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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? 800 tests using a significance level of \(5 \%\).

Short Answer

Expert verified
Approximately 40 tests will incorrectly find significance at a 5% level when the null hypothesis is true.

Step by step solution

01

Understanding the problem

In this scenario, 800 hypothesis tests are conducted and we are asked to find out how many of the tests will incorrectly find significance, given that the true null hypothesis is true. Since the significance level of the tests performed is \(5\%\), it means that there is a \(5\%\) chance of rejecting a true null hypothesis. This is known as a Type I error.
02

Calculate the number of tests that will incorrectly find significance

To calculate the number of tests that would find false significance, you would use the given significance level and multiply it by the number of tests performed. With a significance level of \(5\%\), or \(0.05\) when expressed as a decimal, and 800 tests performed, the calculation is: \(800 * 0.05 = 40\).
03

Conclusion

So, if the null hypothesis is true in each case, we can expect that approximately 40 of the 800 tests performed will incorrectly find significance at a \(5\%\) level.

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Key Concepts

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

Type I Error
A Type I error occurs when we incorrectly reject the null hypothesis when it is actually true. This kind of error can be quite misleading because it suggests that there is an effect or difference when none actually exists.

To put it in simple terms, imagine you're trying to decide if a coin is unfairly biased towards heads. You flip the coin several times; if you make a Type I error, you'd conclude the coin is biased, even if it's perfectly fair.

In the context of our problem, a Type I error occurs if we declare significance in a hypothesis test when the null hypothesis is, in fact, true. If we conduct 800 tests, even with an honest null hypothesis, about 40 tests might wrongly suggest a significant effect due to the Type I error at a significance level of 5%.

It's essential to understand that this error is part of statistical testing, and while we can minimize it by adjusting our significance level, we can never completely eliminate it.
Significance Level
The significance level, often denoted by the Greek letter \(\alpha\), represents the probability of making a Type I error in a hypothesis test. It is the threshold at which we decide whether to reject the null hypothesis.

Commonly used levels include 0.05, 0.01, and 0.10. In this exercise, a 5% significance level is used, meaning that we are allowing an acceptability of a 5% chance of incorrectly rejecting the null hypothesis.

To visualize this, consider your significance level as a line drawn in the sand. If the results from your hypothesis test fall beyond this line, you reject the null hypothesis. With a 5% significance level, out of 100 tests run when the null hypothesis is true, you would expect about 5 of them to mistakenly show significant results.

The lower the significance level, the less risk you take of making a Type I error, but at the same time, the potential for making another type of error, called a Type II error, increases. This makes choosing the right significance level a balance between these two types of errors.
Null Hypothesis
The null hypothesis, denoted as \(H_0\), is a fundamental concept in hypothesis testing. It is the default assumption that there is no effect or no difference; any observed effect is attributed to sampling variability or randomness.

For example, if we are testing a new drug, the null hypothesis might state that the drug has no effect on patients compared to a placebo. The goal of hypothesis testing is to evaluate data and decide if there is enough evidence to reject the null hypothesis in favor of an alternative hypothesis, which suggests there is an effect.

In hypothesis tests, the null hypothesis serves as the "status quo." We only reject it if we have strong statistical evidence against it.

In our exercise, we assume the null hypothesis is true for each of the 800 tests. This baseline allows us to calculate how many results may incorrectly indicate significance because, by definition, a Type I error only occurs when there is a true null hypothesis. Understanding the null hypothesis helps us grasp the importance of what we are testing against and why errors like Type I can occur.

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Most popular questions from this chapter

Exercise 4.26 discusses a sample of households in the US. We are interested in determining whether or not there is a linear relationship between household income and number of children. (a) Define the relevant parameter(s) and state the null and alternative hypotheses. (b) Which sample correlation shows more evidence of a relationship, \(r=0.25\) or \(r=0.75 ?\) (c) Which sample correlation shows more evidence of a relationship, \(r=0.50\) or \(r=-0.50 ?\)

Giving a Coke/Pepsi taste test to random people in New York City to determine if there is evidence for the claim that Pepsi is preferred.

Using the definition of a p-value, explain why the area in the tail of a randomization distribution is used to compute a p-value.

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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}

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