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Chapter 9: Problem 86

Make the following hypothesis tests about \(p\). a. \(H_{0}: p=.57, \quad H_{1}: p \neq .57, \quad n=800, \quad \hat{p}=.50, \quad \alpha=.05\) b. \(H_{0}: p=.26, \quad H_{1}: p<.26, \quad n=400, \quad \hat{p}=.23, \quad \alpha=.01\) c. \(H_{0}: p=.84, \quad H_{1}: p>.84, \quad n=250, \quad \hat{p}=.85, \quad \alpha=.025\)

Short Answer

Expert verified
a. Do not reject \(H_0\). b. Do not reject \(H_0\). c. Reject \(H_0\) in favor of \(H_1\).

Step by step solution

01

Define Hypotheses and Parameters

For each part of the problem, the null hypothesis \(H_0\), the alternative hypothesis \(H_1\), the sample size \(n\), the estimated proportion \(\hat{p}\), and the significance level \(\alpha\) have been given. The null hypothesis represents the status quo, while the alternative hypothesis represents what we are testing for. We are given that the sample size and sample proportions used to estimate the true parameter \(p\). The alpha level is the threshold below which we reject the null hypothesis.
02

Compute Test Statistic

The test statistic for a proportion hypothesis test is a Z-score. The Z-score is calculated by the formula: \(Z = \frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}\). In this formula, \(p_0\) is the predicted proportion under the null hypothesis, \(\hat{p}\) is the sample proportion, and \(n\) is the sample size.
03

Compare Test Statistic to Critical Value

For a two-tailed test (such as part a), we reject the null hypothesis if the Z score is greater than +1.96 or less than -1.96. For a one-tailed test where \(H_1: p < p_0\) (such as part b), we reject the null hypothesis if the Z score is less than -2.33. For a one-tailed test where \(H_1: p > p_0\) (such as part c), we reject the null hypothesis if the Z score is greater than +1.96.
04

Decision and Conclusion

If the test statistic falls in the rejection region, we reject the null hypothesis in favor of the alternative hypothesis. If not, we do not reject the null hypothesis. Whether we reject or not reject the null hypothesis gives us our final conclusion.

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Key Concepts

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

Null Hypothesis
The null hypothesis, denoted as \(H_0\), is a fundamental concept in hypothesis testing. It represents a statement of no effect or no difference that we aim to test against. In our examples from the exercise, the null hypotheses claim certain proportions for a population parameter \(p\).
Examples include:
  • \(H_0: p = 0.57\) which suggests that the population proportion \(p\) equals 0.57.
  • \(H_0: p = 0.26\) indicating the population proportion \(p\) is 0.26.
  • \(H_0: p = 0.84\) stating the population proportion \(p\) is 0.84.
It is important to note that the null hypothesis is assumed true until evidence suggests otherwise. This assumption allows statisticians to use data to test whether the evidence strongly contradicts \(H_0\). If it does, we proceed to reject the null hypothesis.
Alternative Hypothesis
Opposite to the null hypothesis is the alternative hypothesis, represented as \(H_1\). This hypothesis indicates a specific change or effect that the researcher wants to detect. Each scenario in the exercise showcases different alternative hypotheses:
  • In part a, \(H_1: p eq 0.57\) suggests that the population proportion is not equal to 0.57.
  • In part b, \(H_1: p < 0.26\) means we suspect the population proportion is less than 0.26.
  • In part c, \(H_1: p > 0.84\) which implies we are testing if the population proportion exceeds 0.84.
The choice of alternative hypothesis affects how the statistical test is conducted. If it includes "\(eq\)", it indicates a two-tailed test, otherwise a one-tailed test is used depending on whether \(p\) is hypothesized to be less than or greater than \(p_0\).
Proportions Test
The proportions test is a type of statistical test used to determine if there is a significant difference between observed sample proportions \(\hat{p}\) and the population proportion under the null hypothesis \(p_0\). In such tests, the Z-test is typically employed for large samples to calculate the test statistic.
This is expressed mathematically as: \[Z = \frac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}\] where:
  • \(\hat{p}\) is the sample proportion.
  • \(p_0\) is the assumed population proportion from the null hypothesis.
  • \(n\) is the sample size.
Using this test statistic, we can determine how far off the sample proportion is from the claimed population proportion. The result informs us if the deviation is significant enough to question the validity of the null hypothesis.
Significance Level
The significance level, denoted as \(\alpha\), is a threshold set by researchers that defines the cutoff for rejecting the null hypothesis. It represents the probability of making a Type I error, which is rejecting a true null hypothesis. Typical values for \(\alpha\) are 0.05, 0.01, and 0.025, corresponding to 5%, 1%, and 2.5% probabilities of incorrectly rejecting \(H_0\).
  • In part a, \(\alpha = 0.05\), indicating a 5% risk of mistakenly rejecting \(H_0: p = 0.57\).
  • In part b, \(\alpha = 0.01\), which shows a 1% risk for \(H_0: p = 0.26\).
  • In part c, \(\alpha = 0.025\), underlines a 2.5% risk associated with \(H_0: p = 0.84\).
Once the Z-score is computed, it's compared to critical values derived from \(\alpha\). The null hypothesis is rejected if the Z-score falls into the critical region beyond these thresholds.

