A farming company wants to know if a new fertilizer has improved crop yield or not. Historic data shows the average yield of the farm is 20 tonne per acre. They decide to test a new organic fertilizer on a smaller sample of farms and observe the new yield is 20.175 tonne per acre with a standard deviation of 3.02 tonne for 12 different farms. Did the new fertilizer work?

To determine if the new fertilizer has statistically improved crop yield, we can perform a one-sample \( t \)-test comparing the sample mean yield to the historical average yield.

### Hypothesis Testing:

1. **Null Hypothesis (H₀):**
   - The new fertilizer does not improve crop yield, so the mean yield is the same as the historical average.
   - \( H_0: \mu = 20 \) tonne per acre.

2. **Alternative Hypothesis (H₁):**
   - The new fertilizer improves crop yield, so the mean yield is greater than the historical average.
   - \( H_1: \mu > 20 \) tonne per acre (one-tailed test).

3. **Given Data:**
   - Population mean (historical average), \( \mu = 20 \) tonne per acre.
   - Sample mean, \( \bar{x} = 20.175 \) tonne per acre.
   - Sample standard deviation, \( s = 3.02 \) tonne per acre.
   - Sample size, \( n = 12 \).

4. **Test Statistic:**
   The formula for the \( t \)-test statistic is:

   \[
   t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
   \]

   Plugging in the values:

   \[
   t = \frac{20.175 - 20}{\frac{3.02}{\sqrt{12}}}
   \]

   Let's calculate the \( t \)-value.

### Step-by-Step Calculation:

1. **Calculate the difference:**
   \[
   20.175 - 20 = 0.175
   \]

2. **Calculate the standard error:**
   \[
   \frac{3.02}{\sqrt{12}} = \frac{3.02}{3.464} \approx 0.872
   \]

3. **Calculate the \( t \)-statistic:**
   \[
   t = \frac{0.175}{0.872} \approx 0.201
   \]

The calculated \( t \)-statistic is approximately 0.201.

5. **Degrees of Freedom:**
   \[
   df = n - 1 = 12 - 1 = 11
   \]

6. **Find the Critical Value or P-value:**
   - For a one-tailed test at a significance level of 0.05 and 11 degrees of freedom, the critical \( t \)-value is approximately 1.796 (from \( t \)-distribution tables).
   - Since the calculated \( t \)-value (0.201) is much less than 1.796, we fail to reject the null hypothesis.

### Conclusion:
The \( t \)-statistic is not significant at the 0.05 level. Therefore, **we do not have enough evidence to conclude that the new fertilizer has improved crop yield**.

Would you like any further details or clarifications?

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### Follow-up Questions:

1. Would you like to conduct a two-tailed test instead of a one-tailed test?
2. Is there a specific significance level you want to use (e.g., 0.01 instead of 0.05)?
3. Do you have additional data for other variables that might affect yield?
4. Would you like to visualize the results with a plot or graph?
5. Do you need to calculate the power of the test to ensure it is adequately sensitive?

### Tip:
When testing for improvements, ensure that the sample size is large enough to detect even small changes, especially if the standard deviation is relatively large compared to the mean difference.

Explore statistical hypothesis testing to determine if a new organic fertilizer has improved crop yield. Includes a detailed one-sample t-test calculation comparing sample yield data with historical averages. Key data points: historical average yield of 20 tonne per acre, sample mean yield of 20.175 tonne per acre with a standard deviation of 3.02 tonne. Suitable for advanced college-level students and professionals.

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