A biologist looked at the relationship between number of seeds a plant produces and the percent of thoseseeds that sprout. The results of the survey are shown below.Seeds Produced 68 56 61Sprout Percent 60.6 69.2= 04864.7 64.6#0a. Find the correlation coefficient:b. The null and alternative hypotheses for correlation are:H₂: ?H₁: 2The p-value is:6364.140Round to 2 decimal places.63 6265.1 68.4 84(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 contextof the study.O There is statistically significant evidence to conclude that there is a correlation between thenumber of seeds that a plant produces and the percent of the seeds that sprout. Thus, theregression line is useful.O There is statistically insignificant evidence to conclude that a plant that produces more seedswill have seeds with a lower sprout rate than a plant that produces fewer seeds.O There is statistically significant evidence to conclude that a plant that produces more seedswill have seeds with a lower sprout rate than a plant that produces fewer seeds.O There is statistically insignificant evidence to conclude that there is a correlation between thenumber of seeds that a plant produces and the percent of the seeds that sprout. Thus, the useof the regression line is not appropriate.(Round to two decimal places)f. The equation of the linear regression line is:d. ²=e. Interpret ²:O There is a 65% chance that the regression line will be a good predictor for the percent of seedsthat sprout based on the number of seeds produced.O 65% of all plants produce seeds whose chance of sprouting is the average chance of sprouting.O Given any group of plants that all produce the same number of seeds, 65% of all of theseplants will produce seeds with the same chance of sprouting.O There is a large variation in the percent of seeds that sprout, but if you only look at plantsthat produce a fixed number of seeds, this variation on average is reduced by 65%.(Please show your answers to two decimal places)g. Use the model to predict the percent of seeds that sprout if the plant produces 55 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:O For every additional seed that a plant produces, the chance for each of the seeds to sprouttends to decrease by 0.62 percent.i. Interpret the y-intercept in the context of the question:O The slope has no practical meaning since it makes no sense to look at the percent of the seedsthat sprout since you cannot have a negative number.O As x goes up, y goes down.O The average sprouting percent is predicted to be 103.39.O If plant produces no seeds, then that plant's sprout rate will be 103.39.O The y-intercept has no practical meaning for this study.O The best prediction for a plant that has 0 seeds is 103.39 percent.

The data on seed produced and sprout percent is given below,Sr noSeed produced (x)Sprout percent(y)16860.625669.236164.744864.656364.166365.176268.484084Let x denote the seed produced.and y denotes the Sprout percentsample size n=8level of significance = 0.05a) The correlation coefficient r is calculate as,r= now Sr noxyxyx2y216860.64120.846243672.3625669.23875.231364788.6436164.73946.737214186.0944864.63100.823044173.1656364.14038.339694108.8166365.14101.339694238.0176268.44240.838444678.5684084336016007056sum461540.730783.92716736901.63r The correlation coefficient between the number of seeds that a plant produces and the percentage of the seeds that sprout is -0.81.b) The test hypothesis is,H0: =0Vs H1: 0the test statistics are,t= The value of the test statistic is -3.3833p-value is,p-value = P(T>t) where T follows t-distribution with n-2 dfp-value = P(T>-3.3833) = P(T<3.3833) = T.DIST.2T(3.3833,6) .........Use excel function(=TDIST(x ,df, tail)) = 0.0148The p-value is 0.0148c) level of significance =0.05Decision: If the p-value is less than then reject H0, otherwise reject to fail H0p-value < reject to fail H0 at level 0.05.There is statistically significant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percentage of the seeds that sprout. Thus, the regression line is useful.Option(1) is correct.d) The value of r2 is,r2 = (-0.8066)2 = 0.6506 0.65The value of r2 is 0.65e) Interpretation:There is a large variation in the percentage 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 65%.Option (4) is correct.{option (1),(2), and (3) say the chance of variation does not say exact variation hence option (4) is correct}f) The regression equation is,y= a+bxwhere Y is the dependent variable represented by variable sprout percent.a is the intercept coefficient.b is the slope coefficientx is the independent variable represented by the variable seed produced.now find the value of a and b.The fitted regression equation is, = 103.39-0.62xg) Find the value of when x=55103.39-0.62 55 = 69.29 69The percent of seeds that sprout is 69 if the plant produces 55 seeds.h) The interpretation of slope coefficient.The value of slope coefficient b= -0.62For every additional seed that a plant produces, the chance for each of the seeds to sprout tends to decrease by 0.62 percent.Option (1) is correcti) The interpretation of the intercept coefficient.The best prediction for a plant that has 0 seeds is 103.39 percent.Option (4) is correcta) The correlation coefficient between the number of seeds that a plant produces and the percentage of the seeds that sprout is -0.81.b) The test hypothesis is,H0: =0 Vs H1: 0The p-value is 0.0148c) Option(1) is correct.There is statistically significant evidence to conclude that there is a correlation between the number of seeds that a plant produces and the percentage of the seeds that sprout. Thus, the regression line is useful.d) The value of r2 is 0.65e)Option (4) is correct.There is a large variation in the percentage 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 65%.f) The fitted regression equation is, 103.39-0.62xg) The percent of seeds that sprout is 69 if the plant produces 55 seeds.h) Option (1) is correct.For every additional seed that a plant produces, the chance for each of the seeds to sprout tends to decrease by 0.62 percent.i) Option (4) is correct.The best prediction for a plant that has 0 seeds is 103.39 percent.

