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You are testing the null hypothesis \(p=0.4\) and will reject this hypothesis if \(z \star\) is less than -2.05. a. If the null hypothesis is true and you observe \(z \star\) equal to \(-2.12,\) which of the following will you do? (1) Correctly fail to reject \(H_{o} .\) (2) Correctly reject \(H_{o} .(3)\) Commit a type I error. (4) Commit a type II error. b. What is the significance level for this test? c. What is the \(p\) -value for \(z \star=-2.12 ?\)

Short Answer

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a. You will commit a Type I error because you reject the null hypothesis when it is true. b. The significance level for this test is 2.02%. c. The p-value for \(z^*=-2.12\) is 0.017.

Step by step solution

01

Decision about the Null Hypothesis

Compare the observed \(z^*\) (-2.12) with the critical value (-2.05). Since -2.12 is less than -2.05, reject the null hypothesis \(H_{o}\). As the true state of nature is that the null hypothesis is true (as given), this leads to a Type I error.
02

Calculation of Significance Level

The significance level is the probability of committing a Type I error, i.e., rejecting the null hypothesis when it is true. It corresponds to the values less than the critical value -2.05 in a standard normal distribution. In this case, use a standard normal table or a z-score calculator to find the area to the left of -2.05, which comes out to be 0.0202 or 2.02%.
03

Calculation of the p-value

The p-value is the smallest significance level that would lead to rejection of the null hypothesis, given the observed data. It is the probability of obtaining a result at least as extreme as the observed \(z^*\), assuming the null hypothesis is true. In this case, it would be the area to the left of \(z^*=-2.12\) in a standard normal distribution. Using a z-score calculator or standard normal table, this area comes out to be 0.017 which is the p-value for \(z^*=-2.12\).

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Key Concepts

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

Null Hypothesis
In hypothesis testing, the null hypothesis, denoted as \(H_0\), is a statement made for the sake of testing. It usually implies no effect or no difference in the context of the study. For instance, if you are checking whether a new treatment is effective, the null hypothesis might state that the treatment has no effect. It's the foundation on which the test is built.

When we talk about "rejecting the null hypothesis," it means the data provides sufficient evidence to conclude that the null hypothesis is unlikely to be true. Conversely, "failing to reject the null hypothesis" implies there isn't enough evidence to discount it.

A critical aspect of hypothesis testing is that you never "prove" the null hypothesis; you only determine whether you have enough evidence to reject it. In our exercise, the null hypothesis, \(p=0.4\), implies that the proportion parameter equals 0.4.
Type I Error
A Type I error occurs when you reject the null hypothesis when it is actually true. This can be seen as a "false positive" result. It's one of the terms often used and it reflects mistakes made in the decision-making process.

Consider the context of a courtroom, where a Type I error would be analogous to convicting an innocent person. In our exercise, by rejecting the null hypothesis based on the \(z^*\) value that we calculated, we would make a Type I error if in reality, the null hypothesis was indeed true.

The risk of committing this type of error is represented by the significance level of the test. Being aware of the possibility of Type I errors is crucial since it can lead to unwarranted conclusions which may have practical implications.
Significance Level
The significance level, often denoted as \(\alpha\), is the threshold for decision-making in hypothesis testing. It's the probability of rejecting a true null hypothesis, which corresponds to the likelihood of making a Type I error.

In simpler terms, if you decide on a 5% significance level, you are accepting a 5% risk of incorrectly rejecting the null hypothesis when it is true.

From our exercise, determining the significance level meant calculating the probability of observing a \(z^*\) value even smaller than the critical value \(-2.05\). This came out to be 2.02%, which tells us that there is a 2.02% chance of making a Type I error by rejecting \(H_0\) when it's true. It's a vital component in the critical view of the testing process.
p-value
The p-value is an essential component in hypothesis testing. It quantifies the evidence against the null hypothesis. Essentially, the p-value is the probability of obtaining a result that is at least as extreme as the one you got, assuming the null hypothesis is true.

A smaller p-value suggests stronger evidence in favor of rejecting the null hypothesis. For instance, a p-value of 0.017 in our exercise signifies that there's a 1.7% chance of obtaining such a \(z^*\) value or more extreme, under the assumption that the null hypothesis is true.

The decision to reject or fail to reject the null hypothesis is often based on comparing the p-value to the significance level. If the p-value is smaller than the defined significance level, the null hypothesis is rejected, indicating that the observed data is inconsistent with the null hypothesis.

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Most popular questions from this chapter

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How important is the assumption "The sampled population is normally distributed" to the use of Student's \(t\) -distribution? Using a computer, simulate drawing 100 samples of size 10 from each of three different types of population distributions, namely, a normal, a uniform, and an exponential. First generate 1000 data values from the population and construct a histogram to see what the population looks like. Then generate 100 samples of size 10 from the same population; each row represents a sample. Calculate the mean and standard deviation for each of the 100 samples. Calculate \(t \neq\) for each of the 100 samples. Construct histograms of the 100 sample means and the 100 t \(\star\) values. (Additional details can be found in the Student Solutions Manual.) For the samples from the normal population: a. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. b. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the rectangular or uniform population: c. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. d. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. For the samples from the skewed (exponential) population: e. Does the \(\bar{x}\) distribution appear to be normal? Find percentages for intervals and compare them with the normal distribution. f. Does the distribution of \(t \star\) appear to have a \(t\) -distribution with df \(=9 ?\) Find percentages for intervals and compare them with the \(t\) -distribution. In summary: g. In each of the preceding three situations, the sampling distribution for \(\bar{x}\) appears to be slightly different from the distribution of \(t \star .\) Explain why. h. Does the normality condition appear to be necessary in order for the calculated test statistic \(t \star\) to have a Student's \(t\) -distribution? Explain.

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