Students of statistics often have the impression that this course is all about cold, hard facts. Nothing could be further from the truth: statistics is a direct extension of philosophy of science. The numbers we calculate here only have relevance to real-world research questions thanks to some complex philosophical arguments. While most of these are outside the scope of the present course, we want you to be familiar with the main concepts before moving on.
6.1 Deduction and Induction in Hypothesis Testing
Hypothesis testing has roots in the philosophy of logic, or correct reasoning. In logic, arguments consist of a set of premises, which can be true or false, that together lead to a conclusion, which can also be true or false. The most famous example is:
Premise: All men are mortal
Premise: Socrates is a man
Conclusion: Therefore, Socrates is mortal.
This is a deductive argument, which has the property that if the premises are true, then the conclusion must also be true. Deductive arguments are also often thought of as arguments from the general to the specific. In this case, the general rule “all men are mortal” gives rise to the specific claim that one specific man, Socrates, is also mortal.
The counterpart to deductive reasoning is inductive reasoning, which proceeds from specific observations or claims to general rules. The most famous example is:
Premise: All swans I have ever seen are white
Conclusion: Therefore, all swans are white.
Unlike the deductive argument above, however, inductive reasoning does not guarantee that true premises always produce true conclusions. Even if it is true that I have only seen white swans, the conclusion is false - black swans exist. David Hume introduced another famous example:
Premise: The sun has risen in the east every morning up until now.
Conclusion: The sun will also rise in the east tomorrow.
Here, too, the conclusion is not supported by the premise. This might make us feel uncomfortable - which sane person would reject the conclusion that the sun will rise in the east tomorrow? This discomfort illustrates that people are naturally inclined to reason inductively. As scientists, however, we should be very cautious to remember that this way of reasoning is not guaranteed to produce true conclusions.
Inductive and deductive reasoning are both used in statistical hypothesis testing. Deduction is used when we derive a specific prediction (hypothesis) from a general theory. For example, we could say that:
Premise: More time spent studying causes better grades.
Conclusion: In my dataset, time spent studying should correlate positively with grades.
If the premise is true, then we would expect to observe the corresponding pattern in our data.
Induction comes into play when we draw general conclusions from observed data.
Premise: I observed a positive correlation between time spent studying and grades in my dataset
Conclusion: Therefore, time spent studying causes better grades.
This conclusion does not logically follow from the premise. The problem is not resolved by removing the word “causes”:
Premise: I observed a positive correlation between time spent studying and grades in my dataset
Conclusion: Therefore, time spent studying correlates positively with grades in the general population of students.
It follows that we can never conclusively “prove” general conclusions from specific observations. The philosopher David Hume wrote extensively about this “problem of induction”: How can we justify the assumption that unobserved cases will follow the same patterns as observed ones? Hume argued that this assumption cannot be logically justified. Our sense that generalization is justified might be based on intuition, but not logic. The problem of induction challenges the very foundation of science. We cannot escape the use of induction when seeking to learn general insights from specific observations, but Hume showed that induction lacks a purely rational justification.
6.2 Falsificationism
It follows from the problem of induction that it is impossible to definitively prove a theory to be true. No matter how much evidence I have observed that supports a theory (white swans), all it takes is one refuting observation (black swan) to reject is. Karl Popper sought to avoid the problem of induction by devising a scientific method that relies exclusively on deduction: falsificationism. Popper demarcated the distinction between “pseudo-science” and science by arguing that “[scientific] statements […] must be capable of conflicting with […] observations” (Popper 1962, 39). The core business of science, according to Popper, should be to try to reject theories. Note that - while Popper’s work has been heavily criticized (for good reasons), his work is very influential in social science. Therefore, Popper is a good starting point for our course - even though he probably should not be the endpoint for students with a genuine interest in philosophy of science.
