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Female primates visibly display their fertile window, often with red or pink coloration. Do humans also do this? A study \(^{18}\) looked at whether human females are more likely to wear red or pink during their fertile window (days \(6-14\) of their cycle \()\). They collected data on 24 female undergraduates at the University of British Columbia, and asked each how many days it had been since her last period, and observed the color of her shirt. Of the 10 females in their fertile window, 4 were wearing red or pink shirts. Of the 14 females not in their fertile window, only 1 was wearing a red or pink shirt. (a) State the null and alternative hypotheses. (b) Calculate the relevant sample statistic, \(\hat{p}_{f}-\hat{p}_{n f}\), for the difference in proportion wearing a pink or red shirt between the fertile and not fertile groups. (c) For the 1000 statistics obtained from the simulated randomization samples, only 6 different values of the statistic \(\hat{p}_{f}-\hat{p}_{n f}\) are possible. Table 4.7 shows the number of times each difference occurred among the 1000 randomizations. Calculate the p-value.

Short Answer

Expert verified
The null hypothesis is that the proportions are equal, and the alternative hypothesis is that the proportion is greater in the fertile group. The sample statistic for the difference in proportions is 0.329. The exact p-value cannot currently be calculated without the distribution of simulated statistics, but would be the proportion of simulated statistics greater than or equal to 0.329.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis (H0) is that the probability of wearing red or pink is the same whether a woman is in her fertile window or not. In mathematical terms, \( H_{0}: p_{f} - p_{nf} = 0 \), where \( p_{f} \) is the proportion of women in their fertile window wearing red or pink, and \( p_{nf} \) is the proportion of non-fertile women wearing red or pink. The alternative hypothesis (Ha) is that the probabilities are not the same, or in other words the probability of wearing red or pink is higher during the fertile window. Mathematically, \( H_{a}: p_{f} - p_{nf} > 0 \).
02

Calculate the Relevant Sample Statistic

To calculate the sample statistic, we first calculate the sample proportions. For the fertile group, \( \hat{p}_{f} = \frac{4}{10} = 0.4 \). For the non-fertile group, \( \hat{p}_{nf} = \frac{1}{14} ≈ 0.071 \). The sample statistic is then the difference between these two proportions: \( \hat{p}_{f}-\hat{p}_{nf} = 0.4 - 0.071 = 0.329 \).
03

Calculate the p-value

The p-value is derived from the simulated randomization samples. From the provided information, it's given that only 6 different values are possible. For a valid p-value calculation, it's necessary to know the observed frequency of each possible value. Without this information, the exact p-value cannot be computed. However, if the observed frequency distribution was provided, to calculate the p-value, the proportion of simulated statistics greater than or equal to the observed statistic 0.329 would be found. That proportion represents the p-value.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is crucial when conducting statistical hypothesis testing. The null hypothesis, symbolized as H0, represents a statement of no effect or no difference. In the context of the exercise, it asserts that the likelihood of females wearing red or pink during their fertile window is the same as during other times, mathematically denoted as H0: pf - pnf = 0.

The alternative hypothesis, represented by Ha, is the hypothesis that researchers want to test for. It indicates there is an effect or a difference. In this case, it proposes that females are more likely to wear red or pink during their fertile window compared to other times, expressed as Ha: pf - pnf > 0. Testing these hypotheses helps determine whether there's sufficient evidence to reject the null hypothesis in favor of the alternative.
Sample Statistic Calculation
The sample statistic is a numerical value calculated from the data that provides insight into the hypotheses being tested. It's a point estimate that reflects a population parameter.

In our exercise, we calculate the sample proportions first. For females in their fertile window (fertile group), the proportion who wore red or pink is \( \hat{p}_{f} = \frac{4}{10} = 0.4 \). For females not in their fertile window (non-fertile group), the proportion is \( \hat{p}_{nf} = \frac{1}{14} ≈ 0.071 \). To assess the difference hypothesized, we compute the difference between these two proportions: \( \hat{p}_{f}-\hat{p}_{nf} = 0.4 - 0.071 = 0.329 \).

This sample statistic represents the observed difference between the two sample proportions and serves as the basis for our hypothesis test.
P-Value Calculation
The p-value is a crucial component in hypothesis testing, as it helps us to determine the strength of the evidence against the null hypothesis. It is the probability of observing a sample statistic as extreme as the test statistic, assuming the null hypothesis is true.

