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 "attribute" 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" in Appendix B. Use a 0.05 significance level to test the claim that there is a difference between female attribute ratings and male attribute ratings. $$\begin{array}{|l|l|l|l|l|l|l|l|l|l|} \hline \text { Rating of Male by Female } & 29 & 38 & 36 & 37 & 30 & 34 & 35 & 23 & 43 \\ \hline \text { Rating of Female by Male } & 36 & 34 & 34 & 33 & 31 & 17 & 31 & 30 & 42 \\ \hline \end{array}$$

Step by step solution

01

State the Hypotheses

Formulate the null hypothesis (H_0) and the alternative hypothesis (H_A). H_0: There is no difference between female attribute ratings and male attribute ratings. H_A: There is a difference between female attribute ratings and male attribute ratings.

02

Calculate the Differences

Compute the difference for each pair of ratings by subtracting the male rating from the female rating. List the differences: \[D = [29-36, 38-34, 36-34, 37-33, 30-31, 34-17, 35-31, 23-30, 43-42]\]

03

Calculate the Mean and Standard Deviation of Differences

Calculate the sample mean (\bar{D}) and the sample standard deviation (s_D) of the differences. \[\bar{D} = \frac{\text{Sum of differences}}{\text{Number of differences}} = \frac{\text{-7 + 4 + 2 + 4 - 1 + 17 + 4 - 7 + 1}}{9} = 1.89\] Calculate (s_D) using the standard deviation formula.

04

Conduct the t-Test

Use the t-test formula: \[t = \frac{\bar{D}}{s_D/\frac{ \text{sqrt}(n)}}\]where \bar{D} is the mean difference, (s_D) is the standard deviation of differences, and (n) is the number of pairs. Compare the calculated t-value against the critical t-value from the t-distribution table at \frac{\text{df}}{2} and 0.05 significance level.

05

Decision and Conclusion

Based on the t-test result, determine whether to reject or fail to reject the null hypothesis. If the t-value is greater than the critical value, reject the null hypothesis and conclude there is a significant difference between the ratings. If the t-value is less, fail 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.

null hypothesis

In any statistical test, the null hypothesis (\text{H}_0) is a statement that there is no effect or no difference. It is the assumption we start with and is usually what we aim to test against. For the paired sample t-test in this example, the null hypothesis states:
\( \text{H}_0 \text{: There is no difference between female attribute ratings and male attribute ratings.} \)
This hypothesis assumes that any observed difference in ratings is due to random chance and not a real, significant difference. It always features an equality sign (\( = \)).

alternative hypothesis

The alternative hypothesis (\text{H}_A) is what we are trying to prove. It is the statement we will accept if we have enough evidence to reject the null hypothesis. For the paired sample t-test, the alternative hypothesis is:
\( \text{H}_A \text{: There is a difference between female attribute ratings and male attribute ratings.} \)
This hypothesis suggests that the observed differences in ratings are real and not due to random chance. Notice how it does not include an equality sign, instead showing a difference (\( eq \)). This means we are conducting a two-tailed test where differences in both directions (positive and negative) are considered.

mean difference

The mean difference is a key part of the paired sample t-test. It is the average of the differences between paired observations. Here’s how you do it:
1. Calculate the difference for each pair:
\,\,\( \text{Differences} = [29-36, 38-34, 36-34, 37-33, 30-31, 34-17, 35-31, 23-30, 43-42] \)
2. Sum the differences:
\,\,\( \text{Sum of differences} = -7 + 4 + 2 + 4 - 1 + 17 + 4 - 7 + 1 = 17 \)
3. Divide by the number of pairs (n = 9):
\,\,\( \text{Mean Difference} (\bar{D}) = \frac{17}{9} \thickapprox 1.89 \).

standard deviation

The standard deviation (\text{s}_D) measures the variability of the differences from their mean:
1. Find each difference from the mean:
\,\,\( \text{Differences} - \bar{D} \)
2. Square each result:
\,\,\( (-7 - 1.89)^2, (4 - 1.89)^2,... \)
3. Sum these squared values:
\,\,\( = 228.72 \)
4. Divide the sum by (n-1) and take the square root:
\,\,\( s_D = \frac{228.72}{8} \approx 5.34 \).
The standard deviation tells us how spread out the differences are around the mean difference. Larger values indicate more variability.

t-test

The t-test checks if the mean difference is statistically significant. The formula is:
\,\( t = \frac{\bar{D}}{s_D / \text{sqrt}(n)} \).
For our example:
1. \,\,\( \bar{D} \thickapprox 1.89 \)
2. \,\,\( s_D \thickapprox 5.34 \)
3. Number of pairs, \( n = 9 \)
Input values:
\,\( t = \frac{1.89}{5.34 / \text{sqrt}(9)} \thickapprox 1.26 \).
Compare \( t \) with critical value from t-distribution table at 8 degrees of freedom and 0.05 significance level (two-tailed test). Reject \( \text{H}_0 \) if \( t \) > 2.306 or < -2.306.
Here \( t = 1.26 \), so we don’t reject \( \text{H}_0 \): evidence insufficient to claim significant difference between ratings.