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A researcher is investigating which of two fertilizers, F1 or F2, is more effect...

Jun 21, 2024

A researcher is investigating which of two fertilizers, F1 or F2, is more effective in improving the yield of oil palm. To compare the fertilizer, six plantations located in different states with different soil characteristics were used for testing. For each plantation, the experimental plot was then divided into two plots of equal area. Fertilizer F1 was randomly assigned to one plot in each field and F2 to another plot next to it. The yield of oil palm was measured, and the data are given in the following table. States Sarawak Perak Selangor Johor Sabah Melaka Fertilizer F1 135.2 83.6 45.0 71.7 68.1 97.4 Fertilizer F2 131.1 82.4 44.2 70.6 67.8 94.9 Conduct an appropriate statistical test based on the design chosen above to test if oil palm treated with Fertilizer 1 produces significantly different yield than Fertilizer 2. Use alpha 0.05. Show your calculations to receive full credits (Start from hypothesis conclusion)

Solution by Steps

step 1

Formulate the null hypothesis (

H_0

) and the alternative hypothesis (

H_a

)

$H_0$: There is no significant difference in the yield of oil palm between Fertilizer F1 and Fertilizer F2.

$H_a$: There is a significant difference in the yield of oil palm between Fertilizer F1 and Fertilizer F2.

step 2

Calculate the differences in yield for each state

\begin{align*} \text{Sarawak} & : 135.2 - 131.1 = 4.1 \\ \text{Perak} & : 83.6 - 82.4 = 1.2 \\ \text{Selangor} & : 45.0 - 44.2 = 0.8 \\ \text{Johor} & : 71.7 - 70.6 = 1.1 \\ \text{Sabah} & : 68.1 - 67.8 = 0.3 \\ \text{Melaka} & : 97.4 - 94.9 = 2.5 \\ \end{align*}

step 3

Calculate the mean (

\bar{d}

) and standard deviation (

s_d

) of the differences

\begin{align*} \bar{d} & = \frac{4.1 + 1.2 + 0.8 + 1.1 + 0.3 + 2.5}{6} = 1.6667 \\ s_d & = \sqrt{\frac{\sum (d_i - \bar{d})^2}{n-1}} \\ & = \sqrt{\frac{(4.1-1.6667)^2 + (1.2-1.6667)^2 + (0.8-1.6667)^2 + (1.1-1.6667)^2 + (0.3-1.6667)^2 + (2.5-1.6667)^2}{5}} \\ & = 1.346 \end{align*}

step 4

Calculate the t-statistic using the formula:

\begin{align*} t & = \frac{\bar{d}}{s_d / \sqrt{n}} \\ & = \frac{1.6667}{1.346 / \sqrt{6}} \\ & = 3.01 \end{align*}

step 5

Determine the degrees of freedom (

df

) and the critical t-value from the t-distribution table at

\alpha = 0.05

\begin{align*} df & = n - 1 = 6 - 1 = 5 \\ t_{\text{critical}} & = 2.571 \quad (\text{two-tailed test}) \end{align*}

step 6

Compare the calculated t-statistic with the critical t-value to make a decision

Since $3.01 > 2.571$, we reject the null hypothesis.

step 7

Conclude that there is a significant difference in the yield of oil palm between Fertilizer F1 and Fertilizer F2 at the 0.05 significance level

Answer

There is a significant difference in the yield of oil palm between Fertilizer F1 and Fertilizer F2 at the 0.05 significance level.

Key Concept

Paired t-test

Explanation

A paired t-test is used to compare the means of two related groups. In this case, the yields from the same plantations treated with two different fertilizers are compared.