A grass seed company conducts a study to determine the relationship between the density of seeds planted (in pounds per 500 sq ft)and the quality of the resulting lawn.Eight similar plots of land are selected and each is planted with a particular density of seed.One month later the quality of each lawn is rated on a scale of 0 to 100.The regression analysis is given below.Find a 95% confidence interval for the slope of the regression line. Dependent variable is: Lawn Quality
R-squared = 36.0%
S = 9.073602 with 8 - 2 = 6 degrees of freedom $$\begin{array} { l l l l l }
\text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \
\text { Constant } & 33.14815 & 7.510757 & 4.413423 & 0.004503 \
\text { Seed Density } & 4.537037 & 2.469522 & 1.837213 & 0.115825
\end{array}$$
A)(-0.26,9.34)
B)(-1.51,10.58)
C)(14.77,51.53)
D)(27.10,39.18)
E)(-1.30,10.38)

Question

A grass seed company conducts a study to determine the relationship between the density of seeds planted (in pounds per 500 sq ft)and the quality of the resulting lawn.Eight similar plots of land are selected and each is planted with a particular density of seed.One month later the quality of each lawn is rated on a scale of 0 to 100.The regression analysis is given below.Find a 95% confidence interval for the slope of the regression line. Dependent variable is: Lawn Quality
R-squared = 36.0%
S = 9.073602 with 8 - 2 = 6 degrees of freedom $$\begin{array} { l l l l l } 
\text { Variable } & \text { Coefficient } & \text { SE(Coeff) } & \text { t-ratio } & \text { P-value } \
\text { Constant } & 33.14815 & 7.510757 & 4.413423 & 0.004503 \
\text { Seed Density } & 4.537037 & 2.469522 & 1.837213 & 0.115825
\end{array}$$
A)(-0.26,9.34)
B)(-1.51,10.58)
C)(14.77,51.53)
D)(27.10,39.18)
E)(-1.30,10.38)

Quizplus · Accepted Answer

The 95% confidence interval for the slope is calculated using the formula: coefficient ± (t-critical value * standard error). With 6 degrees of freedom, the t-critical value is approximately 2.447. Thus, the interval is 4.537037 ± (2.447 * 2.469522), which results in the interval (-1.51, 10.58).

A Grass Seed Company Conducts a Study to Determine the Relationship