The following data resulted from an experiment studying the effect of nitrogen f

Question

The following data resulted from an experiment studying the effect of nitrogen fertilizer on lettuce. Data are taken from Kuehl (2000). Amount of fertilizer Heads of lettuce per plot 0 lbs/acre 104 114 90 140 200 lbs/acre 131 148 154 163 Assume that the number of heads of lettuce in the plots is approximately Normally distributed.

(a) Let 2 1 represent the variance of the number of heads of lettuce under 0 lbs/acre of fertilizer and let 2 2 represent that under 200 lbs/acre. For each of the following sets of hypotheses get the p-value based on the data in the table. i. H0: 2 1 2 2 versus H1: 2 1 > 2 2 ii. H0: 2 1 2 2 versus H1: 2 1 < 2 2 iii. H0: 2 1 = 2 2 versus H1: 2 1 6= 2 2

(b) Do you believe that the variances 2 1 and 2 2 are different? Why/why not?

(c) Let µ1 represent the mean of number of heads of lettuce under 0 lbs/acre of fertilizer and let µ2 represent that under 200 lbs/acre. For each of the following sets of hypotheses get the p-value based on the data in the table, assuming that 2 1 = 2 2 . i. H0: µ1 µ2 versus H1: µ1 > µ2 ii. H0: µ1 µ2 versus H1: µ1 < µ2 iii. H0: µ1 = µ2 versus H1: µ1 6= µ2

SOLUTIONS WITH R IS BETTER!!!!

1. The following data resulted from an experiment studying the effect of nitrogen fertilizer on lettuce Data are taken from Kuehl (2000) Amount of fertilizer Heads of lettuce per plot 0 lbs/acre 200 lbs/acre 104 114 90 140 131 148 154 163 Assume that the number of heads of lettuce in the plots is approximately Normally distributed (a) Let represent the variance of the number of heads of lettuce under 0 lbs/acre of fertilizer and let represent that under 200 lbs/acre. For each of the following sets of hypotheses get the p-value based on the data in the table (b) Do you believe that the variances and are different? Why/why not? (c) Let represent the mean of number of heads of lettuce under 0 lbs/acre of fertilizer and let 2 represent that under 200 lbs/acre. For each of the following sets of hypotheses get the p-value based on the data in the table, assuming that - (d) Do you believe the m eans 1 and 2 are different? Why/why not?

Explanation / Answer

> #(a)

> #(i)

> x=c(104,114,90,140)

> y=c(131,148,154,163)

> var.test(x,y,alternative ="less" )

F test to compare two variances

data: x and y

F = 2.4469, num df = 3, denom df = 3, p-value = 0.7592

alternative hypothesis: true ratio of variances is less than 1

95 percent confidence interval:

0.00000 22.69886

sample estimates:

ratio of variances

2.446886

>

> #(ii)

> var.test(x,y,alternative = "greater")

F test to compare two variances

data: x and y

F = 2.4469, num df = 3, denom df = 3, p-value = 0.2408

alternative hypothesis: true ratio of variances is greater than 1

95 percent confidence interval:

0.2637689 Inf

sample estimates:

ratio of variances

2.446886

>

> #(iii)

> var.test(x,y,alternative = "two.sided")

F test to compare two variances

data: x and y

F = 2.4469, num df = 3, denom df = 3, p-value = 0.4817

alternative hypothesis: true ratio of variances is not equal to 1

95 percent confidence interval:

0.1584855 37.7779261

sample estimates:

ratio of variances

2.446886

(b) for (i) and (ii) case, ²1 and ²2 are different and for (iii) case, ²1 = ²2  because p-value is greater than 0.05 for all cases.

> #(c)

> x=c(104,114,90,140)

> y=c(131,148,154,163)

> #(i)

> t.test(y,x,alternative = "less")

Welch Two Sample t-test

data: y and x

t = 2.9545, df = 5.1012, p-value = 0.9845

alternative hypothesis: true difference in means is less than 0

95 percent confidence interval:

-Inf 62.12496

sample estimates:

mean of x mean of y

149 112

> #(ii)

> t.test(y,x,alternative = "greater")

Welch Two Sample t-test

data: y and x

t = 2.9545, df = 5.1012, p-value = 0.01548

alternative hypothesis: true difference in means is greater than 0

95 percent confidence interval:

11.87504 Inf

sample estimates:

mean of x mean of y

149 112

> #(iii)

> t.test(y,x,alternative = "two.sided")

Welch Two Sample t-test

data: y and x

t = 2.9545, df = 5.1012, p-value = 0.03095

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

4.998869 69.001131

sample estimates:

mean of x mean of y

149 112

(d) if we take the level of significance=0.05

then in c(i), the p-value is 0.98 which is greater than 0.05 then, in this case, we fail to reject the null hypothesis. However, for all the three cases mu1 and mu2 are different. (because p-value is less than 0.05 in both cases).

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