/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 163 Do Ovulating Women Affect Men's ... [FREE SOLUTION] | 91Ó°ÊÓ

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Do Ovulating Women Affect Men's Speech? Studies suggest that when young men interact with a woman who is in the fertile period of her menstrual cycle, they pick up subconsciously on subtle changes in her skin tone, voice, and scent. A study introduced in Exercise \(\mathrm{B} .23\) suggests that men may even change their speech patterns around ovulating women. The men were randomly divided into two groups with one group paired with a woman in the fertile phase of her cycle and the other group with a woman in a different stage of her cycle. The same women were used in the two different stages. For the men paired with a less fertile woman, 38 of the 61 men copied their partner's sentence construction in a task to describe an object. For the men paired with a woman at peak fertility, 30 of the 62 men copied their partner's sentence construction. The experimenters hypothesized that men might be less likely to copy their partner during peak fertility in a (subconscious) attempt to attract more attention to themselves. Use the normal distribution to test at a \(5 \%\) level whether the proportion of men copying sentence structure is less when the woman is at peak fertility.

Short Answer

Expert verified
Based on the test, we can not conclude that men are less likely to copy their partner's sentence structure when the woman is at peak fertility.

Step by step solution

01

State the hypothesis

The null hypothesis H0 is that the proportion of men copying sentence structure when the woman is at peak fertility is the same as when she is less fertile. That is, p = p0. The alternative hypothesis Ha is that the proportion of men copying sentence structure is less when the woman is at peak fertility. That is, p < p0.
02

Compute test statistic

The sample proportions are p1 = 30/62 (peak fertility) and p2 = 38/61 (less fertile), therefore the test statistic Z = (p1 - p2 )/sqrt((p2 * (1 - p2 ))/n1 + (p1 * (1 - p1))/n2 ) = -1.44, where n1 is the sample size for peak fertility and n2 is for less fertile.
03

Conclude

Since this is a one-tailed test at a 5% level of significance, we reject the null hypothesis H0 if the calculated Z (-1.44) is less than the critical value Z0.05= -1.645. In our case, -1.44 > -1.645, so we fail to reject the null hypothesis. We can not conclude that men are less likely to copy their partner's sentence structure at peak fertility.

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Key Concepts

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

Null Hypothesis
Understanding the null hypothesis is critical in hypothesis testing. It's the claim we assess through statistical analysis. In the context of our exercise involving men's speech patterns around ovulating women, the null hypothesis (\(H_0\)) states that there is no difference in the proportion of men copying sentence structure whether the women are at peak fertility or not. This hypothesis serves as the starting assumption, which we either reject or fail to reject based on the evidence provided by the test statistics.
Alternative Hypothesis
On the flip side, we have the alternative hypothesis (\(H_a\)), which directly opposes the null hypothesis. In this study, the alternative hypothesis posits that the proportion of men who mimic sentence structure is indeed lower when interacting with women at peak fertility compared to less fertile phases. If the data sufficiently supports this claim, we may reject the null hypothesis in favor of the alternative.
Test Statistic
Moving forward, let's break down the idea of a test statistic. It's a standardized value calculated from sample data during a hypothesis test. Here, we're using it to determine how far our sample result lies from the null hypothesis. The test statistic in question, denoted as 'Z', is calculated by considering the difference between our sample proportions and measuring that against the expected variability in those proportions.
Normal Distribution
In hypothesis testing, the normal distribution often comes into play due to the Central Limit Theorem. This bell-shaped curve represents the distribution of a test statistic under the null hypothesis. We use it to find the probability of observing a test statistic as extreme as ours. If that probability is low enough, it suggests that our sample provides evidence against the null hypothesis.
Level of Significance
The level of significance, usually denoted by \(\alpha\), is the threshold we set for deciding whether to reject the null hypothesis. In the exercise, it's set at 5%, which means there's a 5% risk of rejecting the null hypothesis when it's actually true. This level is used to determine the critical value of the test statistic, which is the cut-off point between rejecting and not rejecting the null hypothesis.
Sample Proportion
Key to our analysis is the sample proportion, which reflects the fraction of the sample that meets a certain criterion – for instance, the proportion of men who change their speech patterns. From our exercise, we have two proportions: \(p1 = 30/62\) for men paired with ovulating women, and \(p2 = 38/61\) for the other group. We're testing the hypothesis based on the difference between these two proportions.
One-tailed Test
Lastly, we have the one-tailed test, which is used when we're investigating if there's a decrease or increase—in our case, a decrease—in the proportion of men copying sentence structure around ovulating women. This contrasts with a two-tailed test, where we would be interested in any significant difference, either increase or decrease. Our decision to reject or fail to reject the null hypothesis is based on whether the test statistic falls in the critical region of this one-tail.

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Most popular questions from this chapter

Exercise 6.19 discusses the headline "Domestic cats kill many more wild birds in the United States than scientists thought," and estimates the proportion of domestic cats that hunt outside. A separate study \(^{23}\) used KittyCams to record all activity of \(n=55\) domestic cats that hunt outdoors. The video footage showed that the mean number of kills per week for these cats was 2.4 with a standard deviation of \(1.51 .\) Find and interpret a \(99 \%\) confidence interval for the mean number of kills per week by US household cats that hunt outdoors.

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