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

Short Answer

Expert verified
If null hypothesis is true in every case and 100 tests are conducted with \(5\%\) significance level, we would incorrectly find significance in nearly 5 tests.

Step by step solution

01

Understand Type I error and significance level

Firstly, understand the concept of Type I error and significance level in hypothesis testing. A Type I error occurs when a true null hypothesis is rejected, and the significance level of a test is the probability of committing a Type I error. It is represented by the Greek letter alpha (\(\alpha\)).
02

Apply significance level to calculate number of incorrect tests

Given that a significance level of \(5\%\) is used, this means that if the null hypothesis is true, we may incorrectly reject it \(5\%\) of the time simply due to sample variability. Applying this to the 100 tests conducted, we multiply the number of tests by the significance level to find the number of tests that we would expect to incorrectly find significance: \(100 \times 5\% = 5\).
03

Provide final answer

Therefore, if the null hypothesis is true in every case and we perform 100 tests with a significance level of \(5\%\), we would expect to incorrectly reject the null hypothesis in approximately 5 tests.

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Key Concepts

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

Type I Error
When we talk about a Type I error in the realm of statistics, we're addressing a specific kind of mistake. Imagine you've taken a buzzer to respond to true or false questions, and even though the statement is true, you hit false. This is analogous to a Type I error, where a correct null hypothesis is wrongly rejected.

In a statistical hypothesis test, the null hypothesis, denoted as H0, is essentially our default assumption. It's the status quo, such as 'this medicine has no effect' or 'the new program doesn't change test scores'. When we perform a test, we're seeing if the data provides enough evidence to dismiss this assumption. However, there's always a chance of getting misled by random chance, causing us to think we found proof when there was none – this is the notorious Type I error.

Referring to the exercise, with 100 hypothesis tests each at a 5% significance level, our expectation is that about 5 would lead us down the wrong path, giving us a false signal of significance when there's really nothing there. It's a reminder of the fallibility of statistical conclusions – they always come with a degree of uncertainty.
Significance Level
The significance level, typically denoted by \( \alpha \), is the threshold we set to determine when to declare statistical significance. Think of it like setting a filter on your email to catch spam – the filter catches most spam, but occasionally, a genuine message slips into the junk folder. The significance level is like the sensitivity of this filter, controlling our cautiousness in labeling results 'significant'.

Generally, a 5% significance level, or \( \alpha = 0.05 \), is the conventional standard. This means we're willing to accept a 5% chance of making a Type I error – of mistakenly hitting the buzzer for false. In the exercise, with 100 tests at this \( \alpha \) level, the math is straightforward. It's like planning that out of 100 emails, you're okay with incorrectly sending 5 to spam. It provides a balance, allowing for meaningful detection of effects while controlling the rate at which we make these Type I errors.
Null Hypothesis
At the core of any hypothesis test lies the null hypothesis (H0), which is the skeptic's starting point. It represents the statement or condition that indicates no effect or no difference. Suppose we're back in school, and there's a rumor that a new teaching method will improve grades. The null hypothesis is the equivalent of saying, 'This method won't make any difference.' It's what we aim to challenge with our data.

In the exercise scenario, the null hypothesis is assumed to be true for all 100 tests. This serves as a baseline from which we measure any deviation as either due to a true effect or simply random chance. Whenever we conduct a test, we seek evidence strong enough to cast doubt on the null hypothesis. If found, we can reject it in favor of an alternative hypothesis, which might claim, 'Yes, this method does boost grades.' But herein lies the possibility of committing a Type I error, mistaking randomness for actual evidence, as discussed earlier.
Statistical Significance
Have you ever experienced a moment when you're certain something special is happening, like when a shy friend speaks up and you think, ‘This is significant’? In statistics, we’re also seeking to identify such moments, but we need more than a hunch – we require evidence. Statistical significance is the conclusion that an observed effect is unlikely to be due to chance alone.

This is determined by p-values and the pre-determined significance level, with the widespread benchmark of 0.05 setting the standard for claiming significance. To say a result is statistically significant is to assert we have enough evidence to believe something noteworthy is occurring. In the context of our exercise, when we reach a significant result, it should mean we're at least 95% confident there's a real effect at play. However, it's vital to remember that statistical significance doesn't mean certainty – there's always room for a small proportion of those surprises, the Type I errors we're keen to keep in check.

