Chapter 11.2, Problem 29E

The paper “The Truth About Lying in Online Dating Profiles” (Proceedings, Computer-Human Interaction [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 calculate the values in the accompanying table.

You can 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.)

1. 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 α = 0.05.
2. 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. (Hint: See Example 11.8.)
3. c. Use the two-sample t test of Section 11.1 to test H₀: μ_m – μ_f = 0 versus H_a: μ_m – μ_f > 0, where μ_m is the mean height difference (profile – actual) for male online daters and μ_f, is the mean height difference (profile – actual) for female online daters.
4. d. Explain why a paired t test was used in Part (a) but a two-sample t test was used in Part (c).

To determine

Check whether there is convincing evidence that the average male online daters overstate their height in online dating profile or not.

Answer to Problem 29E

There is convincing evidence that the average male online daters overstate their height in online dating profile.

Explanation of Solution

Calculation:

Total of 40 men and 40 women are sampled. The sample size, standard deviation, and mean for diference between profile height and actual height, for men and women are given.

Assumption for conducting the test:

• The sample difference should be a random sample.
• The population distribution for the mean differences should follow approximately normal distribution.
• The samples are paired.

Here, the samples are paired. It is reasonable to assume that the two samples of men and women acts as the representative for all the male online daters and female online daters.

Let ${\mu }_1$ be the mean profile height for men.

Let ${\mu }_2$ be the mean actual height for men.

Here, ${\mu }_d$ is the mean difference between profile height and actual height, for men.

That is, ${\mu }_d={\mu }_1-{\mu }_2$ .

Hypotheses:

Null hypothesis:

$H_0:{\mu }_d=0$

That is, the mean difference between profile height and actual height, for men is zero.

Alternative hypothesis:

$H_a:{\mu }_d>0$

That is, the mean difference between profile height and actual height is greater than zero.

Test statistic:

Software procedure:

Step by step procedure to obtain the test statistic by using MINITAB software is given below:

• Choose Stat > Basic Statistics > Paired t.
• Choose summarized data.
• In choose sample size as 40, means as 0.57 and standard deviation as 0.81.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select greater than.
• Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

From the above MINITAB output the test statistic is 4.451 and the P-value is 0.

Decision rule:

• If P-value is less than or equal to the level of significance, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is less than the level of significance.

That is, $P\text{-value}\left(=0\right)<0.05$ .

Therefore, reject the null hypothesis.

It can be concluded that there is convincing evidence that the average male online daters overstate their height in online dating profile.

To determine

Construct a 95% confidence interval for the difference in mean profile height and mean actual height for female online daters.

Interpret the interval.

Answer to Problem 29E

The 95% confidence interval for the difference in mean profile height and mean actual height for female online daters is $\underset{\_}{\left(-0.210,0.270\right)}$.

Explanation of Solution

Calculation:

Assumptions for conducting the hypothesis  test:

• The sample difference should be a random sample.
• The population distribution for the mean differences should follow approximately normal distribution.
• The samples are paired.

Here, samples are paired. The selected samples were representatives of population. Since the sample size is large $\left(40\left(n\right)>30\right)$ , the assumption of normality is satisfied. Therefore, assumptions are satisfied.

Confidence interval:

Software procedure:

Step by step procedure to obtain the confidence interval by using MINITAB software is as follows:

• Choose Stat > Basic Statistics > Paired t.
• Choose summarized data.
• In choose sample size as 40, means as 0.03 and standard deviation as 0.75.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Click OK in all dialogue boxes.

Output using the MINITAB software is given below:

From the output, the 95% confidence interval is $\left(-0.210,0.270\right)$.

Interpretation:

One can be 95% confident that the difference in mean profile height and mean actual height for female online daters lies between –0.210 and 0.270.

To determine

Conduct a two sample t-test.

Answer to Problem 29E

There is convincing evidence that the mean height difference is higher for males than females.

Explanation of Solution

Calculation:

In this context, two sample t-test is used for the comparison.

Assumption for the two sample t-test:

• The random samples should be collected independently.
• The sample sizes should be large. That is, each sample size is at least 30. Or the populations are approximately normally distributed.

Requirement check:

Here, samples are selected at random from the population. Each sample has size of 40, which is greater than 30.

Therefore, the assumptions are satisfied.

Let ${\mu }_m$ be the mean height difference for male.

Let ${\mu }_f$ be the mean height difference for female.

Hypotheses:

Null hypothesis:

$H_0:{\mu }_m-{\mu }_f=0$

That is, the mean height difference is same for both males and females.

Alternative hypothesis:

$H_a:{\mu }_m-{\mu }_f>0$

That is, the mean height difference is higher for males than females.

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain the P-value and test statistic by using MINITAB software:

• Choose Stat > Basic Statistics > 2 sample t.
• Choose Summarized data.
• In sample 1, enter Sample size as 40, Mean as 0.57, Standard deviation as 0.03.
• In sample 2, enter Sample size as 40, Mean as 0.03, Standard deviation as 0.75.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select greater than.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Therefore, the P-value is 0.001 and the test statistic is 3.094.

Decision rule:

• If P-value is less than or equal to the level of significance, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is less than the level of significance.

That is, $P\text{-value}\left(=0.001\right)<0.05$

Therefore, reject the null hypothesis.

Hence, it can be concluded that there is convincing evidence that the mean height difference is higher for males than females.

To determine

Explain the reason for paired t test was used in Part (a) two sample t-test was used in Part (c).

Explanation of Solution

Calculation:

In this context, the result obtained in Part (a) compares the mean actual and profile height for males. That is, the male profile and actual heights are paired. Therefore, the paired t test was used in Part (a).

Now, in this case, the result obtained in Part (c) compares the mean height difference for male and female. That is, the samples of male and female are independent. Therefore, the two sample t-test was used in Part (c).