A porester studving diameter growth of red pine belleves that the mean diameter growth will be different from the known theatment growth of 1.25in/ year if a fertilization treatment is applied to the stand. He condua his experiment collects data from 32 plots \& gets sample mean of 1.4 in per year The population standard deviation For this is .46 .
The Correct Answer and Explanation is :
This scenario describes a hypothesis test for comparing the mean diameter growth of red pine trees, in the presence of a fertilization treatment, to the known treatment growth of 1.25 inches per year. Let’s break down the process.
Step 1: Define the Hypotheses
In hypothesis testing, we first define the null hypothesis (H₀) and the alternative hypothesis (H₁). Here:
- H₀: μ = 1.25 (the mean diameter growth is equal to 1.25 inches per year).
- H₁: μ ≠ 1.25 (the mean diameter growth is different from 1.25 inches per year).
Step 2: Identify Known Values
We have the following:
- Population standard deviation (σ) = 0.46 inches
- Sample mean (x̄) = 1.4 inches
- Sample size (n) = 32 plots
- Hypothesized population mean (μ₀) = 1.25 inches
Step 3: Conduct the Z-Test
Since the population standard deviation is known, we can perform a Z-test for the mean. The formula for the Z-test statistic is:
[
Z = \frac{x̄ – μ₀}{\frac{σ}{\sqrt{n}}}
]
Plugging in the values:
[
Z = \frac{1.4 – 1.25}{\frac{0.46}{\sqrt{32}}}
]
Let’s calculate this.
Step 4: Determine the Critical Value and Make the Decision
The critical value for a two-tailed test at the 0.05 significance level is ±1.96. If the calculated Z-value is beyond this range, we reject the null hypothesis.
I’ll compute the Z-test statistic and make the final conclusion.
Let’s go ahead and calculate this now.
The calculated Z-test statistic is approximately 1.845.
Step 5: Compare the Z-Score with the Critical Value
The critical Z-value for a two-tailed test at a significance level of 0.05 is ±1.96. Since 1.845 is less than 1.96, it does not fall in the rejection region. Therefore, we fail to reject the null hypothesis.
Conclusion:
There is not enough evidence to conclude that the fertilization treatment has a significant effect on the mean diameter growth of red pine trees compared to the known treatment growth of 1.25 inches per year.
Explanation:
In hypothesis testing, the Z-test is used when the population standard deviation is known, and we are comparing a sample mean to a population mean. The Z-test statistic helps us determine how far our sample mean is from the population mean in terms of standard deviations. If the calculated Z-value is greater than the critical value (1.96), we reject the null hypothesis. In this case, because the Z-value of 1.845 is less than 1.96, we fail to reject the null hypothesis, meaning the fertilization treatment did not significantly change the growth rate.
Let me also generate the visualization for this.
Here is the visualization showing the Z-test diagram with the critical value regions shaded. The Z-value of 1.845 is within the non-rejection region, which confirms the decision to fail to reject the null hypothesis. Let me know if you need further clarification!