6.3 Falsificationism and Hypothesis Testing
The idea of falsificationism has been very influential in applied statistics in the social sciences, particularly in the practice of “Null-Hypothesis Significance Testing” (NHST). In NHST, researchers proceed as follows:
Develop a testable proposition about a population parameter; for example:
“On average, my students understand the course material. Their mean grade is \(\mu \geq 6\)”.
“On average, there is a positive association between hours studied and grade, \(\rho \geq 0\)”.
Develop a second hypothesis whose sole purpose is to be rejected, to pay lip service to falsificationism. Call this the “null hypothesis”. The “null hypothesis” is often taken to be the exact opposite of the researcher’s true belief, or “alternative hypothesis”:
“On average, my students DO NOT understand the course material. Their mean grade is \(\mu < 6\)”.
“On average, there is NO positive association between hours studied and grade, \(\rho < 0\)”.
Execute a procedure to make a decision to reject (falsify) or not reject the null hypothesis (next chapters).
If the null hypothesis is rejected, act as if this finding supports the alternative hypothesis.
As argued by Andrew Gelman in this blog post, this approach only pays lip service to falsificationism. A true falsificationist would put their true belief (alternative hypothesis) to the test. Fake falsificationism is creating a meaningless, “straw man” null-hypothesis, whose sole purpose is to be rejected.
6.4 Moving Forward
Where does this leave us? The most important point is that there are important limitations to commonly used methods, including null-hypothesis significance testing. There is no real satisfying solution. Just keep in mind that no statistical test can give evidence in support of a theory or hypothesis; neither a null nor an alternative hypothesis.
6.5 Causality
Another crucial philosophical issue relevant for statistics is the question of causality. Scientists are often interested in causal questions. We assume causality, for example, any time we want to act on knowledge derived from scientific research. For example, say that I do find a strong correlation between hours studied and grade obtained. If, based on this finding, you increase your study hours in order to improve your grade - then you are assuming causality. The same applies for governments making evidence-based policy, companies adjusting sales strategies based on customer analytics, or drugs that replenish a particular neurotransmitter that has been found lacking in patients with a specific diagnosis.
Despite the fact that causality is so important in scientific research, it is rarely defined. Contemporary definitions are typically based on counterfactuals: A is a cause of B if B would not have happened if A had not happened. This definition has important limitations, but it is sufficient for our course (Halpern, 2015).
Most people have heard the phrase “correlation does not imply causation”. What does this mean? One important misunderstanding is that the correlation coefficient is an inappropriate statistic for investigating causal research questions. This is not the case. This phrase warns us that, just because observe a statistical association between variables X and Y (for example, a correlation of \(r = .43\)), that does not mean that X caused Y. Importantly, observing this correlation is consistent with a causal effect of X on Y - but also with other explanations. For example, maybe Y caused X, or maybe a third variable caused both X and Y, and that is why they are correlated.
The problem is related to the previous section: observing a pattern in data that is consistent with a causal association between X and Y cannot conclusively prove that X caused Y, no matter what statistic you use to describe the pattern.
So where does causality come from? The short answer is: from theory or methodology. Causality can be assumed on theoretical grounds, or established using a randomized controlled experiment. In such an experiment, researchers randomly assign participants to either an experimental condition (e.g., receiving a drug, instruction, treatment, et cetera), or a control condition (e.g., receiving a placebo, no instruction, non-effective treatment, et cetera). The random assignment should, theoretically, result in two groups with no systematic differences that could explain between-group differences in the outcome of interest. Of course, due to pure chance, it could happen that there are systematic differences (more men in one group, taller people in one group, et cetera). But there is no procedure that has a better chance of resulting in comparable groups than random assignment.
In the social sciences, experiments are not always feasible or ethical, so researchers often use observational data. Does this preclude all causal claims? It does not. It is perfectly legitimate to present a theory that predicts a causal effect in the Introduction section of your scientific writing, and then present empirical data that show a pattern consistent with that effect. For example, if I assume that hours studied causes improved grades, a correlation of \(r = .43\) is consistent with that assumption. An alternative explanation might be that having receiving high grades in the past has motivated some students to study harder. Just make sure not to take the observed data as evidence for a causal effect.
