91Ó°ÊÓ

Use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal. Listed below are "attractiveness" ratings made by participants in a speed dating session. Each attribute rating is the sum of the ratings of five attributes (sincerity, intelligence, fun, ambition, shared interests). The listed ratings are from Data Set 18 "Speed Dating." Use a 0.05 significance level to test the claim that there is a difference between female attractiveness ratings and male attractiveness ratings. $$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline \text { Rating of Male by Female } & 4.0 & 8.0 & 7.0 & 7.0 & 6.0 & 8.0 & 6.0 & 4.0 & 2.0 & 5.0 & 9.5 & 7.0 \\ \hline \text { Rating of Female by Male } & 6.0 & 8.0 & 7.0 & 9.0 & 5.0 & 7.0 & 5.0 & 4.0 & 6.0 & 8.0 & 6.0 & 5.0 \\ \hline \end{array}$$

Step by step solution

01

Calculate the Differences

Given the ratings of males by females \[ [4.0, 8.0, 7.0, 7.0, 6.0, 8.0, 6.0, 4.0, 2.0, 5.0, 9.5, 7.0] \] and the ratings of females by males \[ [6.0, 8.0, 7.0, 9.0, 5.0, 7.0, 5.0, 4.0, 6.0, 8.0, 6.0, 5.0] \]. Calculate the differences (rating of male by female - rating of female by male) for each pair. The differences are: \[ [-2.0, 0.0, 0.0, -2.0, 1.0, 1.0, 1.0, 0.0, -4.0, -3.0, 3.5, 2.0] \]

02

Calculate the Mean and Standard Deviation of Differences

Calculate the mean (\bar{d}) and standard deviation (\text{s}d) of the differences. Mean of differences: \[ \bar{d} = \frac{-2.0 + 0.0 + 0.0 - 2.0 + 1.0 + 1.0 + 1.0 + 0.0 - 4.0 - 3.0 + 3.5 + 2.0}{12} = -0.125 \] Calculate the standard deviation of differences: \[ s_d \approx 2.216 \]

03

State the Hypotheses

State the null hypothesis (H_0) and the alternative hypothesis (H_1). \[ H_0: \mu_d = 0 \] (There is no difference between female and male attractiveness ratings) \[ H_1: \mu_d eq 0 \] (There is a difference between female and male attractiveness ratings)

04

Calculate the Test Statistic

Use the formula to find t: \[ t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}} \] Where \( \bar{d} = -0.125 \), \( \mu_d = 0 \), \( s_d \approx 2.216 \), and \( n = 12 \). \[ t = \frac{-0.125 - 0}{2.216 / \sqrt{12}} \approx -0.197 \]

05

Determine the Critical t-Value

Using a significance level of 0.05 with \( n-1 = 11 \) degrees of freedom, find the critical t-value from the t-distribution table, which is approximately \( \pm 2.201 \).

06

Make the Decision

Compare the calculated t-value with the critical t-value. If \(|t| < 2.201\), fail to reject \(H_0\). Here, \( |-0.197| < 2.201 \), so we fail to reject \( H_0 \).

07

Conclusion

Since we fail to reject the null hypothesis, we conclude that there is not enough evidence to support the claim that there is a difference between female and male attractiveness ratings.

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a crucial method used in statistics to determine if there is enough evidence to support a specific claim about a population parameter. In this exercise, we want to test if there’s a significant difference between female and male attractiveness ratings.
We start by formulating two hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)).

• \(H_0\) states there is no difference in ratings (\(\mu_d = 0\)).
• \(H_1\) states there is a difference in ratings (\(\mu_d eq 0\)).

After defining these hypotheses, we calculate a test statistic to determine which hypothesis is supported by the data. If the test statistic falls within a critical region, we reject the null hypothesis in favor of the alternative hypothesis.

Significance Level

The significance level, denoted as \(\alpha\), is the probability of rejecting the null hypothesis when it is actually true. It measures the risk of making a Type I error.
In this exercise, we use a significance level of 0.05, which means we are accepting a 5% chance of incorrectly rejecting the null hypothesis.

For example, with \(\alpha = 0.05\) in our problem, we compare our test statistic to the critical values from the t-distribution table corresponding to 95% confidence level. If our test statistic falls within these critical values, we do not reject the null hypothesis.

t-Distribution

The t-distribution is used in statistics when the sample size is small and/or the population standard deviation is unknown. It is similar to the normal distribution but has thicker tails, providing more allowance for variability in small samples.

In this problem, because we have a small sample size (n = 12), we use the t-distribution. We calculate our test statistic using the t-formula: \[ t = \frac{\bar{d} - \mu_d}{s_d / \sqrt{n}} \]
We then compare this calculated t-value to the critical t-value (found in t-tables) to determine whether to reject the null hypothesis.

Mean Differences

Calculating the mean difference is a key step in a paired sample t-test. Here, we subtract the rating of each female by a male from the rating of each male by a female.
These differences help us understand if there’s a consistent trend in one set of ratings being higher or lower than the other.

For example, the differences in our problem are: \[ [-2.0, 0.0, 0.0, -2.0, 1.0, 1.0, 1.0, 0.0, -4.0, -3.0, 3.5, 2.0] \]
We then compute the mean of these differences (\(\bar{d}\)), which in our case is -0.125. This mean difference helps in calculating the t-statistic for the hypothesis test.

Standard Deviation

The standard deviation (\(s_d\)) of the differences measures the amount of variability or spread in our data.
It is calculated based on the differences between each pair of ratings. A higher standard deviation indicates that the data points are spread out over a wider range of values.

In this exercise, the standard deviation of our differences is approximately 2.216. This value is used in the formula to calculate the t-statistic, which will help us decide whether to accept or reject the null hypothesis.
Understanding the standard deviation is crucial as it influences the width of confidence intervals and the test's sensitivity.