1. Construct a scatter plot using excel for the given data. Determine whether there is a positive linear correlation, negative linear correlation, or no linear correlation. Complete the table and find the correlation coefficient r.
The data for x and y is shown below.
y / 65 / 80 / 60 / 88 / 66 / 78 / 85 / 90 / 90 / 71
a: Scatter plot with regression line
b: Type of correlation (positive linear correlation, negative linear correlation, or no linear correlation)
The scatter plot shows that as x increases y also increases. Thus there exist a positive linear correlation between x and pressure.
c: Complete the table and find the correlation coefficient r rounded to 4 decimals.
x / y / xy / x^2 / y^23 / 65 / 195 / 9 / 4225
5 / 80 / 400 / 25 / 6400
2 / 60 / 120 / 4 / 3600
8 / 88 / 704 / 64 / 7744
2 / 66 / 132 / 4 / 4356
4 / 78 / 312 / 16 / 6084
4 / 85 / 340 / 16 / 7225
5 / 90 / 450 / 25 / 8100
6 / 90 / 540 / 36 / 8100
3 / 71 / 213 / 9 / 5041
42 / 773 / 3406 / 208 / 60875
Use the last row of the table to show the column totals.
From the given data we have
=10, = 42, =773, =3406, = 208 and = 60875
We have,
=
= 0.8465
2. Construct a scatter plot and include the regression line on the graph using excel for the given data. Determine whether there is a positive linear correlation, negative linear correlation, or no linear correlation. Complete the table and find the correlation coefficient r.
In an area of the Midwest, records were kept on the relationship between the rainfall (in inches) and the yield of wheat (bushels per acre).
Yield, y / 50.5 / 46.2 / 58.8 / 59.0 / 82.4 / 49.2 / 31.9 / 76.0 / 78.8
a: Scatter plot with regression line
b: Type of correlation (positive linear correlation, negative linear correlation, or no linear correlation)
The scatter plot shows that as Rain fall increases Yield also increases. Thus there exist a positive linear correlation between Rain fall and yield.
c: Complete the table and find the correlation coefficient r rounded to 4 decimals.
10.5 / 50.5 / 530.25 / 110.25 / 2550.25
8.8 / 46.2 / 406.56 / 77.44 / 2134.44
13.4 / 58.8 / 787.92 / 179.56 / 3457.44
12.5 / 59 / 737.5 / 156.25 / 3481
18.8 / 82.4 / 1549.12 / 353.44 / 6789.76
10.3 / 49.2 / 506.76 / 106.09 / 2420.64
7.0 / 31.9 / 223.3 / 49 / 1017.61
15.6 / 76 / 1185.6 / 243.36 / 5776
16.0 / 78.8 / 1260.8 / 256 / 6209.44
112.9 / 532.8 / 7187.81 / 1531.39 / 33836.58
Use the last row of the table to show the column totals.
From the given data we have
=9, = 112.9, =532.8, =7187.81, = 1531.39 and
= 33836.58
We have,
=
= 0.98082
3. Using the r calculated in problem 2c test the significance of the correlation coefficient using a = 0.01 and the claim rho = 0. Use the 7-steps hypothesis test shown at the end of this project. (Note: Round the computed t to 3 decimals.)
Let ρ be the population correlation coefficient.
1. H0: ρ = 0
Ha: ρ ≠ 0
2. a = 0.01
3. We have,
=
= 13.3129
4. Since a =0.01, from Student’s t table with (n-2) = 7 degrees of freedom, the critical value is t0 = 3.499
5. Rejection region:
Reject Ho if t < -3.499 or t > 3.499
6. Decision:
Here, t = 13.3129 > 3.499
So we reject the null hypothesis Ho.
7. Interpretation:
Thus the correlation between Rain fall and Yield is statistically significant at 1% level of significance.
[If the alternative hypothesis is Ha: ρ > 0, use the following
1. H0: ρ = 0
Ha: ρ > 0
2. a = 0.01
3. We have,
=
= 13.3129
4. Since a =0.01, from Student’s t table with (n-2) = 7 degrees of freedom, the critical value is t0 = 2.998
5. Rejection region:
Reject Ho if t > 2.998
6. Decision:
Here, t = 13.3129 > 2.998
So we reject the null hypothesis Ho.
7. Interpretation:
Thus there exist a positive correlation between Rain fall and Yield at 1% level of significance.]
Linear Regression
4. In an area of the Midwest, records were kept on the relationship between the rainfall (in inches) and the yield of wheat (bushels per acre).
Yield, y / 50.5 / 46.2 / 58.8 / 59.0 / 82.4 / 49.2 / 31.9 / 76.0 / 78.8
Use the table developed in #2
a. Find the equation of the regression line for the given data. Round the line values to the nearest two decimal places.
Let y denote the Yield and x denote the Rain fall. Assume that x and y are linearly related. Let be the suggested linear relationship.
By the method of least squares the estimates of and are given by,
and
From the given data we have
=9, = 112.9, =532.8, =7187.81, = 1531.39 and
= 33836.58
We have,
= 4.3791
= 4.2668
Thus the fitted regression equation is
Yield = 4.2668 + 4.3791 Rain fall
b. Using the equation found in part a, predict the bushel yield when the rainfall is 11 inches. Round to the nearest bushel.
When the rainfall is 11 inches, the predicted value of Yield is
Yield = 4.2668 + 4.3791*11
= 52.4367