What is required to argue that X has a causal effect on Y? Several philosophers have addressed this issue; most notably Hume and Mill (see Morabia, 2013). The necessary conditions for causality are sometimes summarized as association, temporal precedence, and non-spuriousness. Below, each is supported with quotes from Hume (1902)1:
Association: Cause and effect must be associated (this could be statistical association)
“When one particular species of event has always […] been conjoined with another, we […] call the one object, Cause; the other, Effect.” (VII, II, 59)
“familiar objects or events to be constantly conjoined together” (V, I, 35) There must be a “constant union” between cause and effect, and they must be “contiguous in space and time” (Hume 1739,16, pp. 173–175).
Temporal precedence: The cause must occur before the effect.
observe a continual succession of objects, and one event following another (V, I, 35)
“[when] the same object is always followed by the same event; we then begin to entertain the notion of cause and connexion.” (VII, II, 61)
Non-spuriousness: All alternative explanations of the effect are excluded.
“Their conjunction may be arbitrary and casual. There may be no reason to infer the existence of one from the appearance of the other.” (describing spuriousness; V, I, 35)
“we may define a cause to be an object, followed by another, and where all the objects similar to the first are followed by objects similar to the second. Or in other words where, if the first object had not been, the second never had existed.” (VII, II, 61)
This latter definition resembles the aforementioned counterfactual definition of causality. Note that this definition does not require randomized experimentation, but randomized experiments do help us meet all three criteria. The field of causal inference focuses on developing methods that can estimate causal effects (like you would get from a randomized controlled experiment) from observational data (Pearl, 2009).
7 Lecture
TO DO
8 Formative Test
A formative test helps you assess your progress in the course, and helps you address any blind spots in your understanding of the material. If you get a question wrong, you will receive a hint on how to improve your understanding of the material.
Complete the formative test ideally after you’ve seen the lecture, but before the lecture meeting in which we can discuss any topics that need more attention
Question 1
What is the main difference between deductive and inductive reasoning?
Question 2
Which of the following is an example of inductive reasoning?
Question 3
What is the role of deduction in statistical hypothesis testing?
Question 4
What was Karl Popper’s proposed solution to the problem of induction?
Question 5
In Null-Hypothesis Significance Testing (NHST), what role does the null hypothesis play?
Question 6
What does the phrase ‘correlation does not imply causation’ mean?
Question 7
According to a common interpretation of Hume, which three conditions are necessary for causality?
Question 8
How do randomized controlled experiments support causal inference?
Question 1
Deductive reasoning leads to necessarily true conclusions if the premises are true, while inductive reasoning involves probable conclusions based on patterns.
Question 2
Assuming that the sun will rise tomorrow because it always has is an example of induction, and it is not logically justified.
Question 3
Deduction is used to derive derive specific, testable predictions from general theories, such as expecting a correlation between study time and grades if studying improves performance.
Question 4
Popper argued that science should focus on rejecting falsifiable hypotheses rather than confirming them, avoiding the pitfalls of induction.
Question 5
The null hypothesis is a ‘straw man’ hypothesis, only intended to be rejected. This interpretation of Popper’s falsificationism actually never submits theories to the test, and aligns more closely with confirmationism.
Question 6
No statistical finding can prove causation, so a correlation (or other type of association) between two variables doesn’t confirm a causal link.
Question 7
A common interpretation of Hume is that for a cause-effect relationship, the variables must be associated, the cause must precede the effect, and other explanations must be ruled out.
Question 8
Random assignment maximizes the probability that the treatment and control groups do not differ on any confounding variables, thus ensuring nonspuriousness. Furthermore, a well-designed experiment ensures association (if there is an effect) and temporal precedence.
Discuss with your portfolio group the logic behind hypothesis testing, and how it relates to your personal (and group’s) research interests.
Consider the following three research descriptions. Formulate H0 and H1 in words. Discuss your answers with your group members.
