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Use concepts on hypothesis testing and sampling distribution for one and two means.

The image describes a hypothesis testing problem where a botanist wants to determine if an experimental fertilizer has significantly increased plant height. Let's break down the problem step-by-step, using hypothesis testing for one mean:

### Problem Summary:
- **Claim**: The botanist wants to test if the mean increase in plant height is different from 15 inches.
- **Sample Size (n)**: 12 plants.
- **Sample Mean (\(\bar{x}\))**: 14.33 inches.
- **Sample Standard Deviation (s)**: 1.37 inches.
- **Population Mean (μ)**: 15 inches (this is the mean increase expected without the fertilizer, i.e., the null hypothesis value).
  
### Step-by-Step Hypothesis Test:
1. **State the Null and Alternative Hypotheses**:
   - **Null Hypothesis (\(H_0\))**: \( \mu = 15 \) (the mean increase is 15 inches, no significant effect from the fertilizer).
   - **Alternative Hypothesis (\(H_a\))**: \( \mu \neq 15 \) (the mean increase is different from 15 inches, indicating a possible effect from the fertilizer).

2. **Select a Significance Level (α)**:
   - Typically, \( \alpha = 0.05 \) is chosen unless stated otherwise.

3. **Test Statistic**:
   Since the sample size is small (n = 12), we use a **t-test**:
   \[
   t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}
   \]
   Where:
   - \( \bar{x} = 14.33 \) (sample mean),
   - \( \mu = 15 \) (hypothesized population mean),
   - \( s = 1.37 \) (sample standard deviation),
   - \( n = 12 \) (sample size).

   Plugging in the values:
   \[
   t = \frac{14.33 - 15}{\frac{1.37}{\sqrt{12}}}
   \]
   \[
   t \approx \frac{-0.67}{0.3954} \approx -1.69
   \]

4. **Degrees of Freedom (df)**:
   For a t-test, degrees of freedom are \( df = n - 1 = 12 - 1 = 11 \).

5. **P-value or Critical t-value**:
   Using a t-distribution table (for \( df = 11 \) and a two-tailed test at \( \alpha = 0.05 \)):
   - The critical t-value is approximately ±2.201.

6. **Conclusion**:
   - The calculated \( t \)-value is -1.69, which does **not** exceed the critical value of ±2.201.
   - Thus, we **fail to reject the null hypothesis** at the 5% significance level.
   - This means that the data does not provide strong enough evidence to conclude that the fertilizer had a significant effect on plant height.

### Final Conclusion:
The botanist **cannot conclude** that the fertilizer was effective in increasing plant height, as the sample data does not show a statistically significant difference from the expected 15 inches.

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Would you like more details or explanations on any step? 

Here are some related questions to expand your understanding:
1. What is the purpose of using a t-test instead of a z-test in this scenario?
2. How does changing the sample size affect the result of a hypothesis test?
3. What would happen if the significance level was set to 0.01 instead of 0.05?
4. How do you interpret a P-value, and how is it different from critical values?
5. What assumptions must be met for this t-test to be valid?

**Tip**: In hypothesis testing, always ensure the sample size is large enough or normally distributed for accurate t-test results.

A botanist wants to know if the mean increase in plant height after use of an experimental fertilizer is equal to 15 inches. She collected a random sample of 12 plants and noted that the mean increase in plant height was 14.33 ± 1.37 inches. Can she conclude that the fertilizer was effective?

This problem involves testing if the mean increase in plant height with a fertilizer is significantly different from 15 inches. Using a t-test with a sample of 12 plants, the botanist finds a sample mean of 14.33 inches and a sample standard deviation of 1.37 inches.

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