To test whether the new variety of wheat differs in resistance to diseases compared to the old variety, we can set up a hypothesis test for proportions. Here we will compare the infection rate of the new variety to the known infection rate of 15% for the old variety.
Step 1: Define the Hypotheses
- Null Hypothesis (\(H_0\)): The proportion of infected plants for the new variety (\(p\)) is equal to the old variety's proportion, i.e., \(p = 0.15\).
- Alternative Hypothesis (\(H_a\)): The proportion of infected plants for the new variety (\(p\)) is different from the old variety's proportion, i.e., \(p \neq 0.15\).
Step 2: Collect and Summarize the Data
From the field experiment:
- Number of new plants, \(n = 120\)
- Number of infected new plants, \(x = 20\)
Calculate the sample proportion (\(\hat{p}\)) of infected plants in the new variety: \[ \hat{p} = \frac{x}{n} = \frac{20}{120} = 0.1667 \]
Step 3: Calculate the Test Statistic
This is a two-tailed test, and the test statistic for proportions is calculated using the formula: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1 - p_0)}{n}}} \] where:
- \(\hat{p}\) is the sample proportion,
- \(p_0\) is the population proportion under the null hypothesis (0.15),
- \(n\) is the sample size.
Now, plug in the values: \[ z = \frac{0.1667 - 0.15}{\sqrt{\frac{0.15(1 - 0.15)}{120}}} \] Calculating the standard error: \[ \sqrt{\frac{0.15 \times 0.85}{120}} = \sqrt{\frac{0.1275}{120}} \approx \sqrt{0.0010625} \approx 0.0326 \]
Now calculate the test statistic: \[ z = \frac{0.1667 - 0.15}{0.0326} \approx \frac{0.0167}{0.0326} \approx 0.512 \]
Step 4: Determine the Critical Value and Make a Decision
For a two-tailed test at the 1% significance level (\(\alpha = 0.01\)), we need to find the critical values from the standard normal distribution.
The critical z-values for a two-tailed test at \(\alpha = 0.01\) are approximately \(-2.576\) and \(2.576\).
Step 5: Conclusion
- If \(|z| > 2.576\), we reject the null hypothesis.
- If \(|z| \leq 2.576\), we do not reject the null hypothesis.
In our case: \[ z \approx 0.512 \] Since \(0.512\) does not exceed \(2.576\), we fail to reject the null hypothesis.
Final Decision
At the 1% level of significance, there is not enough evidence to conclude that the new variety of wheat differs in resistance to diseases compared to the old variety.