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The eating habits of \(n=12\) bats were examined in the article "Foraging Behavior of the Indian False Vampire Bat" (Biotropica [1991]: 63-67). These bats consume insects and frogs. For these 12 bats, the mean time to consume a frog was \(\bar{x}=21.9\) minutes. Suppose that the standard deviation was \(s=7.7\) minutes. Is there convincing evidence that the mean supper time of a vampire bat whose meal consists of a frog is greater than 20 minutes? What assumptions must be reasonable for the one-sample \(t\) test to be appropriate?

Short Answer

Expert verified
In this problem, we tested the hypothesis that the mean supper time of a vampire bat whose meal consists of a frog is greater than 20 minutes. We performed a one-sample t-test and calculated a test statistic of \(t \approx 0.986\) and a p-value of \(\approx 0.172\). Comparing the p-value to a significance level of \(\alpha = 0.05\), we fail to reject the null hypothesis. Therefore, there isn't enough convincing evidence to suggest that the mean supper time of a vampire bat whose meal consists of a frog is greater than 20 minutes, based on the given data. The assumptions for this test were that the observations were independent and the distribution of the sample mean was nearly normal.

Step by step solution

01

Assumptions

Before we perform the t-test, let's check the assumptions required for the test to be valid. 1. The observations should be independent: We can assume that each bat's eating habits are independent of the others. 2. The distribution of the sample mean should be nearly normal: With a sample size of \(n=12\), we cannot guarantee the normality of the distribution due to the small sample size. However, for the purpose of this exercise, we'll assume it to be nearly normal. Now that the assumptions have been considered, let's proceed with the one-sample t-test.
02

State the Hypotheses

The null hypothesis (\(H_0\)) is that the mean supper time for a vampire bat consuming a frog is 20 minutes or less. The alternative hypothesis (\(H_a\)) is that the mean supper time is greater than 20 minutes: \(H_0: \mu \leq 20\) \(H_a: \mu > 20\)
03

Calculate the Test Statistic

To calculate the test statistic, we use the formula: \(t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\) Where: \(\bar{x}\) = sample mean (21.9 minutes) \(\mu_0\) = null hypothesis mean (20 minutes) \(s\) = sample standard deviation (7.7 minutes) \(n\) = sample size (12) Plugging in the values, we get: \(t = \frac{21.9 - 20}{7.7/\sqrt{12}} \approx 0.986\)
04

Determine the P-Value

Since our alternative hypothesis is that the mean supper time is greater than 20 minutes, we have to look at the right tail of the t-distribution to find the p-value. Using a t-table or calculator (with 11 degrees of freedom), we find that: P-value \(\approx 0.172\)
05

Make a Decision and Interpret the Result

We need to compare the p-value with a significance level (\(\alpha\)) to make a decision about the null hypothesis. Let's use a common value, \(\alpha = 0.05\). Since the p-value \((0.172) > \alpha (0.05)\), we fail to reject the null hypothesis. In conclusion, there isn't enough convincing evidence to suggest that the mean supper time of a vampire bat whose meal consists of a frog is greater than 20 minutes, based on the given data.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a method used in statistics to determine whether there is enough evidence in a sample of data to infer that a certain condition is true for the entire population.

In our example involving the feeding behavior of bats, hypothesis testing allows us to assess whether the average time it takes for a bat to consume a frog is statistically greater than a specified value. The process starts with the formulation of two competing hypotheses: the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). The null hypothesis typically represents the status quo or a specific value to be tested against, while the alternative hypothesis represents what we are seeking evidence for.

The next step is to calculate a test statistic that measures how far our sample statistic, such as the sample mean, is from the null hypothesis value. From this, we can determine the p-value, which gives us the probability of observing the sample data, or something more extreme, if the null hypothesis is true. If this p-value is less than a predetermined significance level, such as 0.05, we reject the null hypothesis in favor of the alternative. If not, we fail to reject the null hypothesis, indicating there isn't sufficient evidence to support the alternative hypothesis.
Statistical Significance
Statistical significance is a determination about the non-randomness of the results obtained from hypothesis testing. When we say that a result is statistically significant, we mean that it is unlikely to have occurred by chance given the null hypothesis is true.

The level of statistical significance is typically denoted by \(\alpha\), which is the threshold value that we compare with the p-value. The choice of \(\alpha\) often depends on the field of study, but common values are 0.05, 0.01, or 0.10.

