Question

The toxicity of two sludge fertilizer treatments are compared. One hundred tomato plants were randomly assigned to pots containing sludge processed by either the “new” or “old” treatment. The tomatoes harvested from the plants were evaluated to determine if the nickel was at a toxic level. Let π_New be the true population proportion of all plants treated with the “New” treatment that will be toxic and let π_Old be the true population proportion of all plants treated with the “Old” treatment that will be toxic. For the tests considered in parts B,C and E below we will test H0: π_New = π_Old.

	Toxic	Non-Toxic	Total
New	5	45	50
Old	9	41	50
Total	14	86	100

A. Calculate the estimated proportion of toxic plants for EACH treatment.

B. Use the two-sample “Z test” (prop.test function in R) to compare the proportion of Toxic plants for the two treatments. Use correct = TRUE (default). Give your (chi-square) test statistic and p-value.

C. Use the chi-square test for contingency tables (chisq.test function in R) to compare the proportion of Toxic plants for the two treatments. Use correct = TRUE (default). Give your test statistic and p-value. Note: You should find that these results exactly match the results above.

D. Calculate the expected cell counts from the chi-square test.

Answer 1

A. Calculate the estimated proportion of toxic plants for EACH treatment.

For new , p=5/50 = 0.10

For old , p=9/50 = 0.18

B. Use the two-sample “Z test” (prop.test function in R) to compare the proportion of Toxic plants for the two treatments. Use correct = TRUE (default). Give your (chi-square) test statistic and p-value.

R code:

x<-c(5,9)

y<-c(50,50)

prop.test(x,y)

R outpit:

        2-sample test for equality of proportions with continuity correction

data: x out of y

X-squared = 0.74751, df = 1, p-value = 0.3873

alternative hypothesis: two.sided

95 percent confidence interval:

-0.23510963 0.07510963

sample estimates:

prop 1 prop 2

0.10   0.18

Calculated chi square = 0.74751, df = 1, p-value = 0.3873.

C. Use the chi-square test for contingency tables (chisq.test function in R) to compare the proportion of Toxic plants for the two treatments. Use correct = TRUE (default). Give your test statistic and p-value. Note: You should find that these results exactly match the results above.

R code:

x<-c(5,45)

y<-c(9,41)

mydata <- data.frame(x,y)

chisq.test(mydata)

        Pearson's Chi-squared test with Yates' continuity correction

data: mydata

X-squared = 0.74751, df = 1, p-value = 0.3873

Chi square test statistic = 0.7475, P=0.3873

D. Calculate the expected cell counts from the chi-square test.

Expected Frequencies
	Toxic	Non-Toxic	Total
New	7	43	50
Old	7	43	50
Total	14	86	100