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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

Effect of Splitting the Bill Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54 , while the 24 people splitting the bill had a higher mean cost of 50.92 Israeli shekels with a standard deviation of 14.33. The raw data can be found in SplitBill and both distributions are reasonably bell-shaped. Use this information to find and interpret a \(95 \%\) confidence interval for the difference in mean meal cost between these two situations.

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution above 2.1 if the samples have sizes \(n_{1}=12\) and \(n_{2}=12\).

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 \(H_{a}: \mu_{1}>\mu_{2}\) using the paired data in the following table: $$ \begin{array}{lllllllllll} \hline \text { Situation } & 1 & 125 & 156 & 132 & 175 & 153 & 148 & 180 & 135 & 168 & 157 \\ \text { Situation } & 2 & 120 & 145 & 142 & 150 & 160 & 148 & 160 & 142 & 162 & 150 \\ \hline \end{array} $$

Using Data 5.1 on page \(375,\) we find a significant difference in the proportion of fruit flies surviving after 13 days between those eating organic potatoes and those eating conventional (not organic) potatoes. Exercises 6.166 to 6.169 ask you to conduct a hypothesis test using additional data from this study. \(^{40}\) In every case, we are testing $$\begin{array}{ll}H_{0}: & p_{o}=p_{c} \\\H_{a}: & p_{o}>p_{c}\end{array}$$ where \(p_{o}\) and \(p_{c}\) represent the proportion of fruit flies alive at the end of the given time frame of those eating organic food and those eating conventional food, respectively. Also, in every case, we have \(n_{1}=n_{2}=500 .\) Show all remaining details in the test, using a \(5 \%\) significance level. Effect of Organic Raisins after 15 Days After 15 days, 320 of the 500 fruit flies eating organic raisins are still alive, while 300 of the 500 eating conventional raisins are still alive.

For each scenario, use the formula to find the standard error of the distribution of differences in sample means, \(\bar{x}_{1}-\bar{x}_{2}\) Samples of size 25 from Population 1 with mean 6.2 and standard deviation 3.7 and samples of size 40 from Population 2 with mean 8.1 and standard deviation 7.6

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