Question

A biologist looked at the relationship between number of seeds a plant produces and the percent of those seeds that sprout. The results of the survey are shown below.

Seeds Produced	64	60	65	58	43	45	53	58	62
Sprout Percent	54	49	58.5	61	64.5	64.5	60.5	64	50

1. Find the correlation coefficient: r=r=    Round to 2 decimal places.
2. The null and alternative hypotheses for correlation are:
   H0:H0: ? ρ μ r   == 0
   H1:H1: ? μ ρ r   ≠≠ 0
   The p-value is:    (Round to four decimal places)
   
3. Use a level of significance of α=0.05α=0.05 to state the conclusion of the hypothesis test in the context of the study.
   • There is statistically significant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the regression line is useful.
   • There is statistically insignificant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the use of the regression line is not appropriate.
   • There is statistically insignificant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds.
   • There is statistically significant evidence to conclude that a plant that produces more seeds will have seeds with a lower sprout rate than a plant that produces fewer seeds.
4. r2r2 =  (Round to two decimal places)
5. Interpret r2r2 :
   • There is a 45% chance that the regression line will be a good predictor for the percent of seeds that sprout based on the number of seeds produced.
   • There is a large variation in the percent of seeds that sprout, but if you only look at plants that produce a fixed number of seeds, this variation on average is reduced by 45%.
   • 45% of all plants produce seeds whose chance of sprouting is the average chance of sprouting.
   • Given any group of plants that all produce the same number of seeds, 45% of all of these plants will produce seeds with the same chance of sprouting.
6. The equation of the linear regression line is:   
   ˆyy^ =  + xx   (Please show your answers to two decimal places)   
   
7. Use the model to predict the percent of seeds that sprout if the plant produces 56 seeds.
   Percent sprouting =  (Please round your answer to the nearest whole number.)   
   
8. Interpret the slope of the regression line in the context of the question:
   • The slope has no practical meaning since it makes no sense to look at the percent of the seeds that sprout since you cannot have a negative number.
   • As x goes up, y goes down.
   • For every additional seed that a plant produces, the chance for each of the seeds to sprout tends to decrease by 0.52 percent.
   
   
9. Interpret the y-intercept in the context of the question:
   • If plant produces no seeds, then that plant's sprout rate will be 87.53.
   • The average sprouting percent is predicted to be 87.53.
   • The best prediction for a plant that has 0 seeds is 87.53 percent.
   • The y-intercept has no practical meaning for this study.

Answer 1

correlation coefficient r=

Sxy/(√Sxx*Syy) =

-0.67

test stat t=	r*(√(n-2)/(1-r2))=	-2.4012
P value    =		0.0474

There is statistically significant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percent of the seeds that sprout. Thus, the regression line is useful.

Coeffficient of determination R^2 =SSR/SST=

0.45

There is a large variation in the percent of seeds that sprout, but if you only look at plants that produce a fixed number of seeds, this variation on average is reduced by 45%.

y^ =87.53+(-0.52)*x

predicted val=87.528+56*-0.515=

59

For every additional seed that a plant produces, the chance for each of the seeds to sprout tends to decrease by 0.52 percent.

The y-intercept has no practical meaning for this study.