Researchers want to know whether it matters for test performance if an exam is completed on a computer or using paper and pencil. Hence, the research question reads: Is there an effect of the type of administration (computer or paper and pencil) on the test performance?
What would be the H0 and HA for this study?
This appears to be an undirected hypothesis about a mean difference for two independent samples, without a clearly specified alternative hypothesis. Thus, we could state:
Researchers want to know whether the alcohol consumption among Dutch students differs from the alcohol consumption in the general Dutch population. Using CBS statistics, they know that in the general population the average alcohol consumption is 5.6 glasses a week. The question is whether the average alcohol consumption among students is different from this national average.
What would be the H0 and H1 for this study?
This appears to be an undirected hypothesis about the difference between a mean and a hypothesized value, without a clearly specified alternative hypothesis. Thus, we could state:
\(H_0: \mu = 5.6\)\(H_A: \mu \neq 5.6\)
Researchers want to study whether social isolation is associated with income.
What would be the H0 and H1 for this study?
This appears to be an undirected hypothesis about an association between two variables, without a clearly specified alternative hypothesis. We could thus state:
\(H_0: \rho = 0\)\(H_A: \rho \neq 0\)
Formulating the hypothesis is an important very first step in hypotheses testing. Continue with the next assignment, in which we will go through the steps of a hypothesis test.
9.2 Assignment 2: Test Statistics, Alpha and Significance
In this assignment we will go through the steps of a hypothesis test.
While going through the steps we will come across the most important concepts related to hypothesis testing.
For the next steps, we consider the following situation:
Suppose we are interested in the personality profile of musicians; that is, we want to know whether, on average, personality characteristics of musicians differ from those of the general population. For now, we’ll only focus on Openness. We pretend that we have collected data among 25 musicians using a validated scale for which previous research has shown that in the general population the scores are normally distributed with mean 50 and SD 15. It is our task to test whether the mean of Openness for musicians differs from the mean in the general population. To keep things simple, we assume that in the population of musicians the SD is the same as in the general population; that is, we assume that LaTeX: \(\sigma = 15\).
Let openness be the variable of interest. Let \(\mu_{musicians}\) represent the mean openness in the population of musicians. The hypothesis test amounts to testing:
\(H0: \mu_{musicians} = 50\)
\(H1: \mu_{musicians} \neq 50\)
Now, when we do the hypothesis test, we seek for evidence against the null hypothesis. More specifically, our testing procedure starts with the assumption that H0 represents the truth and as long as we don’t have convincing evidence that our assumption is false we stick to that assumption.
The question is, however, when do we have convincing evidence against H0?
Finding evidence against H0 works as follows:
If H0 is true, we expect mean values close to H0. And, if we observe a mean value that is much different from the value under H0, we have convincing evidence against H0. If this happens, we reject H0 as representing the truth and accept the alternative hypothesis, H1.
Hypothesis testing fits Popper’s philosophy of falsification. He introduced this well-known analogy to explain the logic of falsificationism:
Suppose we assume that all Swans are white, \(H_0: Swans = white\)
We would then not expect to observe black ones.
If we do observe black swans, our initial hypothesis is called into question.
The number of white swans we see (= observations consistent with the hypothesis) does not provide evidence for \(H_0\), because there could always be a black swan out there we haven’t observed yet.
So, the next questions are:
What are the sample values we can expect under H0? When is evidence “convincing” enough? To answer the first question we have to go back to sampling distributions!
For the second question, we need a criterion. We have to realize that even if H0 is true, sample values can be far off just by sampling fluctuations (i.e., by chance). The common criterion is: if the observed value is among the 5% most unlikely samples under H0 (i.e. if H0 is true), we reject the null hypothesis.
Let’s go back to our example about musicians.
Let X be openness. Under H0 we assume that X is normally distributed with mean 50 and SD equal to 15.
What are the mean and standard deviation of the sampling distribution of the mean under H0 given that the sample size is 25? And what do we call the standard deviation of the sampling distribution?