In the exercise given, the p-value would be calculated from the simulated randomization samples provided by the study. Unfortunately, specific values from the simulation were not supplied, but generally, one would tally the number of statistics from the simulation that are grater than or equal to the observed sample statistic, in this case 0.329. This count is then divided by the total number of simulations, 1000, to obtain the p-value.

This p-value tells us how likely it is to observe a difference in proportions as large or larger than the one we calculated, given that there is actually no difference (i.e., under the null hypothesis). A low p-value indicates that such an observation would be very unlikely, thus providing evidence against the null hypothesis. A common threshold for 'low' is 0.05, but this can vary depending on the field of study and specific circumstances.

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Most popular questions from this chapter

Mating Choice and Offspring Fitness: MiniExperiments Exercise 4.153 explores the question of whether mate choice improves offspring fitness in fruit flies, and describes two seemingly identical experiments yielding conflicting results (one significant, one insignificant). In fact, the second source was actually a series of three different experiments, and each full experiment was comprised of 50 different mini-experiments (runs), 10 each on five different days. (a) Suppose each of the 50 mini-experiments from the first study were analyzed individually. If mating choice has no impact on offspring fitness, about how many of these \(50 \mathrm{p}\) -values would you expect to yield significant results at \(\alpha=0.05 ?\) (b) The 50 p-values, testing the alternative \(H_{a}\) : \(p_{C}>p_{N C}\) (proportion of flies surviving is higher in the mate choice group) are given below: $$ \begin{array}{lllllllllll} \text { Day 1: } & 0.96 & 0.85 & 0.14 & 0.54 & 0.76 & 0.98 & 0.33 & 0.84 & 0.21 & 0.89 \\ \text { Day 2: } & 0.89 & 0.66 & 0.67 & 0.88 & 1.00 & 0.01 & 1.00 & 0.77 & 0.95 & 0.27 \\ \text { Day 3: } & 0.58 & 0.11 & 0.02 & 0.00 & 0.62 & 0.01 & 0.79 & 0.08 & 0.96 & 0.00 \\ \text { Day 4: } & 0.89 & 0.13 & 0.34 & 0.18 & 0.11 & 0.66 & 0.01 & 0.31 & 0.69 & 0.19 \\ \text { Day 5: } & 0.42 & 0.06 & 0.31 & 0.24 & 0.24 & 0.16 & 0.17 & 0.03 & 0.02 & 0.11 \end{array} $$ How many are actually significant using \(\alpha=0.05 ?\) (c) You may notice that two p-values (the fourth and last run on day 3 ) are 0.00 when rounded to two decimal places. The second of these is actually 0.0001 if we report more decimal places. This is very significant! Would it be appropriate and/or ethical to just report this one run, yielding highly statistically significant evidence that mate choice improves offspring fitness? Explain. (d) You may also notice that two of the p-values on day 2 are 1 (rounded to two decimal places). If we had been testing the opposite alternative, \(H_{a}:\) \(p_{C}

Do You Own a Smartphone? A study \(^{19}\) conducted in July 2015 examines smartphone ownership by US adults. A random sample of 2001 people were surveyed, and the study shows that 688 of the 989 men own a smartphone and 671 of the 1012 women own a smartphone. We want to test whether the survey results provide evidence of a difference in the proportion owning a smartphone between men and women. (a) State the null and alternative hypotheses, and define the parameters. (b) Give the notation and value of the sample statistic. In the sample, which group has higher smartphone ownership: men or women? (c) Use StatKey or other technology to find the pvalue.

We are conducting many hypothesis tests to test a claim. In every case, assume that the null hypothesis is true. Approximately how many of the tests will incorrectly find significance? 40 tests using a significance level of \(10 \%\).

For each situation described, indicate whether it makes more sense to use a relatively large significance level (such as \(\alpha=0.10\) ) or a relatively small significance level (such as \(\alpha=0.01\) ). Testing to see whether taking a vitamin supplement each day has significant health benefits. There are no (known) harmful side effects of the supplement.

Do iPads Help Kindergartners Learn: A Series of Tests Exercise 4.147 introduces a study in which half of the kindergarten classes in a school district are randomly assigned to receive iPads. We learn that the results are significant at the \(5 \%\) level (the mean for the iPad group is significantly higher than for the control group) for the results on the HRSIW subtest. In fact, the HRSIW subtest was one of 10 subtests and the results were not significant for the other 9 tests. Explain, using the problem of multiple tests, why we might want to hesitate before we run out to buy iPads for all kindergartners based on the results of this study.

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