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Chapter 9: Problem 87

According to the U.S. Census Bureau, \(11 \%\) of children in the United States lived with at least one grandparent in 2009 (USA TODAY, June 30,2011 ). Suppose that in a recent sample of 1600 children, 224 were found to be living with at least one grandparent. At a \(5 \%\) significance level, can you conclude that the proportion of all children in the United States who currently live with at least one grandparent is higher than .11? Use both the \(p\) -value and the critical-value approaches.

Short Answer

Expert verified
Yes, based on the result of the hypothesis test, it can be concluded that the proportion of all children in the United States living with at least one grandparent is significantly higher than 0.11 at the 5% significance level.

Step by step solution

01

Formulate the Null and Alternative Hypotheses

The null hypothesis \(H_0\) assumes that there is no change in the population proportion, i.e., \(p = 0.11\). On the other hand, the alternative hypothesis \(H_1\) posits a change in the population proportion - according to the problem statement, the claim is that the proportion is now higher, thus \(H_1: p > 0.11\).
02

Compute the Test Statistics

The test statistics for hypothesis test for population proportion is given by \((\hat{p} - p_0)/ (\sqrt{p_0 (1 - p_0) / n})\), where \( \hat{p} \) represents the sample proportion, \( p_0 \) is the population proportion under the null hypothesis and \( n \) is the sample size. Here, \( \hat{p} = 224/1600 = 0.14, p_0 = 0.11\) and \( n = 1600\). Substituting these values leads to \( Z = (0.14 - 0.11) / (\sqrt{0.11 * 0.89 / 1600}) = \) around 4.83.
03

Calculate the P-value

The P-value is the probability of obtaining a result as extreme or more extreme than the observed data, under the assumption that the null hypothesis is true. For a Z-score of 4.83 and for a one-sided test, the P-value is practically zero (less than 0.00001, because 4.83 is off the chart).
04

Conclusion

At 5% significance level (i.e., if P-value < 0.05, we reject the null), both P-value and the critical value approaches lead to the rejection of the null hypothesis \( H_0 \), implying acceptance of the alternative hypothesis. Therefore, it can be concluded that the proportion of all children in the United States who currently live with at least one grandparent is significantly higher than 0.11.

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Key Concepts

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

Population Proportion
When we refer to the population proportion, we are talking about the fraction or percentage of a certain characteristic within a population. In hypothesis testing, this is denoted as \( p \). It represents the true proportion of the population that has the attribute of interest. For this exercise, the population proportion is originally stated as 0.11, meaning 11% of children were living with a grandparent according to past data.

The sample proportion, on the other hand, is the portion of this characteristic observed in a sample drawn from the population. We denote it as \( \hat{p} \). In the sample taken from 1600 children, 224 were living with at least one grandparent. Calculating the sample proportion gives us \( \hat{p} = \frac{224}{1600} = 0.14 \).

Understanding the difference between population and sample proportions is crucial in hypothesis testing. By comparing them, we can determine if there has been a change or difference regarding this characteristic in the population.
P-value
The P-value acts as a gauge for the strength of the evidence against the null hypothesis. Essentially, it tells us the probability of observing a statistic at least as extreme as the one seen, assuming the null hypothesis is true.

In this problem, a Z-score of 4.83 was calculated. This corresponds to a P-value that is less than 0.00001 in a one-sided test. The P-value being so low indicates that observing such a sample proportion by random chance under the null hypothesis is extremely unlikely.

A low P-value, generally less than the significance level \( \alpha \), which is 0.05 in this exercise, leads us to reject the null hypothesis. This outcome suggests significant evidence for the alternative hypothesis that a higher proportion of children are now living with a grandparent.
Critical Value
The critical value is a threshold in hypothesis testing used to determine whether to reject the null hypothesis. It depends on the significance level, often denoted by \( \alpha \). For a 5% significance level in a one-sided test, the critical value on the Z-table is approximately 1.645.

In hypothesis testing, if the calculated test statistic exceeds the critical value, the null hypothesis is rejected. In our exercise, with a Z-score of 4.83 exceeding the critical value of 1.645, it indicates that the sample data provides strong evidence against the null hypothesis.

The critical value approach correlates with the P-value method. Both approaches validate whether the null hypothesis holds, but the critical value offers a visual comparison against the test statistic, simplifying decision-making when testing hypotheses.
Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement of no effect or no difference. It posits that any observed effect in a sample arises solely from random chance, rather than a genuine population effect.

In this exercise, the null hypothesis states that the proportion \( p \) of children living with at least one grandparent is equal to 0.11 (or 11%), which echoes the proportion from previous data. Null hypotheses are typically tested against alternative hypotheses, which propose that there is a real, significant difference in the population.

Rejecting the null hypothesis鈥攁s done in this situation鈥攊mplies that the data provides enough evidence to support a true difference, here suggesting an increase in the number of children living with grandparents. However, rejection of \( H_0 \) doesn't prove the alternative hypothesis but rather strengthens the evidence supporting it.

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Most popular questions from this chapter

Consider the null hypothesis \(H_{0}: \mu=5 .\) A random sample of 140 observations is taken from a population with \(\sigma=17\). Using \(\alpha=.05\), show the rejection and nonrejection regions on the sampling distribution curve of the sample mean and find the critical value(s) of \(z\) for the following. \(\begin{array}{lll}\text { a. a right-tailed test } & \text { b. a left-tailed test } & \text { c. a two-tailed test }\end{array}\)

Consider the null hypothesis \(H_{0}: \mu=12.80 .\) A random sample of 58 observations is taken from this population to perform this test. Using \(\alpha=.05\), show the rejection and nonrejection regions on the sampling distribution curve of the sample mean and find the critical value(s) of \(t\) for the following. a. a right-tailed test \(\underline{\text { b. a left-tailed test }}\) c. a two-tailed test

According to a book published in \(2011,45 \%\) of the undergraduate students in the United States show almost no gain in learning in their first 2 years of college (Richard Arum et al., Academically Adrift, University of Chicago Press, Chicago, 2011 ). A recent sample of 1500 undergraduate students showed that this percentage is \(38 \%\). Can you reject the null hypothesis at a \(1 \%\) significance level in favor of the alternative that the percentage of undergraduate students in the United States who show almost no gain in learning in their first 2 years of college is currently lower than \(45 \%\) ? Use both the \(p\) -value and the critical-value approaches.

Make the following tests of hypotheses. a. \(H_{0}: \mu=25, \quad H_{1}: \mu \neq 25, \quad n=81, \quad \bar{x}=28.5, \quad \sigma=3, \quad \alpha=.01\) b. \(H_{0}=\mu=12, \quad H_{1}: \mu<12, \quad n=45, \quad \bar{x}=11.25, \quad \sigma=4.5, \quad \alpha=.05\) c. \(H_{0}=\mu=40, \quad H_{1}: \mu>40, \quad n=100, \quad \bar{x}=47, \quad \sigma=7, \quad \alpha=.10\)

Write the null and alternative hypotheses for each of the following examples. Determine if each is a case of a two-tailed, a left-tailed, or a right-tailed test. a. To test if the mean number of hours spent working per week by college students who hold jobs is different from 20 hours b. To test whether or not a bank's ATM is out of service for an average of more than 10 hours per month c. To test if the mean length of experience of airport security guards is different from 3 years d. To test if the mean credit card debt of college seniors is less than \(\$ 1000\) e. To test if the mean time a customer has to wait on the phone to speak to a representative of a mailorder company about unsatisfactory service is more than 12 minutes
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