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Most popular questions from this chapter

A past study claimed that adults in America spent an average of 18 hours a week on leisure activities. A researcher wanted to test this claim. She took a sample of 12 adults and asked them about the time they spend per week on leisure activities. Their responses (in hours) are as follows. \(\begin{array}{lllllllllll}13.6 & 14.0 & 24.5 & 24.6 & 22.9 & 37.7 & 14.6 & 14.5 & 21.5 & 21.0 & 17.8 & 21.4\end{array}\) Assume that the times spent on leisure activities by all adults are normally distributed. Using the \(10 \%\) significance level, can you conclude that the average amount of time spent on leisure activities has changed?

As reported on carefair.com (November 15,2006\(), 40 \%\) of women aged 30 years and older would rather get Botox injections than spend a week in Paris. In a recent survey of 400 women aged 65 years and older, 108 women would rather get Botox injections than spend a week in Paris. Using a \(10 \%\) significance level, perform a test of hypothesis to determine whether the current percentage of women aged 65 years or older who would rather get Botox injections than spend a week in Paris is less than \(40 \%\). Use both the \(p\) -value and the critical-value approaches.

A random sample of 200 observations produced a sample proportion equal to \(.60 .\) Find the critical and observed values of \(z\) for each of the following tests of hypotheses using \(\alpha=.01\). a. \(H_{0}: p=.63\) versus \(H_{1}: p<.63\) b. \(H_{0}: p=.63\) versus \(H_{1}: p \neq .63\)

According to the most recent Bureau of Labor Statistics survey on time use in the United States, the average U.S. man spends \(67.20\) minutes per day eating and drinking. Suppose that a survey of 43 Norwegian men resulted in an average of \(81.10\) minutes per day eating and drinking [Note: This value is consistent with the data in a report by the Organization for Economic Cooperation and Development (Source: http://economix.blogs.nytimes.com/2009/05/05/obesity-and- the-fastness-of-food/)]. Assume that the population standard deviation for all Norwegian men is \(18.30\) minutes. a. Find the \(p\) -value for the test of hypothesis with the alternative hypothesis that the average daily time spent eating and drinking by all Norwegian men is higher than \(67.20\) minutes. What is your conclusion at \(\alpha=.05\) ? b. Test the hypothesis of part a using the critical-value approach. Use \(\alpha=.01\).

Perform the following tests of hypotheses for data coming from a normal distribution. a. \(H_{0}: \mu=94.80, H_{1}: \mu<94.80, n=12, \quad \bar{x}=92.87, s=5.34, \quad \alpha=.10\) b. \(H_{0}: \mu=18.70, \quad H_{1}: \mu \neq 18.70, n=25, \quad \bar{x}=20.05, s=2.99, \quad \alpha=.05\) c. \(H_{0}: \mu=59, \quad H_{1}: \mu>59, \quad n=7, \quad \bar{x}=59.42, s=.418, \quad \alpha=.01\)
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