to driving distance (x). A Minitab
printout with prediction and confidence intervals for a driving
distance of x = 200 is shown below.Minitab Output
Predicted Values For New Observations
New
Obs Fit SE Fit 95% CI 95% PI
1 68.253 0.426
(42.265,
45.362)
(38.258,
49.956)
Values of Predictors for New
Observations
New
Obs
Distanc
e
1 200

What is a practical interpretation of the results on the printout?
Well, we would say we are 95% confident that the actual driving
accuracy for a golfer driving the ball 200 yards is between the limits
of the prediction interval.
Key words: One golfer or in this case, “a golfer” goes with the
Prediction Interval
“All golfers” goes with the confidence interval. Don’t let them fool
you.

What is the 95% confidence interval? 95% prediction interval?
Minitab Output
Predicted Values For New Observations
New
Obs Fit SE Fit 95% CI 95% PI
1 68.253 0.426
(42.265,
45.362)
(38.258,
49.956)
Values of Predictors for New Observations
New
Obs
Distanc
e
1 200

Again, if you are interested in knowing the average driving distance
of
“All golfers” - Use confidence interval (You would be 95% confident
in this case)
“A single golfer” – Use prediction interval (You would be 95%
confident in this case)

Example 4 – A study on the effect of jelly beans on working
crossword puzzles measured the crossword puzzle success (on a 30
point scale) and the number of jelly beans consumed before doing
the puzzle. On the basis of the information provided, the data shown
in the table on the next chart were obtained for 15 people who
participated in the study. Conduct a test to determine if the
crossword puzzle success (y) is linearly related to the number of jelly
beans consumed (x). Use α = 0.10

Data for Example 3 Crossword Puzzle
Success
Jelly Beans
Consumed
18.8 49
19.3 49
19.8 52
19.1 53
20 53
20.3 53
19.2 56
17.4 58
18.7 58
20.3 58
20.8 59
21 60
20.9 62
21.3 62
20.5 61

What would the correct null and alternative hypotheses be?
We are testing to see if they are linearly related, so they would be
H0: β1 = 0
Ha: β1 ≠ 0

Find the test statistic and p value.
These can be tricky when you are not sure the number of decimals
(accuracy) you will need. I’m going to show you how to get more
accuracy if you need it.

Get your data into Minitab

Use Stat >> Regression >> Regression

Session Window
Here is the test statistic t, and
the p value.
BUT
BUT
BUT
Go to next chart…

What if they want the t value to three decimal places? Easy, watch…
t is given as 1.93 which is correct, but
to only two decimal places. The
testing software should be such that a
little tolerance is there, but to get more
accuracy just divide the Coef by the SE
Coef. In other words, look just to the
left of the 1.93 and divide 0.11414 by
0.05899
0.11414/0.05899 = 1.93490422105
blah blah
So, if they wanted three decimals, I

So what is the appropriate conclusion?
Compare the p value 0.075 to alpha of 0.10.
If p is less Pless or Please Reject the null hypothesis, so we accept the
null that
β1 ≠ 0, so if it’s not zero, THEY MUST BE LINEARLY RELATED AT AN ALPHA OF 0.10

This covers the hard stuff I think you will see on the quiz. Be
prepared…

Math 533 week 6 more help

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