(Use what you have learned in the previous lectures. Hint: first make a drawing of the situation, then do the computations).
Sampling distribution:
Mean: \(\mu = 50\)
Standard error ( =SD of sampling distribution!): \(\sigma_{\bar{X}} = \frac{15}{\sqrt{25}}= 3\)
Suppose we want to indicate sample means that are unlikely if H0 would be true. In particular, we want to know how far the sample mean must be from the hypothesized mean to be among the 5% of all possible samples under H0 that are furthest away from the hypothesized means.
What should the value of the sample mean be to fall within the 5% most deviant samples if the sample size is 25?
We are talking about the distribution of the mean; so we need to work with the sampling distribution. We want to know the cut offs that mark the 2.5% highest and 2.5% lowest means. We first have to find the Z-values: they are 1.96 for the highest 2.5%, and (by symmetry) -1.96 marks the 2.5% lowest.
Hence, to be among the 5% of all possible sample means that are most unlikely under H0, the sample mean should be:
larger than 50+1.96 x 3 = 55.88 or smaller than 50-1.96 x 3 = 44.12
Let’s do some more exercises on the Z-test.
Suppose the mean for Openness we found in our sample was 59.
If we use a significance level of 5%, would we reject the null hypothesis?
In the previous step we used cut offs for the sample means to decide about significance. The cut off scores were obtained via the Z-distribution. However, doing all these computations is not necessary (there’s a shortcut!!). In fact, if we know the Z-value for the sample, we can easily find out if the sample is among the 5% of the most unlikely sample means. We only have to compare the value with 1.96 and -1.96 to see whether that is the case.
In this course, we will use Z-values for different purposes. In these specific calculations, Z is used as a Test Statistic. A test statistic quantifies evidence against the null hypothesis. In this case, the Z test statistic expresses how far away from the mean under the null hypothesis the observed mean is, in terms of the number of standard errors.
The Z test-statistic follows the standard normal distribution. The values 1.96 and -1.96 are called the critical values and they mark the 5% most unlikely sample means under H0. In other words, the critical values mark the reject region for H0.
So, if we compute the Z-value for the sample mean, and if that sample value of Z falls in the rejection region, we reject H0 (we found something that is unlikely enough to no longer believe H0 is true). If H0 is rejected we speak of a significant result. See the graph below:
Following these steps to test a mean is one example of performing a “Z-test”!
We can use the Z-test to test hypotheses about the population mean if we know the population \(\sigma\).
This statistic is computed using the mean from the sample, the hypothesized mean under H0 and \(\sigma\).
H0 is rejected at the 5% significance level if z is either larger than 1.96 or smaller than -1.96.
So far, we rejected the null hypothesis if the sample is among the 5% most unlikely sample means under H0. This 5% was called the significance level, and is denoted as \(\alpha = .05\). However, we could just as well choose 1% or .5%.
What would be the critical values for the Z-test if one tests at \(\alpha = .01\)?
What would be the critical values for the Z-test if one tests at \(\alpha = .005\%\)?
For historical reasons, social scientists tend to use \(\alpha = 0.05\) as a default. So in this course, if alpha is not explicitly stated, assume \(\alpha = 0.05\).
When we test hypotheses we reject H0 if the sample we find is unlikely if H0 is true. However, the flip side is that, even though H0 is true, we may find a sample that is much different by chance, and erroneously reject H0. Or, in other words, we could make an error. Rejecting H0 while it is true in reality is called a Type I error!
Consider the following:
If H0 is true, and you test at \(\alpha = 0.05\), what is the probability of committing a Type I error?
What is the link between the \(\alpha\)-level and type I error rate?
If H0 is really true (i.e., H0 should not be rejected), then the probability that the sample mean is among the 5% most unlikely is equal to 5%.
The alpha level specifies the risk of a Type I error. So if one tests at an alpha level of .05, it means that one accepts a risk of 5% to commit a Type I error.