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Most popular questions from this chapter

Exercise 2.19 on page 58 introduces a study examining whether giving antibiotics in infancy increases the likelihood that the child will be overweight. Prescription records were examined to determine whether or not antibiotics were prescribed during the first year of a child's life, and each child was classified as overweight or not at age 12. (Exercise 2.19 looked at the results for age 9.) The researchers compared the proportion overweight in each group. The study concludes that: "Infants receiving antibiotics in the first year of life were more likely to be overweight later in childhood compared with those who were unexposed \((32.4 \%\) versus \(18.2 \%\) at age 12 \(P=0.002) "\) (a) What is the explanatory variable? What is the response variable? Classify each as categorical or quantitative. (b) Is this an experiment or an observational study? (c) State the null and alternative hypotheses and define the parameters. (d) Give notation and the value of the relevant sample statistic. (e) Use the p-value to give the formal conclusion of the test (Reject \(H_{0}\) or Do not reject \(H_{0}\) ) and to give an indication of the strength of evidence for the result. (f) Can we conclude that whether or not children receive antibiotics in infancy causes the difference in proportion classified as overweight?

Polling 1000 people in a large community to determine if there is evidence for the claim that the percentage of people in the community living in a mobile home is greater then \(10 \%\).

An article noted that it may be possible to accurately predict which way a penalty-shot kicker in soccer will direct his shot. \({ }^{27}\) The study finds that certain types of body language by a soccer player \(-\) called "tells"-can be accurately read to predict whether the ball will go left or right. For a given body movement leading up to the kick, the question is whether there is strong evidence that the proportion of kicks that go right is significantly different from one-half. (a) What are the null and alternative hypotheses in this situation? (b) If sample results for one type of body movement give a p-value of 0.3184 , what is the conclusion of the test? Should a goalie learn to distinguish this movement? (c) If sample results for a different type of body movement give a p-value of \(0.0006,\) what is the conclusion of the test? Should a goalie learn to distinguish this movement?

In Exercise 3.129 on page \(254,\) we see that the home team was victorious in 70 games out of a sample of 120 games in the FA premier league, a football (soccer) league in Great Britain. We wish to investigate the proportion \(p\) of all games won by the home team in this league. (a) Use StatKeyor other technology to find and interpret a \(90 \%\) confidence interval for the proportion of games won by the home team. (b) State the null and alternative hypotheses for a test to see if there is evidence that the proportion is different from 0.5 . (c) Use the confidence interval from part (a) to make a conclusion in the test from part (b). State the confidence level used. (d) Use StatKey or other technology to create a randomization distribution and find the p-value for the test in part (b). (e) Clearly interpret the result of the test using the p-value and using a \(10 \%\) significance level. Does your answer match your answer from part (c)? (f) What information does the confidence interval give that the p-value doesn't? What information does the p-value give that the confidence interval doesn't? (g) What's the main difference between the bootstrap distribution of part (a) and the randomization distribution of part (d)?

Flying Home for the Holidays, On Time In Exercise 4.115 on page \(302,\) we compared the average difference between actual and scheduled arrival times for December flights on two major airlines: Delta and United. Suppose now that we are only interested in the proportion of flights arriving more than 30 minutes after the scheduled time. Of the 1,000 Delta flights, 67 arrived more than 30 minutes late, and of the 1,000 United flights, 160 arrived more than 30 minutes late. We are testing to see if this provides evidence to conclude that the proportion of flights that are over 30 minutes late is different between flying United or Delta. (a) State the null and alternative hypothesis. (b) What statistic will be recorded for each of the simulated samples to create the randomization distribution? What is the value of that statistic for the observed sample? (c) Use StatKey or other technology to create a randomization distribution. Estimate the p-value for the observed statistic found in part (b). (d) At a significance level of \(\alpha=0.01\), what is the conclusion of the test? Interpret in context. (e) Now assume we had only collected samples of size \(75,\) but got essentially the same proportions (5/75 late flights for Delta and \(12 / 75\) late flights for United). Repeating steps (b) through (d) on these smaller samples, do you come to the same conclusion?

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