91Ó°ÊÓ

A popular theory is that presidential candidates have an advantage if they are taller than their main opponents. Listed are heights \((\mathrm{cm})\) of presidents along with the heights of their main opponents (from Data Set 15 "Presidents"). a. Use the sample data with a \(0.05\) significance level to test the claim that for the population of heights of presidents and their main opponents, the differences have a mean greater than \(0 \mathrm{~cm}\). b. Construct the confidence interval that could be used for the hypothesis test described in part (a). What feature of the confidence interval leads to the same conclusion reached in part (a)? $$ \begin{array}{|l|c|c|c|c|c|c|} \hline \text { Height (cm) of President } & 185 & 178 & 175 & 183 & 193 & 173 \\\ \hline \text { Height (cm) of Main Opponent } & 171 & 180 & 173 & 175 & 188 & 178 \\ \hline \end{array} $$

Step by step solution

01

- Calculate Differences

First, calculate the differences between the heights of the presidents and their main opponents: 185 - 171 = 14,178 - 180 = -2,175 - 173 = 2,183 - 175 = 8,193 - 188 = 5,173 - 178 = -5.

02

- Find Sample Mean and Standard Deviation

Calculate the mean and standard deviation of the differences. The differences are as follows: 14, -2, 2, 8, 5, -5. The mean is given by \(\bar{x} = \frac{14 + (-2) + 2 + 8 + 5 + (-5)}{6} = \frac{22}{6} \approx 3.67 \). To find the standard deviation, use the formula for sample standard deviation.

03

- Perform the Hypothesis Test

The null hypothesis is that the mean difference is equal to 0, \(H_0: \mu = 0\). The alternative hypothesis is that the mean difference is greater than 0, \(H_1: \mu > 0\). Use a t-test to compare the sample mean to the hypothesized population mean of 0. Calculate the t-score using \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( s \) is the sample standard deviation, and \(n\) is the sample size.

04

- Find the Critical Value

For a significance level of 0.05 and degrees of freedom \(df = n - 1 = 6 - 1 = 5\), find the critical t-value from a t-distribution table, which is approximately 2.015.

05

- Make a Decision

Compare the calculated t-score with the critical t-value. If the t-score is greater than the critical t-value, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.

06

- Construct the Confidence Interval

Construct a 95% confidence interval for the mean difference. The interval is given by \(\bar{x} \pm t_{crit} \cdot \frac{s}{\sqrt{n}}\), where \( t_{crit} \) is the critical value.

07

- Interpret the Confidence Interval

If the confidence interval does not contain 0, it supports rejecting the null hypothesis. Otherwise, it does not provide evidence to reject the null hypothesis.

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Key Concepts

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

t-test

A t-test is a statistical test used to compare the means of two groups. In this example, we want to determine if the means of presidents' heights and their main opponents' heights are significantly different. We use the formula for the t-score: The t-test helps us figure out if the observed data could have occurred by chance or if there is a true difference. If the t-score we calculate is greater than the critical value from the t-distribution table, we reject the null hypothesis. This means we have enough evidence to claim there is a significant difference between the two groups.

confidence interval

A confidence interval is a range of values that is likely to contain the true population mean with a certain level of confidence. In our exercise, we use a 95% confidence interval to estimate the mean difference in heights between presidents and their opponents. This is done using the formula: For our data, the confidence interval helps us determine if the mean difference is significantly different from 0. If the interval does not include 0, it suggests that the difference is statistically significant. This matches our hypothesis test result, where we concluded the mean difference was greater than 0. Thus, the confidence interval further supports our hypothesis test conclusion.

sample mean

The sample mean is the average value calculated from the sample data. In our case, we calculated the sample mean for the differences in heights: The sample mean provides us with an estimate of the average height difference between presidents and their opponents. We use this mean in the t-test and confidence interval calculations to draw conclusions about the population mean.

null hypothesis

The null hypothesis is the initial assumption that there is no effect or no difference between groups. In our example, the null hypothesis (H_0) states that the mean difference in height between presidents and their main opponents is 0: We conduct the t-test to see if the data provide sufficient evidence to reject this null hypothesis. If the t-score exceeds the critical t-value, we reject the null hypothesis, suggesting there is a significant difference in heights.

significance level

The significance level, denoted by alpha (α), is the threshold for determining whether a result is statistically significant. In our problem, we use a 0.05 significance level. This means we accept a 5% chance of concluding there is a difference when there is not. If our t-test results in a probability (p-value) less than 0.05, we reject the null hypothesis. This indicates that the observed differences are unlikely to have occurred by chance alone. In our example, the t-test showed a result significant enough to reject the null hypothesis, meaning the height differences are likely real and not due to random variation.