91Ó°ÊÓ

The owner of a mosquito-infested fishing camp in Alaska wants to test the effectiveness of two rival brands of mosquito repellents, \(\mathrm{X}\) and \(\mathrm{Y}\). During the first month of the season, eight people are chosen at random from those guests who agree to take part in the experiment. For each of these guests, Brand \(\mathrm{X}\) is randomly applied to one arm and Brand \(\mathrm{Y}\) is applied to the other arm. These guests fish for 4 hours, then the owner counts the number of bites on each arm. The table below shows the number of bites on the arm with Brand \(X\) and those on the arm with Brand \(Y\) for each guest. $$ \begin{array}{l|rrrrrrrr} \hline \text { Guest } & \text { A } & \text { B } & \text { C } & \text { D } & \text { E } & \text { F } & \text { G } & \text { H } \\ \hline \text { Brand X } & 12 & 23 & 18 & 36 & 8 & 27 & 22 & 32 \\ \hline \text { Brand Y } & 9 & 20 & 21 & 27 & 6 & 18 & 15 & 25 \\ \hline \end{array} $$ a. Construct a \(95 \%\) confidence interval for the mean \(\mu_{d}\) of population paired differences, where a paired difference is defined as the number of bites on the arm with Brand \(X\) minus the number of bites on the arm with Brand \(\mathrm{Y}\). b. Test at the \(5 \%\) significance level whether the mean number of bites on the \(\operatorname{arm}\) with Brand \(\mathrm{X}\) and the mean number of bites on the arm with Brand \(\mathrm{Y}\) are different for all such guests.

Step by step solution

01

Calculate Paired Differences and Their Mean

Subtract the values in row Y from corresponding values in row X to obtain the paired differences. Then calculate the mean (\(\bar{d}\)) of these differences.

02

Calculate Standard Deviation of Differences

Calculate the standard deviation (s) of the differences.

03

Construct 95% Confidence Interval for Mean Difference

The 95% confidence interval for the mean difference is given by \(\bar{d} \pm t.s/\sqrt{n}\), where t is the value from the t-distribution table with n-1 degrees of freedom (in this case 7) that corresponds to the 95% confidence interval, s is the standard deviation of the differences, and n is the number of pairs.

04

Setup Hypotheses for Testing

Setup the null hypothesis (H0) as mean difference being 0, i.e., no difference between the brands. The alternative hypothesis (Ha) would be mean difference not equal to zero, i.e., the brands are different.

05

Select Significance Level and Calculate Test Statistic

The significance level is given as 5%. The test statistic (t) is calculated as \(t = \bar{d}/(s/\sqrt{n})\), where \(\bar{d}\) is the mean difference, s is the standard deviation of the differences, and n is the number of pairs.

06

Decision Making

Compare the computed test statistic with the critical values from t-distribution table. If the test statistic falls in the rejection region, then reject the null hypothesis, otherwise do not reject it. To check if the test statistic falls in the rejection region, also calculate the p-value. If the p-value is less than the significance level (5% in this case), reject the null hypothesis.

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Key Concepts

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

Confidence Interval

When conducting a paired t-test, a 95% confidence interval is used to estimate the range within which the true mean difference between two related groups, in this case, mosquito bites with Brand X versus Brand Y, lies. A confidence interval provides a set of values that we believe, with a specified probability (95% here), contains the true mean difference of the population. This allows us to make informed guesses about the mean difference between these sample groups.

To calculate the 95% confidence interval, the formula \[ \bar{d} \pm t \cdot \frac{s}{\sqrt{n}} \] is used, where:

• \(\bar{d}\) is the mean of the paired differences.
• \(t\) is a value from the t-distribution table corresponding to a 95% confidence level and \(n-1\) degrees of freedom (where \(n\) is the number of pairs).
• \(s/\sqrt{n}\) is the standard error of the mean difference.

Given this interval, we assess how much the data might support the actual mean difference between the two brands.

Mean Difference

The mean difference, in the context of a paired t-test, is the average of the differences between each pair of observations. Here, it involves subtracting the number of bites when using Brand Y from those when using Brand X for each guest. This mean difference helps in understanding whether there is a consistent disparity in effectiveness between the two brands.

To calculate it:

• Compute the difference for each pair (e.g., Guest A, B, etc.) by subtracting the bites with Brand Y from those with Brand X.
• Sum these differences and then divide by the number of guests (pairs) to get the mean difference \(\bar{d}\).

This measure indicates, on average, how much one brand outperformed the other in terms of mosquito bites.

T-Distribution

The t-distribution is a probability distribution widely used in statistics to estimate population parameters when the sample size is small. In this exercise, the t-distribution helps to determine the critical values necessary to calculate the confidence interval and to conduct hypothesis testing.

The t-distribution is characterized by its degrees of freedom, which for a paired t-test is typically \(n-1\) where \(n\) is the number of paired observations. Since we are using a table or software to get the \(t\)-value for the confidence interval and test statistics, understanding the shape and properties of the t-distribution is essential.

• As sample size increases, the t-distribution approaches the standard normal distribution.
• It accounts for extra variability introduced when using sample data to make inferences about populations.

This distribution is crucial for determining how probable or rare a sample result is under the null hypothesis assumption.

Null Hypothesis

In hypothesis testing, the null hypothesis serves as a statement or assumption that there is no effect or difference. For the paired t-test regarding mosquito bites, the null hypothesis \((H_0)\) states that there is no mean difference in the number of bites between Brand X and Brand Y.

The null hypothesis is assumed to be true until evidence suggests otherwise. The goal of hypothesis testing is to ascertain whether the observed data provide sufficient evidence to reject the null hypothesis. In our case, it would mean proving that there is indeed a difference in effectiveness between the two repellent brands.

• If the calculated test statistic falls into the rejection region based on the significance level, or if the p-value is small enough, we reject \(H_0\).
• The alternative hypothesis \((H_a)\) posits that there is a mean difference — in this instance, suggesting Brands X and Y are differently effective against mosquito bites.

Understanding the null hypothesis forms the statistical basis for inference and decision-making.

P-Value

In a paired t-test, the p-value helps in determining the statistical significance of the observed mean difference. It represents the probability of obtaining a test statistic as extreme as, or more extreme than, the observed result, assuming the null hypothesis is true. A small p-value indicates that the observed data are unlikely under the null hypothesis.

The threshold for the p-value, known as the significance level, is typically set at 5% (\(0.05\)) for this test. This means:

• If the p-value is less than or equal to 0.05, there is strong evidence against the null hypothesis, leading to its rejection.
• If the p-value is greater than 0.05, the evidence is not strong enough, and we fail to reject the null hypothesis.

Using the p-value helps provide an objective measure for deciding whether the difference observed is statistically significant in hypothesis testing.