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Chapter 10: Problem 20

In December \(2001,38 \%\) of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. In a recent poll, 403 of 1122 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Has the proportion of families with children under the age of 18 who eat dinner together seven nights a week decreased? Use the \(\alpha=0.05\) significance level.

Short Answer

Expert verified
Fail to reject the null hypothesis; not enough evidence to conclude a decrease.

Step by step solution

01

State the Hypotheses

Formulate the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_a\)). \(H_0\): \(p = 0.38\) (The proportion has not decreased). \(H_a\): \(p < 0.38\) (The proportion has decreased).
02

Determine the Sample Proportion

Calculate the sample proportion (\(\frac{\text{403}}{\text{1122}}\)):\(p_h = \frac{403}{1122} \ p_h \ \approx 0.359\)
03

Calculate the Standard Error

Use the formula for the standard error of the proportion: \(\text{SE} = \ \sqrt{ \ \frac{0.38 \ \times (1-0.38)}\{1122}} \ \approx 0.0144\)
04

Compute the Test Statistic

Calculate the z-score: \(z = \frac{0.359 - \ 0.38} \ {0.0144} \ = \ -1.46\)
05

Determine the Critical Value

For \ \(\ \alpha=0.05\) in a one-tailed test, the critical value \ is \ \(z_{\alpha} \ = \ -1.645\).
06

Compare the Test Statistic to the Critical Value

Since \ \(-1.46 \ > \ -1.645\), we \ do not reject the null hypothesis.
07

State the Conclusion

There is not enough statistical evidence at the 0.05 significance level to conclude that the proportion of families that eat dinner together seven nights a week has decreased.

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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, often symbolized as \(H_0\), is a statement that assumes no effect or no difference. It's a starting point for statistical testing. In our example, the null hypothesis asserts that the proportion of families who eat dinner together has not decreased. Mathematically, it is expressed as \(H_0: p = 0.38\), indicating that the proportion is still at 38%. The null hypothesis is what we test against, and we aim to provide evidence for or against it using our sample data.
alternative hypothesis
The alternative hypothesis, symbolized as \(H_a\), is what you aim to support with your analysis. It reflects the change or effect you are testing for. In this example, the alternative hypothesis suggests that the proportion of families eating dinner together has decreased. Mathematically, it is represented as \(H_a: p < 0.38\). The alternative hypothesis is always in contrast to the null hypothesis and is accepted if there's strong enough evidence against \(H_0\).
significance level
The significance level, represented by the Greek letter alpha (\(\alpha\)), is the probability of rejecting the null hypothesis when it is actually true. It sets the threshold for how extreme our test statistic needs to be to reject \(H_0\). In this problem, the significance level is \(\alpha = 0.05\), which means that we are willing to accept a 5% chance of making a Type I error, which occurs when we wrongly reject the null hypothesis. This level needs to be chosen before conducting the test.
sample proportion
The sample proportion, denoted as \(\hat{p}\), is the proportion of success observed in the sample. It provides an estimate of the population proportion. In our case, it is calculated by dividing the number of families that eat dinner together by the total number of families surveyed: \(\hat{p} = \frac{403}{1122} \approx 0.359\). This is used in further calculations to assess the validity of the null hypothesis.
standard error
The standard error measures the variability or dispersion of the sample proportion from the true population proportion. It helps to quantify the uncertainty associated with estimating the population proportion. The standard error for a proportion is determined through the formula: \( SE = \sqrt{ \frac{p(1-p)}{n}} \) where \(p \) is the assumed proportion under the null hypothesis, and \(n\) is the sample size. In this exercise, \( SE = \sqrt{\frac{0.38 \times (1 - 0.38)}{1122}} \approx 0.0144 \). This value is crucial for calculating the z-score.
z-score
The z-score is a test statistic that quantifies the number of standard deviations a sample proportion is away from the population proportion under the null hypothesis. It is calculated using the formula: \( z = \frac{\hat{p} - p}{SE} \). For this example, the z-score can be computed as \( z = \frac{0.359 - 0.38}{0.0144} = -1.46 \). This score helps in determining whether the observed sample proportion is significantly different from the assumed population proportion (\( p = 0.38 \)) under the null hypothesis.
critical value
The critical value is a threshold that the test statistic must exceed in order to reject the null hypothesis. In hypothesis testing, we compare the z-score to this critical value. For a significance level of \(\alpha = 0.05\) in a one-tailed test, the critical value is \(z_\alpha = -1.645\). Since our calculated z-score of \( -1.46 \) is greater than \( -1.645 \), we do not reject the null hypothesis. This means there is not enough evidence to conclude that the proportion of families that eat dinner together has decreased.

