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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.

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Key Concepts

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

Single Proportion Test
A single proportion test is used when you want to determine if the proportion of a sample is different from a specified value. In this context, scientists are interested in seeing if lion attacks in Tanzania are more frequent after a full moon compared to before. They use a single proportion test since they have one sample group: the lion attacks between five days before and five days after the full moon.

The goal is to determine if the proportion of attacks after the full moon (p) is greater than 0.5, which would indicate an increase in attacks. This type of test is ideal when dealing with a binary outcome, like in this scenario where the outcome is either before or after the full moon.

To set up the test, define your null hypothesis as the proportion being equal on both sides of the full moon: \( p = 0.5 \). The alternative hypothesis claims that the proportion of attacks after the full moon is greater than before \( p > 0.5 \). These steps form the basis for a solid approach to hypothesis testing with a single sample proportion.
Alternative Hypothesis
In hypothesis testing, the alternative hypothesis (often denoted as \( H_a \)) represents what researchers aim to prove. It is a statement suggesting a difference or effect, and it directly contrasts the null hypothesis \( H_0 \).

For the lion attack scenario, the alternative hypothesis posits that there are more attacks after a full moon than before, suggesting an inequality in the attack distribution between the two periods. In this specific case, the alternative hypothesis is mathematically expressed as \( p > 0.5 \).

The alternative hypothesis is crucial because it guides the direction of the test. It helps determine how the data will be interpreted and what kind of statistical techniques will be used. In essence, the test is looking to gather evidence to support the alternative hypothesis over the null hypothesis.
P-Value Calculation
The p-value is a fundamental concept in hypothesis testing. It helps us quantify how extreme our observed data are under the null hypothesis. Basically, it tells us whether the results from the test are statistically significant or not.

Calculating the p-value involves determining the probability of observing a test statistic as extreme as the one calculated, assuming that the null hypothesis is true. In this scenario, once the z-score is determined using the formula, this score is referenced against standard normal distribution tables or technology to find the p-value.

In practical terms, if the p-value is small (typically less than 0.05), it suggests strong evidence against the null hypothesis, prompting us to reject it. Thus, we support the alternative hypothesis, suggesting that more lion attacks occur after a full moon.
Z-Score
The z-score is a statistical measurement that describes a value's relation to the mean of a group of values. In hypothesis testing for proportions, it helps us understand how far away the observed proportion is from the expected proportion under the null hypothesis.

In this exercise, the z-score is calculated using the formula: \[Z = \frac{p^* - p }{ \sqrt{ \frac{p (1-p)}{n} } }\]where \( p^* \) is the observed proportion of attacks after a full moon, \( p \) is the hypothesized proportion (0.5), and \( n \) is the total number of attacks.

The resulting z-score indicates how many standard deviations the observed proportion is from the expected proportion. A higher z-score indicates that the observed proportion is further from what was expected under the null hypothesis. Ultimately, the z-score is key to finding the p-value and drawing conclusions from the hypothesis test.

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Most popular questions from this chapter

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.

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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? 300 tests using a significance level of \(1 \%\).

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