Properties of the Z-test:
Used to test hypotheses about the mean in a population, assuming \(\sigma\) known.
The test-statistic equals \(z = \frac{\bar{X}-\mu}{\sigma_{\bar{X}}}\)
The test statistic is normally distributed.
9.3 Assignment 3: Z-test
In this assignment we will apply the Z-test.
This assignment first presents an example, followed by two practice questions.
A researcher wants to test \(H_0: \mu = 50\) against \(H_1: \mu \neq 50\)
Data are available from a random sample of 26 respondents. The mean was 53.7. The researcher assumes the SD in the population is 8.5. Perform all steps of the Z-test.
Step 1: Formulate hypotheses
\(H_0: \mu = 50\)\(H_1: \mu \neq 50\)
Step 2: Compute test statistic
Standard error: \(\frac{8.5}{\sqrt{26}}=1.667\)
Test statistic: \(z = \frac{53.7-50}{1.667}=2.212\)
Step 3: Decide about significance
\(\alpha = .05\), so critical values +/- 1.96.
Our test statistic exceeds this critical value.
The sample mean thus falls in the rejection region, and we should conclude that the test is significant so \(H_0\) s rejected.
Step 4: Draw conclusion
We have convincing evidence that the population mean differs from 50.
A researcher wants to test whether the population mean is equal to 80. Data are available from a random sample of 60 respondents. The mean was 74. The researchers assume the SD in the population is 40. Perform and report all steps of the Z-test. What is the resulting p-value?
A researcher wants to test whether the population mean is equal to 500. Data are available from a random sample of 75 respondents. The mean was 546. The researchers assume that the SD in the population is 200. Perform all steps of the Z-test. Use \(\alpha = .01\). Perform and report all steps.
standard error: \(\frac{200}{\sqrt{75}} = 23.094\)
test statistic: \(z = \frac{546-500}{23.094}=1.992\)
Step 3: Decide about significance.
Z does not exceed +/- 2.576. This means that Z does not fall in the reject region when tested at the 1% significance level. The test is not significant.
Step 4: Draw conclusion
\(H_0\) is not rejected.
9.4 Quiz
“The null and alternative hypothesis are deduced from the data.”
“When performing a hypothesis test, we start by assuming \(H_0\) is true.”
“If we reject \(H_0\) with \(\alpha=0.05\), then we will also reject it at \(\alpha=0.10\), assuming all other quantities are held constant.”
The critical values of \(\alpha =0.05\) are +/- 1.96. Hence, if \(H_0\) is rejected it means that z in the sample is larger than 1.96 or smaller than -1.96.”
The critical values of \(\alpha =0.1\) are +/- 1.645. This means that for rejecting \(H_0\) at this alpha level, that z should be larger than 1.645 or smaller than -1.645. That is implied by the fact that it exceeds +/- 1.96.
“If we reject \(H_0\), then \(H_0\) is surely wrong.”
We should always be aware of the possibility of making a Type I error. The probability of making a Type I error is equal to \(\alpha\).
“Increasing the sample size n (and holding all the rest constant) decreases the probability of a Type I error.”
Increasing the sample size n (and holding all the rest constant) does not decrease the probability of a Type I error.
The Type I error is determined by the alpha level.
If our sample is among the 5% most unlikely sample means of all possible sample means with the same size under \(H_0\), whatever that sample size N may be.
Increasing the sample size n (and holding all the rest constant) does not decrease the probability of a Type I error.
9.5 Assignment 4: Z-test and Alpha-levels
In this assignment we will practice some more with the Z-test, meanwhile we will review important concepts of hypothesis testing. In particular, we will look at significance levels.
To test hypotheses, we need to specify the “significance level”, usually denoted by \(\alpha\). The significance level is our decision criterion to reject H0.
The most common choice is .05. But what does this criterion exactly entail?
Discuss with your group what an \(\alpha\) level entails.
If we test at an \(\alpha\) of .05 it means that we are willing to reject H0 in favor of H1 if our sample mean belongs to the 5% most extreme scores (2.5% in each tail) under the null hypothesis.
