Twenty plots were randomly selected in a large field of corn. The plant density
ID: 3218427 • Letter: T
Question
Twenty plots were randomly selected in a large field of corn. The plant density (number of plants per plot) and the mean cob weight (grams of grain per cob) were recorded for each plot. The average plant density across the 20 plots was 128 plants per plot. The average of the mean cob weights computed across the 20 plots was 224 grams of grain per cob. The standard deviation of the 20 plant densities was 32.6 plants per plot. The standard deviation of the 20 mean cob weights was 25.0 grams of grain per cob. The correlation between plant density and mean cob weight was -0.94. We will use linear regression to describe the relationship between plant density and mean cob weight.
a) Which of the two variables would be more naturally considered as the explanatory variable and which would be more naturally considered as the response variable?
b) Compute the equation of the least-squares regression line with mean cob weight as response and plant density as explanatory variable.
c) Interpret the estimate of slope in the context of this problem.
d) Interpret the estimate of intercept in the context of this problem.
f) Estimate the mean cob weight for plots with 135 plants.
g) Conduct a test to see whether the slope of the regression line is significantly different from 0 at the significance level of 0.05.
h) Find a 99% confidence interval for , the slope of the least squares regression line for predicting mean cob weight from plant density.
Explanation / Answer
Here dependent variable is mean cob weight (y) and independent variable is plant density is explanatory variable.
Given that,
n = 20
Xbar = mean of plant density = 128
Ybar = mean of cob weight = 224
Sx = standard deviation of plant density = 32.6
Sy = standard deviation of cob weight = 25.0
r = correlation between x and y = -0.94
Negative sign indicates that there is negative correlatio between x and y.
Slope = r * Sy / Sx
= (-0.94 * 25.0) / 32.6
= -0.72
Intercept = Ybar - slope*Xbar
= 224 - (-0.72)*128 = 316.16
The regression equation is,
y = intercept + slope*x
y = 316.16 -0.72*x
Interpretation of intercept and slope :
If plant density is 0 then cob weight will be 316.16.
One unit change in x (plant density) will be 0.72 unit decrease in mean cob weight.
Now we have to find mean cob weight (y) when x = 135
This we can find using regression equation.
y = 316.16 -0.72*135
y = 218.96
Now we have to test
H0 : B = 0 Vs H1 : B not= 0
where B is population slope for x.
Assume alpha = 0.05
Now the test statistic follows t-distribution.
t = slope / SEb
where SEb is the standard error of the estimate.
SEb = sqrt[(1 - r^2) * SSY / n]
= sqrt [(1 - (-0.94)^2 * 25^2) / 20]
= sqrt [3.6375] = 1.91
t = -0.72 / 1.91 = -0.38
NOw we have to find P-value.
P-value we can find by using EXCEL.
syntax :
=TDIST(x, deg_freedom, tails)
where x is absolute value of test statistic.
deg_freedom = n-2
tails = 2
P-value = 0.7084
P-value> alpha
Accept H0 at 5% level of significance.
COnclusion : Population slope for independent variable may be 0.
We don't get significant result.
h) Find a 99% confidence interval for , the slope of the least squares regression line for predicting mean cob weight from plant density
Here c = confidence level = 99% = 0.99
99% confidence interval for slope is,
b - E < B < b + E
where b is sample slope = -0.72
E is the margin of error
E = tc * SEb
where SEb is the standard error of the estimate = 1.91
tc is the critical value for normal distribution.
tc we can fund by using EXCEL.
syntax :
=TINV(probability, deg_freedom)
probability = 1 - c = 1 - 0.99 = 0.01
deg_freedom = n-2 = 20-2 = 18
tc = 2.878
E = 2.878 * 1.91 = 5.50
Lower limit = b - E = -0.72 - 5.50 = -6.22
upeer limit = b + E = -0.72 + 5.50 = 4.78
99% confidence interval for slope is (-6.22. 4.78)
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