Chapter 7 Homework_ paired t-test

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Jan 9, 2024

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Paired t-test (25 points) Residential properties in Dougal County rarely need phosphorus for good turfgrass growth. As part of an extension education program, Early and Rusty Cuyler asked homeowners to report their phosphorus fertilizer use, in pounds of P 2 O 5 per acre, before the program and then one year later. For each of the following, answer the question, and show the output from the analyses you used to answer the question (Data: BIOL400 phosphorus). Creating a data frame with given data: data <- data.frame( Year = c(rep("year1", 10), rep("year2", 10)), Homeowner = c("a", "b", "C", "di", "e", "f", "g", "h", "i", "j", "a", "b", "C", "d", "e", "f", "g", "h", "i", "j"), P205 = c(0.81, 0.86, 0.79, 0.59, 0.71, 0.88, 0.63, 0.72, 0.76, 0.58, 0.67, 0.83, 0.81, 0.5, 0.71, 0.72, 0.67, 0.67, 0.48, 0.68) ) a. What was mean of the differences in P 2 O 5 before and after the training? (5 points) To calculate the mean difference, we need to subtract the P2O5 values for year 2 from the values for year 1, then take the mean: mean_diff <- mean(data$P205[1:10] - data$P205[11:20]) mean_diff Output : 0.059 b. Is this an increase or a decrease? (2 points) Since the mean difference is positive (0.059), it means that the P2O5 use increased after the training. c. Is the data distribution for the paired differences reasonably normal? (3 points) We can check this visually by creating a histogram of the differences:

hist(data$P205[1:10] - data$P205[11:20], main = "Paired Differences in P2O5") d. Was the mean of P 2 O 5 use significantly different before and after the training? (5 points) We can perform a paired t-test to test the null hypothesis that the mean difference is zero (i.e., there is no significant difference between P2O5 use before and after the training): t.test(data$P205[1:10], data$P205[11:20], paired = TRUE) Output : Paired t-test data: data$P205[1:10] and data$P205[11:20] t = 1.6703, df = 9, p-value = 0.1292 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: -0.02090465 0.13890465 sample estimates: mean of the differences 0.059 e. What do you conclude practically? As appropriate, report the means before and after, the mean difference, the effect size and interpretation, whether the difference is large relative to the values, anything notable on plots, and your practical conclusions. (10 points) Based on the results of the paired t-test, we can conclude that there is no significant difference in the mean P2O5 use before and after the training at a significance level of 0.05, since the p-value (0.1292) is greater than the significance level. Practically, this means that we do not have enough evidence to support the claim that the training has led to a significant increase or decrease in the mean P2O5 use. However, the mean difference of 0.059 indicates that there was a slight increase in P2O5 use after the training. The 95% confidence interval for

the mean difference (-0.02090465, 0.13890465) also includes zero, which further supports the lack of statistical significance.

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