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 sample data are given below where x denotes seed density, and y denotes lawn quality.x 1 1 2 3 3 3 4 5y 30 40 40 40 50 65 50 50The sample linear correlation coefficient is r=0.600. At the 1% significance level, do the data provide sufficient evidence to conclude that seed density and lawn quality are positively linearly correlated?

Answer 1

The correlation between seed density and lawn quality is determined by comparing the sample linear correlation coefficient r=0.600 to the critical value at the 1% significance level for 6 degrees of freedom. If r exceeds this critical value or if the p-value is less than 0.01, there is sufficient evidence to claim a positive linear correlation.

Step-by-step explanation:

To assess if the data provide sufficient evidence to conclude that there is a positive linear correlation between seed density (x) and lawn quality (y), we must analyze the sample linear correlation coefficient (r) and the significance level (alpha). Given the correlation coefficient r=0.600, we need to compare it to the critical value at the 1% significance level for the degrees of freedom (df) which is n - 2, where n is the number of data pairs.

With eight data points (n=8), the degrees of freedom (df) is 6 (n-2=8-2=6). The sample size is relatively small, so we must consult a statistical table or use statistical software to find the critical value for r at the 1% significance level for df=6. Generally, the 1% critical value for r will be higher than at the 5% level, indicating that we need a stronger correlation to reject the null hypothesis of no correlation at the 1% level compared to 5%.

If the computed value of r (0.600) exceeds the critical value at df=6 for the 1% significance level, we conclude there is enough evidence to reject the null hypothesis and accept that there is a significant positive linear correlation. If the computed value of r does not exceed the critical value, we would fail to reject the null hypothesis.

Because a p-value is not directly given here, a statistical test such as the Pearson correlation significance test would usually be conducted to find the p-value, which would then be compared against the significance level. If the p-value is also lower than the 1% significance level, this would be another way to support rejecting the null hypothesis.

Answer 2

I think it -1.50 to 10.58

Step-by-step explanation: