91Ó°ÊÓ

In parts of the eastern United States, whitctail deer are a major nuisance to farmers and homeowners, frequently damaging crops, gardens, and landscaping. A consumer organization arranges a test of two of the leading deer repellents \(\mathrm{A}\) and \(\mathrm{B}\) on the market. Fifty-six unfenced gardens in areas having high concentrations of deer are used for the test. Twenty-nine gardens are chosen at random to receive repellent \(\mathrm{A}\), and the other 27 receive repellent \(\mathrm{B}\). For each of the 56 gardens, the time elapsed between application of the repellent and the appearance in the garden of the first deer is recorded. For repellent \(\mathrm{A}\), the mean time is 101 hours. For repellent \(B\), the mean time is 92 hours. Assume that the two populations of elapsed times have normal distributions with population standard deviations of 15 and 10 hours, respectively. a. Let \(\mu_{1}\) and \(\mu_{2}\) be the population means of elapsed times for the two repellents, respectively. Find the point estimate of \(\mu_{1}-\mu_{2}\). b. Find a \(97 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) c. Test at a \(2 \%\) significance level whether the mean elapsed times for repellents \(A\) and \(B\) are different. Use both approaches, the critical-value and \(p\) -value, to perform this test.

Step by step solution

01

Calculate the point estimate

The point estimate for the difference between two population means, \(\mu_1-\mu_2\), is equal to the difference between the sample means, \(\bar{x}_1 - \(\bar{x}_2\). Given that the sample mean elapsed time for repellent A (\(\bar{x}_1\)) is 101 hours and for repellent B (\(\bar{x}_2\)) is 92 hours, we compute as follows: Point estimate = \(\bar{x}_1 - \(\bar{x}_2 = 101 - 92 = 9 hours.

02

Calculate the 97% confidence interval for the population mean difference

To calculate the confidence interval, we use the formula: \(\bar{x}_1 - \bar{x}_2 \pm z(\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2})^{0.5}\). Here, \(\sigma_1 = 15\), \(\sigma_2 = 10\), \(n_1 = 29\) and \(n_2 = 27\). Z is the z-score from the standard normal distribution for the 97% confidence level which is 2.33. Using these values in the formula gives the confidence interval as follows: 97% confidence interval = \(9 \pm 2.33((\frac{15^2}{29}+\frac{10^2}{27}){0.5}\))

03

Test Hypothesis at 2% Significance level

We test the null hypothesis that the population mean elapsed time for repellent A equals that for repellent B (that is, \(\mu_1-\mu_2 = 0\)) against the alternative hypothesis that they are not equal (\(\mu_1-\mu_2 ≠ 0\)). Adopting both the critical-value and p-value approach lets us test the claim using the formula: \(z=\frac{(\bar{X_1}-\bar{X_2})-(\mu_1-\mu_2) } { \sqrt{ \frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2} } }\). For critical value approach: if the computed z-score is more extreme than the critical z-score of 2 (given that the significance level, α, is 2%), we reject the null hypothesis. For p-value approach: we calculate the probability of obtaining a z-score as extreme as the one computed (the p-value). The p-value is then compared with the significance level, α = 0.02. If the p-value is less than α, we reject the null hypothesis in favour of the alternative hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Confidence Interval

A confidence interval gives you a range of values within which you can be fairly sure the true population parameter lies. In the context of our exercise, we're looking at the difference between the means of two groups.
To compute the confidence interval for the difference between two means, we use the formula: \[\bar{x}_1 - \bar{x}_2 \pm z\left(\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\right)\]Here,

• \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means for repellents A and B.
• \(\sigma_1\) and \(\sigma_2\) are the population standard deviations.
• \(n_1\) and \(n_2\) are the sample sizes.
• \(z\) is the z-score corresponding to the confidence level.

The confidence interval helps you understand not just the point estimate but the range, giving an estimate with a certain degree of confidence (97% in this case) that the true parameter difference is within this interval.

Population Mean

The population mean is the average of a set of values from the entire population. When we can't measure every item in the population, we use the sample mean to estimate the population mean.
In our exercise, the mean time for deer repellents A and B indicates how effective each repellent is in delaying a appearance in the gardens. The mean value is crucial for comparing such effectiveness.

Finding the sample mean involved simply calculating the average time over the gardens treated with each repellent:

• Repellent A's sample mean: 101 hours
• Repellent B's sample mean: 92 hours

By estimating population means with sample means, you can make informed decisions based on sample data.

Normal Distribution

A normal distribution, often called the bell curve, is a common distribution in statistics. It describes how data values are distributed in terms of their mean and standard deviation.
In our context, the elapsed time data for both repellents are assumed to be normally distributed. This assumption is key because it allows us to use z-scores and calculate confidence intervals and hypothesis tests.

• Data falling within one standard deviation of the mean covers about 68% of the data.
• Within two standard deviations, about 95% is included.
• About 99.7% falls within three standard deviations.

Using the normal distribution, you can make predictions about the probability of certain outcomes, critical in hypothesis testing and interval estimation.

Significance Level

The significance level in hypothesis testing measures how willing you are to accept a false positive. It is denoted by \(\alpha\) and often set at 0.05, 0.01, or in our case, 0.02 (2%).
This exercise tests whether the difference in mean times is significant. By setting a significance level, students determine the threshold to reject the null hypothesis (i.e., no difference in means).
For the hypothesis test:

• If the p-value is less than the significance level (\(\alpha = 0.02\)), reject the null hypothesis.
• If the computed z-score is more extreme than the critical z-score value, again reject the null hypothesis.

Choosing the right significance level is crucial. It influences your confidence in the results and any possible errors you may make in your conclusions.