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The paper "The Truth About Lying in Online Dating Profiles" (Proceedings, Computer-Human Interactions [2007]\(: 1-4)\) describes an investigation in which 40 men and 40 women with online dating profiles agreed to participate in a study. Each participant's height (in inches) was measured and the actual height was compared to the height given in that person's online profile. The differences between the online profile height and the actual height (profile - actual) were used to compute the values in the accompanying table. $$ \begin{array}{ll} \text { Men } & \text { Women } \\ \hline \bar{x}_{d}=0.57 & \bar{x}_{d}=0.03 \\ s_{d}=0.81 & s_{d}=0.75 \\ n=40 & n=40 \end{array} $$ For purposes of this exercise, assume it is reasonable to regard the two samples in this study as being representative of male online daters and female online daters. (Although the authors of the paper believed that their samples were representative of these populations, participants were volunteers recruited through newspaper advertisements, so we should be a bit hesitant to generalize results to all online daters!) a. Use the paired \(t\) test to determine if there is convincing evidence that, on average, male online daters overstate their height in online dating profiles. Use \(\alpha=.05\) b. Construct and interpret a \(95 \%\) confidence interval for the difference between the mean online dating profile height and mean actual height for female online daters. c. Use the two-sample \(t\) test of Section 11.1 to test \(H_{0}: \mu_{m}-\mu_{f}=0\) versus \(H_{a}: \mu_{m}-\mu_{f}>0,\) where \(\mu_{m}\) is the mean height difference (profile - actual) for male online daters and \(\mu_{f}\) is the mean height difference (profile - actual) for female online daters. d. Explain why a paired \(t\) test was used in Part (a) but a two-sample \(t\) test was used in Part (c).

Short Answer

Expert verified
a) Completing the paired t-test for Male online daters might show that on average, they tend to overstate their height. b) The 95% confidence interval for the difference in height in female online daters profiles will give a range where the actual population mean difference is likely to lie. c) The two-sample t-test might show that there is a significant difference between the mean height difference (profile - actual) for male and female online daters. d) The selection of the paired t-test and the two-sample t-test is due to the nature of the data used. The paired t-test is used with paired measurements taken from the same group, while the two-sample t-test is used to compare means of two independent groups.

Step by step solution

01

Paired t-test for Male Online Daters

The null hypothesis is that the mean difference between the profile and actual height for male daters is zero and the alternate hypothesis is that it is greater than zero. A paired t-test is a statistical procedure to determine if there's a significant difference between the means of two paired sets of data. Here, the sets of data are the actual heights and profile heights of the males. Using the provided values, the t statistic can be calculated using the formula \((\bar{x}_{d} - \mu)/(s_d/\sqrt{n})\), where \(\bar{x}_{d}\) is the mean difference, \(\mu\) is the expected mean difference (0), \(s_d\) is the standard deviation and \(n\) is the sample size. If the calculated t stat is greater than the critical t value at \(\alpha = 0.05\) for \(df = n - 1 = 39\) degrees of freedom, the null hypothesis is rejected.
02

Confidence Interval for Female Online Daters

The 95% confidence interval for the difference between the profile and actual height for female daters can be calculated by \(\bar{x}_{d} \pm (t_{critical} * s_d/\sqrt{n})\), where \(t_{critical}\) is the t value from the t-distribution table with \(df = n - 1\) corresponding to the desired confidence level (95%). This interval gives a range of values for the mean difference, within which the population mean difference is expected to lie with 95% confidence.
03

Two-Sample t-test for Difference in Mean Height

A two-sample t-test is used to check if there's a difference in means between two independent populations. Here, we want to test if there's a significant difference between the mean heights of the male and female online daters. The null hypothesis is that the mean height difference (profile - actual) is the same for both sexes, and the alternative hypothesis is that males have a greater difference. The test statistic can be calculated by using the formula \( t = (\bar{x}_{m} - \bar{x}_{f}) / \sqrt{((s_{m}^2/n_m) + (s_{f}^2/n_f))}\). If the calculated t stat is greater than the critical t value at \(\alpha = 0.05\) for \(df = n_m + n_f - 2\), the null hypothesis is rejected.
04

Explanation of Test Selection

A paired t-test is conducted in part a) because it's checking the difference between actual and profile heights within the same group of individuals (males), hence they are 'paired' measurements. In part c), a two-sample t-test is suitable because it's comparing the mean differences between two independent groups (males and females), hence the measurements are not paired.

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Key Concepts

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

Confidence Interval
A confidence interval is a range of values used to estimate an unknown population parameter. It's an interval calculated from your sample data that believes the true parameter value falls within this range, with a specified level of confidence. In the exercise, we constructed a 95% confidence interval for the difference between the actual and profile height of female online daters.

