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Determine the \(p\) -value for each of the following hypothesis-testing situations. a. \(H_{o}: p=0.5, H_{a}: p \neq 0.5, z \star=1.48\) b. \(H_{o}: p=0.7, H_{a}: p \neq 0.7, z \star=-2.26\) c. \(H_{o}: p=0.4, H_{a}: p>0.4, z \star=0.98\) d. \(H_{o}: p=0.2, H_{a}: p<0.2, z \star=-1.59\)

Short Answer

Expert verified
a. The p-value is approximately 0.139 \n b. The p-value is approximately 0.024 \n c. The p-value is approximately 0.164 \n d. The p-value is approximately 0.056.

Step by step solution

01

Identify the Hypothesis and Test Statistic

Identify the null hypothesis (\(H_{o}\)), the alternative hypothesis (\(H_{a}\)), and the test statistic \(z^{*}\) in the problem. For example in the first scenario: \n\na. The null hypothesis (\(H_{o}\)) is that \(p = 0.5\), the alternative hypothesis (\(H_{a}\)) is that \(p \neq 0.5\), and the test statistic \(z^{*}\) is \(1.48\).
02

p-value for the two-tailed hypothesis

For a two-tailed scenario like (a) and (b), we determine the p-value by finding the probability that a standard normal random variable is more extreme than \(z^{*}\) in both tails. This is calculated as the area under the curve outside of \(|z^{*}|\), which can be found using a Z-table or a calculator with normal distribution functions. Then, the p-value is twice this area.
03

p-value for the one-tailed hypothesis

For a one-tailed scenario like (c) and (d), we determine the p-value by finding the probability that a standard normal random variable is more extreme than \(z^{*}\) in one tail. This is calculated as the area under the curve either greater than \(z^{*}\) for right-tailed tests, or less than \(z^{*}\) for left-tailed tests. This area can be found using a Z-table or a calculator with normal distribution functions.

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Key Concepts

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

P-value Calculation
The p-value is a crucial concept in hypothesis testing. It helps us determine the significance of our test results. In simple terms, the p-value measures the probability of obtaining a test statistic at least as extreme as the one observed, assuming that the null hypothesis is true. This allows us to decide whether to reject the null hypothesis or not. A smaller p-value indicates stronger evidence against the null hypothesis.

Calculating the p-value involves a few steps:
  • Identify the type of hypothesis test (one-tailed or two-tailed).
  • Locate the test statistic on the standard normal distribution curve.
  • Find the probability associated with that statistic, often using a Z-table or normal distribution calculator.
The end result is a probability value, which you can use to make informed decisions regarding your hypothesis.
Two-tailed Hypothesis
A two-tailed hypothesis test is used when you want to check for deviations in both directions, either lower or higher than a specified value. This type of test applies when you have non-directional alternative hypotheses, such as testing if a parameter is not equal to a certain value.

For example, in the problem scenario where the null hypothesis is that a proportion is equal to 0.5 (\(H_{o}: p=0.5\)) and the alternative hypothesis is not equal to 0.5 (\(H_{a}: p eq 0.5\)), you would use a two-tailed test.

To calculate the p-value for a two-tailed test, you:
  • Find the area under the standard normal curve that lies beyond the absolute value of the test statistic on both sides.
  • Multiply this area by 2, as you are considering both tails.
Using the example of a test statistic of 1.48, you would find the p-value by determining the area of the curve beyond 1.48 and -1.48, then doubling it to reflect both tails.
One-tailed Hypothesis
A one-tailed hypothesis test is appropriate when your alternative hypothesis specifies a direction, either greater than or less than the null hypothesis. This type of test is directional and focuses on one side of the distribution.

For illustration, consider a scenario with the null hypothesis that a proportion is 0.4 (\(H_{o}: p=0.4\)) and the alternative hypothesis is greater than that value (\(H_{a}: p>0.4\)). Here, a one-tailed test is conducted.

To find the p-value for a one-tailed test:
  • Identify whether it's a right-tailed or left-tailed test.
  • Calculate the area under the normal distribution curve that is more extreme than the test statistic.
For a right-tailed test, you look at the area to the right of the test statistic. For a left-tailed test, you look at the area to the left. This area yields the p-value.
Test Statistic Identification
Identifying the test statistic is a fundamental step in hypothesis testing. The test statistic measures how far the sample statistic is from the null hypothesis parameter, in standard error units. In practice, it helps determine how extreme the sample result is.

The test statistic is calculated from your sample data and can be compared against a known distribution, like the standard normal distribution. For population proportions, the test statistic is often denoted by \(z^*\), which is why these tests are sometimes referred to as Z-tests.

To identify the test statistic, you need:
  • The sample proportion.
  • The hypothesized population proportion.
  • The standard error of the sample proportion.
Once these values are known, the test statistic is calculated, providing a basis to determine the p-value and make a decision on the hypothesis. This process is crucial in deciding if the observed data significantly deviates from what was expected under the null hypothesis.

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Most popular questions from this chapter

In the past the standard deviation of weights of certain 32.0 -oz packages filled by a machine was 0.25 oz. A random sample of 20 packages showed a standard deviation of 0.35 oz. Is the apparent increase in variability significant at the 0.10 level of significance? Assume package weight is normally distributed. a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

It has been suggested that abnormal male children tend to be born to older- than-average parents. Case histories of 20 abnormal males were obtained, and the ages of the 20 mothers were as follows: $$\begin{array}{lllllllllll}31 & 21 & 29 & 28 & 34 & 45 & 21 & 41 & 27 & 31 \\\43 & 21 & 39 & 38 & 32 & 28 & 37 & 28 & 16 & 39\end{array}$$ The mean age at which mothers in the general population give birth is 28.0 years. a. Calculate the sample mean and standard deviation. b. Does the sample give sufficient evidence to support the claim that abnormal male children have olderthan-average mothers? Use \(\alpha=0.05 .\) Assume ages have a normal distribution.

The chief executive officer (CEO) of a small business wishes to hire your consulting firm to conduct a simple random sample of its customers. She wants to determine the proportion of her customers who consider her company the primary source of their products. She requests the margin of error in the proportion be no more than \(3 \%\) with \(95 \%\) confidence. Earlier studies have indicated that the approximate proportion is \(37 \%\). a. What is the minimum size of the sample that you would recommend to meet the requirements of your client if you use the earlier results? b. What is the minimum size of the sample that you would recommend to meet the requirements of your client if you ignore the earlier results? c. Is the approximate proportion of value needed in conducting the survey? Explain.

a. What value of chi-square for 5 degrees of freedom subdivides the area under the distribution curve such that \(5 \%\) is to the right and \(95 \%\) is to the left? b. What is the value of the 95 th percentile for the chi-square distribution with 5 degrees of freedom? c. What is the value of the 90th percentile for the chi-square distribution with 5 degrees of freedom?

Just one serving a month of kale or collard greens or more than two servings of carrots a week can reduce the risk of glaucoma by more than \(60 \%,\) according to a UCLA study of 1000 women. Using a \(90 \%\) confidence interval for the true binomial proportion based on this random sample of 1000 binomial trials and an observed proportion of \(0.60,\) estimate the proportion of glaucoma risk reduction in women who eat the recommended servings of kale, collard greens, or carrots.

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