91Ó°ÊÓ

Three different pesticides can be used to control infestation of grapes. It is suspected that pesticide 3 is more effective than the other two. In a particular vineyard, three different plantings of pinot noir grapes are selected for study. The following results on yield are obtained: If \(\mu_{i}\) is the true mean yield after treatment with the \(i\) th pesticide, we are interested in the quantity $$\mu=\frac{1}{2}\left(\mu_{1}+\mu_{2}\right)-\mu_{3}$$ which measures the difference in mean yields between pesticides 1 and 2 and pesticide 3 . If the sample sizes \(n_{i}\) are large, the estimator (say, \(\hat{\mu}\) ) obtained by replacing each individual \(\mu_{i}\) by \(\bar{X}_{i}\) is approximately normal. (a) Find an approximate \(100(1-\alpha) \%\) large-sample confidence interval for \(\mu\). (b) Do these data support the claim that pesticide 3 is more effective than the other two? Use \(\alpha=0.05\) in determining your answer.

Step by step solution

01

Understand the problem statement

We are given the task to analyze the effectiveness of three pesticides on grape yields. The measure of interest \( \mu \) is defined as \( \frac{1}{2}(\mu_{1}+\mu_{2}) - \mu_{3} \). Our aim is to find an approximate large-sample confidence interval for \( \mu \) and determine if pesticide 3 is more effective at the \( \alpha = 0.05 \) significance level.

02

Formula for Confidence Interval of \( \mu \)

The estimator \( \hat{\mu} = \frac{1}{2} (\bar{X}_1 + \bar{X}_2) - \bar{X}_3 \) is approximately normal with mean \( \mu \) and variance \( \sigma^2 = \frac{1}{4} \frac{\sigma_1^2}{n_1} + \frac{1}{4} \frac{\sigma_2^2}{n_2} + \frac{\sigma_3^2}{n_3} \). The normal distribution is used to create the confidence interval.

03

Compute the Confidence Interval

The large-sample \( 100(1-\alpha)\% \) confidence interval for \( \mu \) is given by \( \hat{\mu} \pm Z_{1-\alpha/2} \sqrt{\sigma^2} \), where \( Z_{1-\alpha/2} \) is the corresponding quantile from the standard normal distribution.

04

Test the Effectiveness Hypothesis

To test if pesticide 3 is more effective, we check if 0 is within the confidence interval. If the entire interval for \( \mu \) is greater than 0, pesticide 3 yields significantly less, supporting greater effectiveness. If the interval includes 0, it's inconclusive.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pesticide Effectiveness

Pesticide effectiveness tests are crucial in agriculture to ensure the healthiest crops and best possible yields. In the given exercise, we are evaluating three pesticides to see which one leads to higher grape yields, focusing specifically on whether pesticide 3 is more effective than pesticides 1 and 2. We look at the average yield after applying each pesticide and calculate a value, \( \mu \), that represents the yield difference between the average effects of pesticides 1 and 2 versus pesticide 3. This gives us insight into whether pesticide 3 performs better in controlling grape infestations. The hypothesis is that if \( \mu \) is significantly greater than zero, pesticide 3 is more effective.

Large Sample Theory

Large sample theory applies statistical principles to large data sets to approximate results and make inferences. In this exercise, with pesticide data collected from a large vineyard, large sample theory helps us comfortably assume that the sample mean \( \bar{X}_i \) for each pesticide's yield is an accurate estimator of the true mean \( \mu_i \). Thanks to large sample sizes \( n_i \), we assume that the estimator \( \hat{\mu} \), representing our measure \( \mu \), follows a normal distribution. This assumption is grounded in the central limit theorem, which states that with a large enough sample size, the sampling distribution of the sample mean will become approximately normally distributed.

Hypothesis Testing

Hypothesis testing is a statistical method used to decide whether to support or reject a certain hypothesis based on sample data. In our scenario, we perform hypothesis testing to examine the claim that pesticide 3 is more effective than the other two.
The null hypothesis (\( H_0 \)) states that there is no difference in effectiveness, meaning \( \mu \leq 0 \). The alternative hypothesis (\( H_a \)) suggests that \( \mu > 0 \), implying greater effectiveness for pesticide 3. A significance level of \( \alpha = 0.05 \) helps determine the threshold for rejecting the null hypothesis. If the 95% confidence interval for \( \mu \) does not include 0 and lies entirely above it, we reject \( H_0 \) in favor of \( H_a \), concluding that pesticide 3 is indeed more effective.

Normal Distribution

Normal distribution is a critical concept in statistics, frequently used to model real-world random variables that cluster around a mean. In this exercise, we leverage the properties of normal distribution to calculate confidence intervals. By assessing the approximate normality of our estimator \( \hat{\mu} \), we use it to deduce information regarding the population parameter \( \mu \).
The standard normal distribution provides us with quantiles like \( Z_{1-\alpha/2} \), which are essential for constructing our confidence intervals. This allows us to infer range estimates about \( \mu \), helping to evaluate the effectiveness of the pesticides with a known level of certainty. Understanding normal distribution is key to interpreting and making predictions from the data in a statistically valid way.