/*! 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 70 Female primates visibly display ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 70

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, \( \hat{p}_{nf} = \frac{1}{14} 鈮� 0.071 \). 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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 \( \hat{p}_{nf} = \frac{1}{14} 鈮� 0.071 \). 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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How influenced are consumers by price and marketing? If something costs more, do our expectations lead us to believe it is better? Because expectations play such a large role in reality, can a product that costs more (but is in reality identical) actually be more effective? Baba Shiv, a neuroeconomist at Stanford, conducted a study \(^{25}\) involving 204 undergraduates. In the study, all students consumed a popular energy drink which claims on its packaging to increase mental acuity. The students were then asked to solve a series of puzzles. The students were charged either regular price ( \(\$ 1.89\) ) for the drink or a discount price \((\$ 0.89)\). The students receiving the discount price were told that they were able to buy the drink at a discount since the drinks had been purchased in bulk. The authors of the study describe the results: "the number of puzzles solved was lower in the reduced-price condition \((M=4.2)\) than in the regular-price condition \((M=5.8) \ldots p<.0001 . "\) (a) What can you conclude from the study? How strong is the evidence for the conclusion? (b) These results have been replicated in many similar studies. As Jonah Lehrer tells us: "According to Shiv, a kind of placebo effect is at work. Since we expect cheaper goods to be less effective, they generally are less effective, even if they are identical to more expensive products. This is why brand-name aspirin works better than generic aspirin and why Coke tastes better than cheaper colas, even if most consumers can't tell the difference in blind taste tests."26 Discuss the implications of this research in marketing and pricing.

The same sample statistic is used to test a hypothesis, using different sample sizes. In each case, use StatKey or other technology to find the p-value and indicate whether the results are significant at a \(5 \%\) level. Which sample size provides the strongest evidence for the alternative hypothesis? Testing \(H_{0}: p_{1}=p_{2}\) vs \(H_{a}: p_{1}>p_{2}\) using \(\hat{p}_{1}-\hat{p}_{2}=0.45-0.30=0.15\) with each of the following sample sizes: (a) \(\hat{p}_{1}=9 / 20=0.45\) and \(\hat{p}_{2}=6 / 20=0.30\) (b) \(\hat{p}_{1}=90 / 200=0.45\) and \(\hat{p}_{2}=60 / 200=0.30\) (c) \(\hat{p}_{1}=900 / 2000=0.45\) and \(\hat{p}_{2}=600 / 2000=0.30\)

Data 4.3 on page 265 introduces a situation in which a restaurant chain is measuring the levels of arsenic in chicken from its suppliers. The question is whether there is evidence that the mean level of arsenic is greater than \(80 \mathrm{ppb},\) so we are testing \(H_{0}: \mu=80\) vs \(H_{a}:\) \(\mu>80,\) where \(\mu\) represents the average level of arsenic in all chicken from a certain supplier. It takes money and time to test for arsenic, so samples are often small. Suppose \(n=6\) chickens from one supplier are tested, and the levels of arsenic (in ppb) are: \(\begin{array}{llllll}68, & 75, & 81, & 93, & 95, & 134\end{array}\) (a) What is the sample mean for the data? (b) Translate the original sample data by the appropriate amount to create a new dataset in which the null hypothesis is true. How do the sample size and standard deviation of this new dataset compare to the sample size and standard deviation of the original dataset? (c) Write the six new data values from part (b) on six cards. Sample from these cards with replacement to generate one randomization sample. (Select a card at random, record the value, put it back, select another at random, until you have a sample of size \(6,\) to match the original sample size.) List the values in the sample and give the sample mean. (d) Generate 9 more simulated samples, for a total of 10 samples for a randomization distribution. Give the sample mean in each case and create a small dotplot. Use an arrow to locate the original sample mean on your dotplot.

Translating Information to Other Significance Levels Suppose in a two-tailed test of \(H_{0}: \rho=0\) vs \(H_{a}: \rho \neq 0,\) we reject \(H_{0}\) when using a \(5 \%\) significance level. Which of the conclusions below (if any) would also definitely be valid for the same data? Explain your reasoning in each case. (a) Reject \(H_{0}: \rho=0\) in favor of \(H_{a}: \rho \neq 0\) at a \(1 \%\) significance level. (b) Reject \(H_{0}: \rho=0\) in favor of \(H_{a}: \rho \neq 0\) at a \(10 \%\) significance level. (c) Reject \(H_{0}: \rho=0\) in favor of the one-tail alternative, \(H_{a}: \rho>0,\) at a \(5 \%\) significance level, assuming the sample correlation is positive.

Eating Breakfast Cereal and Conceiving Boys Newscientist.com ran the headline "Breakfast Cereals Boost Chances of Conceiving Boys," based on an article which found that women who eat breakfast cereal before becoming pregnant are significantly more likely to conceive boys. \({ }^{42}\) The study used a significance level of \(\alpha=0.01\). The researchers kept track of 133 foods and, for each food, tested whether there was a difference in the proportion conceiving boys between women who ate the food and women who didn't. Of all the foods, only breakfast cereal showed a significant difference. (a) If none of the 133 foods actually have an effect on the gender of a conceived child, how many (if any) of the individual tests would you expect to show a significant result just by random chance? Explain. (Hint: Pay attention to the significance level.) (b) Do you think the researchers made a Type I error? Why or why not? (c) Even if you could somehow ascertain that the researchers did not make a Type I error, that is, women who eat breakfast cereals are actually more likely to give birth to boys, should you believe the headline "Breakfast Cereals Boost Chances of Conceiving Boys"? Why or why not?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.