For the bat feeding study, a p-value of approximately 0.172 was calculated, which is greater than the common significance level of 0.05. Consequently, we do not have enough evidence to consider the results statistically significant, meaning we cannot conclude that the average supper time for bats eating frogs is greater than 20 minutes.
Sample Mean
The sample mean, often represented by \(\bar{x}\), is the arithmetic average of all the measurements in a sample and serves as an estimate of the population mean.

Calculating the sample mean is straightforward: simply sum up all the measurements in the sample and divide by the number of observations. In the context of our bat study, the sample mean is the average time, \(\bar{x} = 21.9\) minutes, it took for the 12 bats in the study to consume a frog.

We utilize the sample mean in hypothesis testing to compare it against a hypothetical population mean, \(\mu_0\) (in this case, 20 minutes), to see if there is a significant difference or not based on our data.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the individual measurements in our data set tend to differ from the sample mean.

A small standard deviation indicates that the observations tend to be close to the mean, whereas a larger standard deviation indicates that the data is spread out over a wider range of values. In our bat example, the standard deviation is given as \(s = 7.7\) minutes. This means the time it takes for the bats to eat frogs varies by an average of 7.7 minutes from the average feeding time.

When it comes to hypothesis testing, standard deviation plays a crucial role in calculating the test statistic which determines how extreme the sample mean is, assuming the null hypothesis is true. It is also important to note that the standard deviation is used in the denominator of the t-test formula, meaning that the greater the variability in the sample, the less certain we are about our estimate of the population mean, which can affect the outcome of the test.

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Most popular questions from this chapter

The article "Americans' Big Debt Burden Growing, Not Evenly Distributed" (www.gallup.com, retrieved December 14, 2016) reported that for a representative sample of Americans born between 1965 and 1971 (known as Generation \(\mathrm{X}\) ), the sample mean number of credit cards owned was 4.5. Suppose that the sample standard deviation (which was not reported) was 1.0 and that the sample size was \(n=300\). a. Construct a \(95 \%\) confidence interval for the population mean number of credit cards owned by Generation \(\mathrm{X}\) Americans. b. The interval in Part (a) does not include \(0 .\) Does this imply that all Generation X Americans have at least one credit card? Explain.

The authors of the paper "Mean Platelet Volume Could Be Possible Biomarker in Early Diagnosis and Monitoring of Gastric Cancer" (Platelets [2014]: 592-594) wondered if mean platelet volume (MPV) might be a way to distinguish patients with gastric cancer from patients who did not have gastric cancer. MPV was recorded for 31 patients with gastric cancer. The sample mean was reported to be 8.31 femtoliters (fL) and the sample standard deviation was reported to be \(0.78 \mathrm{fL}\). For healthy people, the mean MPV is 7.85 fL. Is there convincing evidence that the mean MPV for patients with gastric cancer is greater than \(7.85 \mathrm{fL} ?\) For purposes of this exercise, you can assume that the sample of 31 patients with gastric cancer is representative of the population of all patients with gastric cancer.

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations. (Hint: See discussion on page \(594 .\) ) a. Upper-tailed test, \(\mathrm{df}=8, t=2.0\) b. Lower-tailed test, \(\mathrm{df}=10, t=-2.4\) c. Lower-tailed test, \(n=22, t=-4.2\) d. Two-tailed test, \(\mathrm{df}=15, t=-1.6\)

A comprehensive study conducted by the National Institute of Child Health and Human Development tracked more than 1000 children from an early age through elementary school (The New York Times, November 1, 2005). The study concluded that children who spent more than 30 hours a week in child care before entering school tended to score higher in math and reading when they were in the third grade. The researchers cautioned that the findings should not be a cause for alarm because the differences were found to be small. Explain how the difference between the mean math score for the child care group and the overall mean for third graders could be small but the researchers could still reach the conclusion that the mean for the child care group is significantly higher than the overall mean.

Give as much information as you can about the \(P\) -value of a \(t\) test in each of the following situations: a. Two-tailed test, \(n=16, t=1.6\) b. Upper-tailed test, \(n=14, t=3.2\) c. Lower-tailed test, \(n=20, t=-5.1\) d. Two-tailed test, \(n=16, t=6.3\)

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