Problem 9 - Basic Statistics for Business and Economics with Student

Question

The Damon family owns a large grape vineyard in western New York along Lake Erie. The grapevines must be sprayed at the beginning of the growing season to protect against various insects and diseases. Two new insecticides have just been marketed: Pernod 5 and Action. To test their effectiveness, three long rows were selected and sprayed with Pernod \(5,\) and three others were sprayed with Action. When the grapes ripened, 400 of the vines treated with Pernod 5 were checked for infestation. Likewise, a sample of 400 vines sprayed with Action were checked. The results are: At the .05 significance level, can we conclude that there is a difference in the proportion of vines infested using Pernod 5 as opposed to Action?

Step-by-Step Solution

Answer

Yes, there is a significant difference in infestation rates between Pernod 5 and Action.

Step 1: State the Hypotheses

We need to determine if there is a significant difference between the infestation rates of vines treated with Pernod 5 and those treated with Action. The null hypothesis \(H_0\) asserts that there is no difference in the proportions of infestation, represented mathematically as \(p_1 = p_2\). The alternative hypothesis \(H_a\) claims that there is a difference \(p_1 eq p_2\), where \(p_1\) and \(p_2\) are the proportions of infested vines for Pernod 5 and Action, respectively.

Step 2: Collect and Organize Data

From the 400 vines treated with Pernod 5, 88 were found to be infested, so \(x_1 = 88\) and \(n_1 = 400\). From the 400 vines treated with Action, 63 were infested, meaning \(x_2 = 63\) and \(n_2 = 400\).

Step 3: Calculate Sample Proportions

The sample proportion for Pernod 5 is \(\hat{p}_1 = \frac{x_1}{n_1} = \frac{88}{400} = 0.22\). For Action, the sample proportion is \(\hat{p}_2 = \frac{x_2}{n_2} = \frac{63}{400} = 0.1575\).

Step 4: Compute the Standard Error

The standard error (SE) for the difference between two proportions is calculated as follows: \\[ SE = \sqrt{ \hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right) } \]Here, \(\hat{p}\) is the pooled proportion, given by \(\hat{p} = \frac{x_1 + x_2}{n_1 + n_2} = \frac{88 + 63}{800} = 0.17625\). Thus, \\[ SE = \sqrt{0.17625 \times (1 - 0.17625) \times \left( \frac{1}{400} + \frac{1}{400} \right)} \approx 0.0289 \]

Step 5: Calculate the Z-value

The Z-value is calculated as:\\[ Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} = \frac{0.22 - 0.1575}{0.0289} \approx 2.161 \]

Step 6: Decision Making

Using a significance level of \(\alpha = 0.05\), we compare the calculated Z-value to the critical Z-value. For a two-tailed test, the critical Z-value is approximately \(\pm 1.96\). Since \(2.161 > 1.96\), we reject the null hypothesis \(H_0\).

Step 7: Conclusion

There is statistically significant evidence at the 0.05 significance level to suggest that there is a difference in the proportion of vines infested between those sprayed with Pernod 5 and those sprayed with Action.

Key Concepts

Proportions

Proportions play a critical role in hypothesis testing for comparing groups. In this exercise, we look at two groups of grapevines, each treated with a different insecticide. The exercise requires us to calculate the proportion of vines that were infested in each group. This is done to compare the effectiveness of the two treatments. A proportion is essentially a part of the whole, calculated by dividing the part by the whole. Here, we divide the number of infested vines by the total number of vines to find the proportions for each group.

• Pernod 5 Group: The proportion of infested vines is calculated as \( \hat{p}_1 = \frac{88}{400} = 0.22 \).


• Action Group: Similarly, for the Action treated vines, the proportion is \( \hat{p}_2 = \frac{63}{400} = 0.1575 \).

These values tell us that 22% of the vines treated with Pernod 5 were infested compared to 15.75% for those with Action. Calculating these proportions helps establish a basis for comparative analysis, which is foundational in hypothesis testing.

Significance Level

The significance level, denoted as \( \alpha \), is a threshold set to determine the strength of the evidence needed to reject the null hypothesis. For this problem, the significance level is set at 0.05. The choice of 0.05 as the significance level is common in many scientific studies and gives a 5% risk of concluding that a difference exists when there is none (Type I error).

• Setting \( \alpha = 0.05 \) means that if the probability of observing our test statistic is less than 5% assuming the null hypothesis is true, we will reject the null hypothesis.


• In essence, it provides a cut-off point for decision-making, minimizing the chances of wrong inference.

Using this significance level, we compare our calculated test statistic (in this case, a Z-value) with the critical value from the statistical distribution (normal distribution for Z-tests). If our statistic falls beyond the significance threshold, it indicates sufficient evidence to support the alternative hypothesis.

Z-value

The Z-value, or Z-score, is a measure of how many standard deviations a data point is from the mean of your data set. In hypothesis testing, the Z-value helps to determine whether the result from comparing two proportions is statistically significant. For this exercise, the Z-value is calculated using the formula:\[ Z = \frac{\hat{p}_1 - \hat{p}_2}{SE} \]where \( \hat{p}_1 \) and \( \hat{p}_2 \) are the sample proportions, and SE is the standard error of the difference between two proportions.

• In this instance, the Z-value is computed as \( \frac{0.22 - 0.1575}{0.0289} \approx 2.161 \).


• This value tells us how much our observed difference deviates from what is expected under the null hypothesis of no difference.

For a significance level of 0.05 in a two-tailed test, a critical Z-value is approximately \( \pm 1.96 \). Since the calculated Z-value (2.161) is greater than 1.96, we reject the null hypothesis, suggesting a statistically significant difference between the two groups.