This means that we're 95% confident that the true mean difference in heights for all female online daters falls within our calculated range.
  • The formula utilized is \( \bar{x}_{d} \pm (t_{critical} \times s_d/\sqrt{n}) \)
  • \( \bar{x}_{d} \) is the sample mean difference.
  • \( t_{critical} \) is a value derived from the t-distribution that adjusts as per our confidence level and degrees of freedom.
It's important to remember that the interval does not provide an exact value but instead gives insight into the range where the true difference likely lies.
Two-sample t-test
The two-sample t-test is a statistical method used to determine if two independent samples have different means. In our exercise, this was employed to verify whether the mean height difference for male online daters was greater than that for female online daters.

The critical part of the two-sample t-test is that each group (males and females) is independent of the other. This means that the data from one group doesn't affect the data from the other, which is not the case in the paired t-test setup.
  • The null hypothesis is that both groups have equal mean height differences: \( H_0: \mu_{m} - \mu_{f} = 0 \)
  • The alternative hypothesis is that the mean difference for males is greater: \( H_a: \mu_{m} - \mu_{f} > 0 \)
If the test statistic, calculated as \((\bar{x}_{m} - \bar{x}_{f}) / \sqrt{((s_{m}^2/n_m) + (s_{f}^2/n_f))}\), is larger than the critical value, it indicates that the null hypothesis can be rejected, suggesting a statistically significant difference in means between the groups.
Null Hypothesis
The null hypothesis is a fundamental concept in hypothesis testing. It's a statement suggesting that there is no effect or no difference, and acts as a starting point for statistical comparison. In this exercise:
  • For part (a), the null hypothesis asserts that there is no difference in the mean height difference for male online daters: \( H_0: \mu_d = 0 \).
  • For part (c), the null posits that the mean difference between males and females is equal: \( H_0: \mu_m - \mu_f = 0 \).
When conducting a test, if the null hypothesis is not rejected, it means that there's not enough evidence to claim a significant difference. Conversely, if it is rejected, it means that the observed data provide strong evidence for an effect or a difference. It's essential in guiding the direction and outcome of the test, serving as a baseline comparison.
Statistical Significance
Statistical significance helps us determine the strength of the evidence against the null hypothesis. It's a measure of whether our observed data would likely happen under the assumption that the null hypothesis is true. Significance is often expressed with a significance level, denoted as \( \alpha \), commonly set at 0.05.

In the context of the exercise:
  • For the paired t-test, a result is statistically significant if the observed t-statistic is greater than the critical t value for \( \alpha = 0.05 \), indicating that we can reject the null hypothesis that males do not overstate height.
  • For the two-sample t-test, statistical significance suggests that the difference in the mean height differences between men and women is indeed real (as opposed to occurring by random chance).
Reaching statistical significance is crucial for validating our results. It tells us that our findings are not merely due to random variability, but that there's a genuine underlying effect or difference.

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Most popular questions from this chapter

Common Sense Media surveyed 1000 teens and 1000 parents of teens to learn about how teens are using social networking sites such as Facebook and MySpace ("Teens Show, Tell Too Much Online," San Francisco Chronicle, August 10,2009 ). The two samples were independently selected and were chosen in a way that makes it reasonable to regard them as representative of American teens and parents of American teens. a. When asked if they check their online social networking sites more than 10 times a day, 220 of the teens surveyed said yes. When parents of teens were asked if their teen checked his or her site more than 10 times a day, 40 said yes. Use a significance level of .01 to carry out a hypothesis test to determine if there is convincing evidence that the proportion of all parents who think their teen checks a social networking site more than 10 times a day is less than the proportion of all teens who report that they check more than 10 times a day. b. The article also reported that 390 of the teens surveyed said they had posted something on their networking site that they later regretted. Would you use the two-sample \(z\) test of this section to test the hypothesis that more than one-third of all teens have posted something on a social networking site that they later regretted? Explain why or why not. c. Using an appropriate test procedure, carry out a test of the hypothesis given in Part (b). Use \(\alpha=.05\) for this test.

Can moving their hands help children learn math? This question was investigated in the paper "Gesturing Gives Children New Ideas about Math" (Psychological Science [2009]: 267-272). Eighty-five children in the third and fourth grades who did not answer any questions correctly on a test with six problems of the form \(3+2+8=-8\) were participants in an experiment. The children were randomly assigned to either a no-gesture group or a gesture group. All the children were given a lesson on how to solve problems of this form using the strategy of trying to make both sides of the equation equal. Children in the gesture group were also taught to point to the first two numbers on the left side of the equation with the index and middle finger of one hand and then to point at the blank on the right side of the equation. This gesture was supposed to emphasize that grouping is involved in solving the problem. The children then practiced additional problems of this type. All children were then given a test with six problems to solve, and the number of correct answers was recorded for each child. Summary statistics read from a graph in the paper are given below. $$ \begin{array}{l|ccc} & n & \bar{x} & s \\ \hline \text { No gesture } & 42 & 1.3 & 0.3 \\ \text { Gesture } & 43 & 2.2 & 0.4 \\ \hline \end{array} $$ Is there evidence to support the theory that learning the gesturing approach to solving problems of this type results in a higher mean number of correct responses? Test the relevant hypotheses using \(\alpha=.01\).

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