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 and summary statistics are given below.Determine a 99% prediction interval for the lawn quality of a lawn sown with a seed density of 1.3. $$\begin{array} { l c l l c }
\text { Variable } & \text { Count } & \text { Mean } & \text { StdDev } & \text { Range } \
\text { Seed Density } & 8 & 2.75 & 0.49099 & 4 \
\text { Lawn Quality } & 8 & 45.625 & 3.712611 & 35
\end{array}$$ 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)1.0 to 77.1
B)3.1 to 75.0
C)21.2 to 56.9
D)29.9 to 48.2
E)30.4 to 73.4

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 and summary statistics are given below.Determine a 99% prediction interval for the lawn quality of a lawn sown with a seed density of 1.3. $$\begin{array} { l c l l c } 
\text { Variable } & \text { Count } & \text { Mean } & \text { StdDev } & \text { Range } \
\text { Seed Density } & 8 & 2.75 & 0.49099 & 4 \
\text { Lawn Quality } & 8 & 45.625 & 3.712611 & 35
\end{array}$$ 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)1.0 to 77.1
B)3.1 to 75.0
C)21.2 to 56.9
D)29.9 to 48.2
E)30.4 to 73.4

Quizplus · Accepted Answer

The 99% prediction interval for the lawn quality of a lawn sown with a seed density of 1.3 is 1.0 to 77.1. This interval is calculated using the formula:prediction interval = mean +/- t-value * standard error of the estimatewhere the mean is the average lawn quality for all of the plots, the t-value is the critical value from the t-distribution for a 99% confidence level and 6 degrees of freedom, and the standard error of the estimate is the standard deviation of the residuals from the regression model.

A Grass Seed Company Conducts a Study to Determine the Relationship