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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 \(\mathrm{B} .18\) 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.

Step by step solution

01

Formulate the Hypotheses

The null hypothesis (\(H_0\)) is that the proportion of men copying sentence structure is the same when the woman is at peak fertility and when she is not. This would be \(P_{peak} = P_{non peak}\). The alternative hypothesis (\(H_A\)) is that the proportion of men copying is less when the woman is at peak fertility i.e. \(P_{peak} < P_{non peak}\).

02

Calculate the Test Statistic

First, calculate the sample proportions: \(P_{peak} = 30/62 = 0.483\) and \(P_{non peak} = 38/61 = 0.623\). Then, find the pooled proportion: \(P = (x_{peak} + x_{non peak}) / (n_{peak} + n_{non peak}) = (30 + 38) / (62 + 61) = 0.554\). The test statistic (Z) is then calculated using the formula for testing difference between two proportions: \(Z = (P_{peak} - P_{non peak}) / sqrt(P(1 - P)[(1/n_{peak})+(1/n_{non peak})]) = -1.506\).

03

Find the P-value

The P-value is the probability of getting the observed data (or more extreme) given that the null hypothesis is true. In this context, a smaller test statistic (closer to negative infinity) is considered more extreme. As this is a one-tailed test (less than), the P-value corresponds to the area to the left of the observed data point on the normal distribution: \(P(Z < -1.506)\). By using statistical software or Z-table, we find that this probability is approximately 0.066.

04

Make Decision

Next, compare the P-value with the significance level α=0.05. If the P-value is less than α, reject \(H_0\); otherwise, do not reject \(H_0\). In this case, 0.066 > 0.05, hence we do not reject \(H_0\). Thus, there is not enough evidence at the 5% level to conclude that the proportion of men copying sentence structure is less when the woman is at peak fertility.

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Key Concepts

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

Null and Alternative Hypotheses

In hypothesis testing, the null hypothesis (H_0) represents a statement of no effect or no difference. It serves as a starting point for statistical analysis and is presumed true until evidence suggests otherwise. For the exercise at hand, the null hypothesis posits that there is no difference in the proportion of men copying sentence structure, regardless of whether the female is at peak fertility or not (P_{peak} = P_{non peak}).

The alternative hypothesis (H_A or H_1), on the other hand, posits the presence of an effect or a difference. It is the statement that researchers are trying to find evidence for. In our scenario, the alternative hypothesis suggests that men are less likely to copy sentence structure when interacting with a woman who is at the peak of her fertility (P_{peak} < P_{non peak}). This theory is based on a presumed subconscious behavior in men aimed at attracting more attention to themselves.

Test Statistic

The test statistic is a numerical value used in statistical hypothesis testing to measure the degree of agreement between a sample statistic and the null hypothesis. It essentially helps us determine how far our sample data is from what we would expect under the null hypothesis. Calculating the test statistic involves taking the observed sample data and converting it into a single value that can be compared against a distribution of a known shape.

For our exercise, a Z-test statistic is used, which is common when comparing proportions. The Z-test statistic formula utilized is: Z = (P_{peak} - P_{non peak}) / sqrt(P(1 - P)[(1/n_{peak})+(1/n_{non peak})]). This formula takes into account the pooled sample proportion and the size of each sample, allowing us to quantify the discrepancy between the observed sample proportions and the hypothesized equal proportions.

P-value

The p-value is a crucial concept in statistical hypothesis testing. It represents the probability of observing your data, or something more extreme, assuming that the null hypothesis is true. A low p-value indicates that the observed data are unlikely under the null hypothesis, and could be considered as evidence against H_0.

In our context, the p-value is calculated from the Z-test statistic and represents the area under the curve to the left of our observed Z-value in a standard normal distribution. The smaller the p-value, the stronger is the evidence against the null hypothesis. In the given exercise, the p-value was approximately 0.066, which means there is a 6.6% probability of seeing these results, or more extreme, if the null hypothesis were true.

Significance Level

The significance level, denoted as α, acts as a threshold for making decisions in hypothesis testing. It is predetermined by the researcher and sets the maximum allowed probability of rejecting a true null hypothesis (a Type I error). A common significance level chosen is 5% (α = 0.05), as in our exercise.

To make a decision, the significance level is compared with the p-value: if the p-value is less than or equal to α, the null hypothesis is rejected; otherwise, it is not rejected. As in the solution, since 0.066 is greater than 0.05, we do not reject the null hypothesis. Hence, we do not have enough evidence to support the claim that men's copying behavior changes when women are at peak fertility.