Math

Statistics

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 68 56 61 48 Sprout Percent 60.6 69.2 64.7 64.6 63 63 62 40 64.1 65.1 68.4 84 a. Find the correlation coefficient: r= b. The null and alternative hypotheses for correlation are: H₂: ? ✓ H₁:20 The p-value is: Round to 2 decimal places. (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. O 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. O 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. O 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. (Round to two decimal places) d. ² e. Interpret ²: There is a 65% chance that the regression line will be a good predictor for the percent of seeds that sprout based on the number of seeds produced. O 65% of all plants produce seeds whose chance of sprouting is the average chance of sprouting. O Given any group of plants that all produce the same number of seeds, 65% of all of these plants will produce seeds with the same chance of sprouting. f. The equation of the linear regression line is: O 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 65%. (Please show your answers to two decimal places) g. Use the model to predict the percent of seeds that sprout if the plant produces 55 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 0.62 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: O The average sprouting percent is predicted to be 103.39. O If plant produces no seeds, then that plant's sprout rate will be 103.39. O The y-intercept has no practical meaning for this study. O The best prediction for a plant that has 0 seeds is 103.39 percent.

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 68 56 61 48 Sprout Percent 60.6 69.2 64.7 64.6 63 63 62 40 64.1 65.1 68.4 84 a. Find the correlation coefficient: r= b. The null and alternative hypotheses for correlation are: H₂: ? ✓ H₁:20 The p-value is: Round to 2 decimal places. (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. O 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. O 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. O 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. (Round to two decimal places) d. ² e. Interpret ²: There is a 65% chance that the regression line will be a good predictor for the percent of seeds that sprout based on the number of seeds produced. O 65% of all plants produce seeds whose chance of sprouting is the average chance of sprouting. O Given any group of plants that all produce the same number of seeds, 65% of all of these plants will produce seeds with the same chance of sprouting. f. The equation of the linear regression line is: O 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 65%. (Please show your answers to two decimal places) g. Use the model to predict the percent of seeds that sprout if the plant produces 55 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 0.62 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: O The average sprouting percent is predicted to be 103.39. O If plant produces no seeds, then that plant's sprout rate will be 103.39. O The y-intercept has no practical meaning for this study. O The best prediction for a plant that has 0 seeds is 103.39 percent.

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018

18th Edition

ISBN:9780079039897

Author:Carter

Publisher:Carter

Chapter4: Equations Of Linear Functions

Section4.5: Correlation And Causation

Problem 23PFA

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