Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Scientists studying lion attacks on humans in Tanzania \(^{32}\) found that 95 lion attacks happened between \(6 \mathrm{pm}\) and \(10 \mathrm{pm}\) within either five days before a full moon or five days after a full moon. Of these, 71 happened during the five days after the full moon while the other 24 happened during the five days before the full moon. Does this sample of lion attacks provide evidence that attacks are more likely after a full moon? In other words, is there evidence that attacks are not equally split between the two five-day periods? Use StatKey or other technology to find the p-value, and be sure to show all details of the test. (Note that this is a test for a single proportion since the data come from one sample.)

Short Answer

Expert verified
The short answer will be based on whether the null hypothesis was rejected or failed to reject after comparing the p-value with the significance level. This will determine whether there is statistical evidence to claim that lion attacks are more likely to occur after a full moon.

Step by step solution

01

Formulate the hypothesis

We start off by stating our null hypothesis (H0) and alternative hypothesis (Ha). For this exercise, the null hypothesis is that the lion attacks are equally likely before and after the full moon that is, \( p = 0.5 \) the alternative hypothesis is that the lion attacks are more likely after the full moon, which could be represented as, \( p > 0.5 \). Here, 'p' denotes the probability of an attack after the full moon.
02

Calculate the observed proportion

Next, compute the observed proportion which is the proportion of attacks after a full moon. This calculation can be done by dividing the attacks after the full moon by the total attacks. There were 71 attacks after full moon out of 95 attacks in total, therefore the observed proportion, \( p^* \), is \( p^* = \frac{71}{95} \).
03

Calculate the test statistic

We need to calculate the test statistic for the null hypothesis. The z-score for this test is given by the formula \( Z = \frac{p^* - p }{ \sqrt{ \frac{p (1-p)}{n} } } \). Here, 'n' is the total number of attacks, 'p' is the assumed probability of an attack after a full moon under the null hypothesis, and \( p^* \) is the observed probability. Substituting the given values you find the calculated test statistic.
04

Compute the p-value

Once the test statistic is calculated, use tables or a calculator to find out the p-value.
05

Interpret the results

Compare the calculated p-value with the significance level (generally 0.05). If the p-value is less than the significance level, then reject the null hypothesis. If the p-value is greater than the significance level, then there is insufficient evidence to reject the null hypothesis. Provide your interpretation based on the result.

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 Vaia!

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

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 \(\alpha=0.01\) ). Using a sample of 10 games each to see if your average score at Wii bowling is significantly more than your friend's average score.

Influencing Voters When getting voters to support a candidate in an election, is there a difference between a recorded phone call from the candidate or a flyer about the candidate sent through the mail? A sample of 500 voters is randomly divided into two groups of 250 each, with one group getting the phone call and one group getting the flyer. The voters are then contacted to see if they plan to vote for the candidate in question. We wish to see if there is evidence that the proportions of support are different between the two methods of campaigning. (a) Define the relevant parameter(s) and state the null and alternative hypotheses. (b) Possible sample results are shown in Table 4.3 . Compute the two sample proportions: \(\hat{p}_{c},\) the proportion of voters getting the phone call who say they will vote for the candidate, and \(\hat{p}_{f},\) the proportion of voters getting the flyer who say they will vote for the candidate. Is there a difference in the sample proportions? (c) A different set of possible sample results are shown in Table 4.4. Compute the same two sample proportions for this table. (d) Which of the two samples seems to offer stronger evidence of a difference in effectiveness between the two campaign methods? Explain your reasoning. $$ \begin{array}{lcc} \hline & \begin{array}{c} \text { Will Vote } \\ \text { Sample A } \end{array} & \text { for Candidate } & \begin{array}{l} \text { Will Not Vote } \\ \text { for Candidate } \end{array} \\ \hline \text { Phone call } & 152 & 98 \\ \text { Flyer } & 145 & 105 \\ \hline \end{array} $$ $$ \begin{array}{lcc} \text { Sample B } & \begin{array}{c} \text { Will Vote } \\ \text { for Candidate } \end{array} & \begin{array}{c} \text { Will Not Vote } \\ \text { for Candidate } \end{array} \\ \hline \text { Phone call } & 188 & 62 \\ \text { Flyer } & 120 & 130 \\ \hline \end{array} $$

Does consuming beer attract mosquitoes? Exercise 4.17 on page 268 discusses an experiment done in Africa testing possible ways to reduce the spread of malaria by mosquitoes. In the experiment, 43 volunteers were randomly assigned to consume either a liter of beer or a liter of water, and the attractiveness to mosquitoes of each volunteer was measured. The experiment was designed to test whether beer consumption increases mosquito attraction. The report \(^{30}\) states that "Beer consumption, as opposed to water consumption, significantly increased the activation \(\ldots\) of \(A n\). gambiae [mosquitoes] ... \((P<0.001)\)." (a) Is this convincing evidence that consuming beer is associated with higher mosquito attraction? Why or why not? (b) How strong is the evidence for the result? Explain. (c) Based on these results, it is reasonable to conclude that consuming beer causes an increase in mosquito attraction? Why or why not?

Polling 1000 people in a large community to determine the average number of hours a day people watch television.

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

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.

Sign-up for free