Question

The following data are average fruit sizes (in grams) on plants that were grown either with or without phosphorus fertilizer over a year.

Fertilized<-c(4.96,6.05,10.93,7.93,3.63,6.2,11.82,4.3,10.16,9.05)
Unfertilized<-c(8.11,5.14,4.1,8.44,10.76,2.94,1.79,2.7,6.6,6.4)

#A. Is there a 'significant' difference between fertilized and unfertilized trees?

#B. What is the standardized effect size of these data

Answer 1

Solution:-

A)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 1.2934
DF = 18
t = [ (x₁ - x₂) - d ] / SE

t = 1.396

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 18 degrees of freedom is more extreme than -1.396; that is, less than -1.396 or greater than 1.396.

Thus, the P-value = 0.18

Interpret results. Since the P-value (0.18) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that there is significance difference between fertilized and unfertilized trees.

B) The standardized effect size of these data is 0.4413.

Effect size = 0.4413