To determine if the new variety of wheat differs in resistance to diseases compared to the old variety, we can use a hypothesis test for proportions.
Step 1: Set Up the Hypotheses
- Null hypothesis (\(H_0\)): The proportion of infected plants in the new variety is equal to the proportion in the old variety, \(p = 0.15\).
- Alternative hypothesis (\(H_a\)): The proportion of infected plants in the new variety is not equal to the proportion in the old variety, \(p \neq 0.15\).
Step 2: Collect the Data
In this case:
- Sample size (\(n\)) = 120 (the number of new plants)
- Number of infected plants in the new variety (\(x\)) = 20
- Sample proportion (\(\hat{p}\)) = \(x/n = 20/120 = 1/6 \approx 0.1667\)
Step 3: Calculate the Test Statistic
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Calculate the standard error (SE) of the sample proportion: \[ SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.15(1 - 0.15)}{120}} = \sqrt{\frac{0.15 \times 0.85}{120}} = \sqrt{\frac{0.1275}{120}} \approx \sqrt{0.0010625} \approx 0.0326 \]
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Calculate the z-test statistic: \[ z = \frac{\hat{p} - p}{SE} = \frac{0.1667 - 0.15}{0.0326} \approx \frac{0.0167}{0.0326} \approx 0.512 \]
Step 4: Determine the Critical Value
Since we are performing a two-tailed test at a significance level of \( \alpha = 0.01\), we look for critical values for \( z \) using a standard normal distribution.
The critical values for \( \alpha = 0.01 \) (two-tailed) correspond to:
- \( z_{0.005} \) and \( -z_{0.005} \), since \( \alpha/2 = 0.005 \).
The critical values are approximately: \[ z_{0.005} \approx 2.576 \quad \text{and} \quad -z_{0.005} \approx -2.576 \]
Step 5: Make a Decision
Compare the calculated test statistic \( z \) with the critical values:
- \( -2.576 < 0.512 < 2.576 \)
Since the test statistic does not fall in the critical region, we fail to reject the null hypothesis.
Conclusion
At the 1% significance level, there is not enough evidence to suggest that the new variety of wheat differs in resistance to diseases compared to the old variety. The critical value for this test is approximately \( \pm 2.576 \).