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Does Red Increase Men's Attraction to Women? Exercise 1.99 on page 44 described a study \(^{46}\) which examines the impact of the color red on how attractive men perceive women to be. In the study, men were randomly divided into two groups and were asked to rate the attractiveness of women on a scale of 1 (not at all attractive) to 9 (extremely attractive). Men in one group were shown pictures of women on a white background while the men in the other group were shown the same pictures of women on a red background. The results are shown in Table 6.14 and the data for both groups are reasonably symmetric with no outliers. To determine the possible effect size of the red background over the white, find and interpret a \(90 \%\) confidence interval for the difference in mean attractiveness rating.

Short Answer

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Detailed interpretation of the answer will depend on the calculated confidence interval, which we don't have due to the absence of numerical data. If the interval contains zero, it suggests the color of the background is not statistically significant in influencing men's ratings. If not, it means there's evidence to suggest that the color red does (positive difference) or does not (negative difference) increase men's attraction, with 90% confidence.

Step by step solution

01

Extract Key Data

First, extract key information from the provided data. This includes getting the sample sizes for both groups (n1 and n2), the mean attractiveness rating for each group (\(\bar{x}_1\) and \(\bar{x}_2\)), and the standard deviation for each group (s1 and s2).
02

Calculate the Standard Error

Calculated the standard error (SE) using the formula \(\mathrm{SE} = \sqrt{\frac{s1^2}{n1} + \frac{s2^2}{n2}}\). This represents the standard deviation of the sampling distribution of the differences between the means.
03

Calculate the t Score

Calculate the t-score using the formula \(t = \frac{(\bar{x}_1 - \bar{x}_2)}{SE}\). This informs us about the number of standard errors that the difference between our sample means is from zero.
04

Determine the Critical Value

Based on p% (here 90%) and the degrees of freedom (df = n1 + n2 - 2), locate the critical t value from the t-distribution table.
05

Formulate Confidence Interval

The confidence interval is calculated as \((\bar{x}_1 - \bar{x}_2) \pm (t_{\text{critical}} \times SE)\). The calculated interval gives the range in which we can be 90% sure that the true mean difference in attractiveness ratings lies.
06

Interpret the Confidence Interval

If the confidence interval contains zero, the difference in attractiveness ratings between the two backgrounds is not statistically significant at the 90% confidence level. If zero is not in the interval, we can claim with 90% confidence that the color red does (or does not, depending on the sign of the difference) increase men's attraction to women.

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Most popular questions from this chapter

In Exercises 6.32 and 6.33, find a \(95 \%\) confidence interval for the proportion two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Proportion of home team wins in soccer, using \(\hat{p}=0.583\) with \(n=120\)

Close Confidants and Social Networking Sites Exercise 6.93 introduces a study \(^{48}\) in which 2006 randomly selected US adults (age 18 or older) were asked to give the number of people in the last six months "with whom you discussed matters that are important to you." The average number of close confidants for the full sample was \(2.2 .\) In addition, the study asked participants whether or not they had a profile on a social networking site. For the 947 participants using a social networking site, the average number of close confidants was 2.5 with a standard deviation of 1.4 , and for the other 1059 participants who do not use a social networking site, the average was 1.9 with a standard deviation of \(1.3 .\) Find and interpret a \(90 \%\) confidence interval for the difference in means between the two groups.

In Exercises 6.203 and \(6.204,\) use Stat Key or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations. Difference in mean commuting time (in minutes) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=29.11,\) and \(s_{1}=20.72\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=21.97,\) and \(s_{2}=14.23\) for St. Louis

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=-2.6, s_{d}=4.1\) \(n_{d}=18\)

Examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DiNP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=10.1\) with \(s_{F}=38.9\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=7.0\) with \(s_{N}=22.8\)

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