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Most popular questions from this chapter

In Problems \(15-22,(a)\) determine the null and alternative hypotheses, (b) explain what it would mean to make a Type I error, and (c) explain what it would mean to make a Type II error. According to the Centers for Disease Control and Prevention, \(19.6 \%\) of children aged 6 to 11 years are overweight. A school nurse thinks that the percentage of 6 - to 11-year-olds who are overweight is different in her school district.

Suppose you wish to determine if the mean IQ of students on your campus is different from the mean IQ in the general population, \(100 .\) To conduct this study, you obtain a simple random sample of 50 students on your campus, administer an IQ test, and record the results. The mean IQ of the sample of 50 students is found to be 107.3 with a standard deviation of \(13.6 .\) (a) Conduct a hypothesis test (preferably using technology) \(H_{0}: \mu=\mu_{0}\) versus \(H_{1}: \mu \neq \mu_{0}\) for \(\mu_{0}=103,104,105,106,107,108,109,110,111,112\) at the \(\alpha=0.05\) level of significance. For which values of \(\mu_{0}\) do you not reject the null hypothesis? (b) Construct a \(95 \%\) confidence interval for the mean IQ of students on your campus. What might you conclude about how the lower and upper bounds of a confidence interval relate to the values for which the null hypothesis is rejected? (c) Suppose you changed the level of significance in conducting the hypothesis test to \(\alpha=0.01\). What would happen to the range of values of \(\mu_{0}\) for which the null hypothesis is not rejected? Why does this make sense?

In \(1994,52 \%\) of parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. A recent survey found that 256 of 800 parents of children in high school felt it was a serious problem that high school students were not being taught enough math and science. Do parents feel differently today than they did in \(1994 ?\) (a) What does it mean to make a Type II error for this test? (b) If the researcher decides to test this hypothesis at the \(\alpha=0.05\) level of significance, determine the probability of making a Type II error if the true population proportion is 0.50. What is the power of the test? (c) Redo part (b) if the true proportion is 0.48 .

The website pundittracker.com keeps track of predictions made by individuals in finance, politics, sports, and entertainment. Jim Cramer is a famous TV financial personality and author. Pundittracker monitored 678 of his stock predictions (such as a recommendation to buy the stock) and found that 320 were correct predictions. Treat these 678 predictions as a random sample of all of Cramer's predictions. (a) Determine the sample proportion of predictions Cramer got correct. (b) Suppose that we want to know whether the evidence suggests Cramer is correct less than half the time. State the null and alternative hypotheses. (c) Verify the normal model may be used to determine the \(P\) -value for this hypothesis test. (d) Draw a normal model with area representing the \(P\) -value shaded for this hypothesis test. (e) Determine the \(P\) -value based on the model from part (d). (f) Interpret the \(P\) -value. (g) Based on the \(P\) -value, what does the sample evidence suggest? That is, what is the conclusion of the hypothesis test? Assume an \(\alpha=0.05\) level of significance.

In Problems 23-29, decide whether the problem requires a confidence interval or hypothesis test, and determine the variable of interest. For any problem requiring a confidence interval, state whether the confidence interval will be for a population proportion or population mean. For any problem requiring a hypothesis test, write the null and alternative hypothesis. An investigator with the Food and Drug Administration wanted to determine whether a typical bag of potato chips contained less than the 16 ounces claimed by the manufacturer.
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