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. 50 67 55 63 43 Seeds Produced Sprout Percent 52 59.2 61 54.6 58 55.8 67 55.2 59 47.2 57 49.8 70.8 a. Find the correlation coefficient: r = Round to 2 decimal places. b. The null and alternative hypotheses for correlation are: Ho: ? = 0 H : ? + 0 The p-value is: (Round to four decimal places) C. Use a level of significance of a = 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 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. 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. d. 72 (Round to two decimal places) e. Interpretrº There is a 80% chance that the regression line will be a good predictor for the percent of seeds that sprout based on the number of seeds produced. Given any group of plants that all produce the same number of seeds, 80% of all of these plants will produce seeds with the same chance of sprouting. 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 80%. 80% of all plants produce seeds whose chance of sprouting is the average chance of sprouting. f. The equation of the linear regression line is: X (Please show your answers to two decimal places) g. Use the model to predict the percent of seeds that sprout if the plant produces 60 seeds. Percent sprouting = (Please round your answer to the nearest whole number.) h. Interpret the slope of the regression line in the context of the question: For every additional seed that a plant produces, the chance for each of the seeds to sprout tends to decrease by 0.74 percent. 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. i. Interpret the y-intercept in the context of the question: The y-intercept has no practical meaning for this study. The best prediction for a plant that has 0 seeds is 98.66 percent. The average sprouting percent is predicted to be 98.66. Ulf plant produces no seeds, then that plant's sprout rate will be 98.66.

Answer 1

S.No y 59.2 54.6 55.8 1 2 3 4 5 6 7 8 9 Total Mean 52 61 58 67 50 67 55 63 43 55.2 59 47.2 57 49.8 70.8 508.6 56.5111 (x-) 28.4444 13.4444 0.4444 93.4444 53.7778 93.4444 5.4444 32.1111 205.4444 526.0000 SSX (7-7) 7.23 3.65 0.51 1.72 6.19 86.70 0.24 45.04 204.17 355.4489 (x-x)(y-7) -14.3407 -7.0074 -0.4741 -12.6741 -18.2519 -90.0074 -1.1407 -38.0296 -204.8074 -386.7333 516 57.333 SSY SXY sample size na 9 X = 57.3333 y = 56.5111 S. (X-X) = Sw (X-X)(y-7)= slope= . = S/S intercept= B. = V-B.X= 526.000 -386.733 -0.7352 COM 98.6646

a)

correlation r='Sxy/(√Sxx*Syy) =

-0.89

b)

null hypothesis: Ho:           ρ'=		0
Alternate Hypothesis: Ha: ρ≠		0

test stat t=	r*(√(n-2)/(1-r2))=	-5.2906
P value    =	0.0011	(from excel:tdist(5.2906,7,2)

c)

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.

d)

r² =0.80

e)

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 80%.

f)

y^ =98.66+(-0.74)*x

g)

predicted val=98.66+60*-0.74=55

(please try 54 if this comes wrong)

h)

For every additional seed that a plant produces, the chance for each of the seeds to sprout tends to decrease by 0.74 percent.

i)

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