If indeed the sample mean is among this 5%, it means that we have observed a sample in a range that is quite unlikely if the null hypothesis would be true and, therefore, justifies rejection of the null hypothesis.
In the previous assignments you already used the critical values for the Z-test for specific alpha levels.
For two-tailed tests, it holds that if the absolute value of Z exceeds the critical value, we may reject \(H_0\).
Let \(Z_\text{crit}\) be the critical value. For the Z-test it holds that:
\(Z_\text{crit} = 1.65\), if \(\alpha = 0.10\) (two-tailed)
\(Z_\text{crit} = 1.96\), if \(\alpha = 0.05\) (two-tailed)
\(Z_\text{crit} = 2.58\), if \(\alpha = 0.01\) (two-tailed)
9.6 Quiz
Researchers want to test whether \(\mu=70\). They assume that \(\sigma = 10\). Researchers found a mean of 72 in a random sample of 40 persons.
True or false:
\(H_0\) can be rejected at one of the three levels discussed above (\(\alpha = .10, .05, .01\).
“If the two-tailed test is significant at the 5% level, it will also be significant at the 1% level (keeping everything else fixed).”
“If the two-tailed test is not significant at the 10% level, it won’t be significant at the 5% level either (keeping everything else fixed).”
“If the two-tailed test is not significant at the 5% level, it could still be significant at the 10% level (keeping everything else fixed).”
“If the two-tailed test is significant at the 1% level, it might not be significant at the 5% level (keeping everything else fixed).”
9.7 Assignment 5: P-values
We will now focus on the interpretation of the p-values and how to use the p-values to decide about significance.
Consider the following situation:
Scores on a test measuring confidence in police are normally distributed in the general population, with \(\mu = 500\) and an \(\sigma = 50\). Researchers want to know if the average confidence level is different for those who have been a victim of crime. They collect data for 60 victims. They find a sample mean of 511. They test \(H_0: \mu = 500\) against \(H_1: \mu \neq 500\), while assuming that the population variance is \(\sigma = 50\).
Compute the p-value. Draw a graph for the two-tailed p-value. Write down in your own words and as precise as possible the interpretation of the p-value in the answer box below. Then, discuss your response with your group.
The p-value represents the proportion of all possible sample means that are further away from our hypothesized mean than the observed sample mean is.
We have the sampling distribution with \(\mu = 500\) and \(\sigma_{\bar{X}} = \frac{50}{\sqrt{60}} = 6.455\).
First, we compute the right-tail area: \(P(\bar{X} > 511) = P(Z > 1.70) = 0.0446\).
Hence, 4.66% of all possible samples is further away from \(H_0\) on the right side.
Second, we compute the left-tail area. These are the sample means that are more than 11 points from the hypothesized mean to the left \(P(\bar{X} < 489) = P(Z < -1.70) = 0.0446\).
Hence, the two-tailed p-value is 0.0892.
Is the test significant at the 5% level?
Is it significant at the 1% level?
Researchers test whether \(\mu = 90\). They assume that \(\sigma=21\). The sample mean was 85. Sample size was 50.
What is the two-tailed p-value?
What is the highest level at which the test is significant?
Researchers test whether \(\mu = 35\). They assume \(\sigma =16\). The sample mean was 38. Sample size was 64.
Compute the two-tailed p-value and indicate which of the following statements is true.
Consider these true- or false statements:
If a two-tailed p-value is .0567 then the test is significant at the 10% level but not at the 5% level.
If a two-tailed test is significant at the 5% level but not at the 1% level, then the two-tailed p-value will be less than 0.01.
A two-tailed p-value of 0.060 indicates that we have 6% chance that the null hypothesis is true.
Morabia, A. (2013). Hume, Mill, Hill, and the Sui Generis Epidemiologic Approach to Causal Inference. American Journal of Epidemiology, 178(10), 1526–1532. https://doi.org/10.1093/aje/kwt223
