A farmer is concerned that changing to an organic fertilizer might affect his crop yield. He subdivides six lots and uses the old fertilizer on one half of each lot and the new fertilizer on the other half. The data below shows the crop yields with the old and new fertilizers. Let the difference be defined as Old – New.

1. Specify the competing hypotheses to determine whether there is any difference between the average crop yields from the different fertilizers.

2. Assuming that differences in crop yields are normally distributed, calculate the value of the test statistic.

Old: 9, 10, 9, 9, 13, 10
New: 12, 9, 12, 10, 10, 13

Null Hypothesis (H₀): There is no significant difference in average crop yields between the old and new fertilizers, meaning the mean difference (3BC_d) is zero.
[tex]H₀: \mu_{d} = 0[/tex]
Alternative Hypothesis (Hₐ): There is a significant difference in the average crop yields between the old and new fertilizers, meaning the mean difference is not zero.
[tex]Hₐ: \mu_{d} \neq 0[/tex]

Step 2: Calculate the Differences

The differences in yields for each lot (Old - New) are:

Lot 1: 9 - 12 = -3
Lot 2: 10 - 9 = 1
Lot 3: 9 - 12 = -3
Lot 4: 9 - 10 = -1
Lot 5: 13 - 10 = 3
Lot 6: 10 - 13 = -3

So, the differences are: -3, 1, -3, -1, 3, -3.

Step 3: Calculate the Mean and Standard Deviation of Differences

Mean difference ([tex]\bar{d}[/tex]):

[tex]\bar{d} = \frac{-3 + 1 - 3 - 1 + 3 - 3}{6} = \frac{-6}{6} = -1[/tex]

Standard deviation (s_d) of differences can be calculated using:

[tex]s_d = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (d_i - \bar{d})^2}[/tex]

Calculating the squared deviations:

For -3: [tex](-3 + 1)^2 = 4[/tex]
For 1: [tex](1 + 1)^2 = 4[/tex]
For -3: [tex](-3 + 1)^2 = 4[/tex]
For -1: [tex](-1 + 1)^2 = 0[/tex]
For 3: [tex](3 + 1)^2 = 16[/tex]
For -3: [tex](-3 + 1)^2 = 4[/tex]

Sum of squared deviations = 4 + 4 + 4 + 0 + 16 + 4 = 32

Standard deviation (s_d):

[tex]s_d = \sqrt{\frac{32}{5}} \approx 2.5298[/tex]

Step 4: Calculate the Test Statistic

The t-test statistic is calculated using:

[tex]t = \frac{\bar{d} - 0}{s_d/\sqrt{n}}[/tex]

where [tex]n[/tex] is the number of differences, which is 6. Now substitute:

[tex]t = \frac{-1 - 0}{2.5298/\sqrt{6}} \approx \frac{-1}{1.033} \approx -0.9688[/tex]

Conclusion

You can compare this test statistic with critical t value from the t distribution table at a chosen significance level (commonly 0.05) with degrees of freedom [tex]n-1[/tex] to decide whether to reject the null hypothesis or not. If [tex]|t|[/tex] is greater than the critical value, then there is a significant difference in the crop yields. If not, you fail to reject the null hypothesis.

The calculated t-statistic helps to determine if the change to an organic fertilizer significantly affects crop yield, beyond just random chance.