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Chapter 15: Problem 8

Age of the Bride A sociologist feels that the median age at which women marry in Cook County, Illinois, is less than the median age of 26.9 throughout the United States (obtained from the U.S. Census Bureau). Based on a random sample of 20 marriage certificates from the county, she obtains the ages shown in the following table: $$ \begin{array}{lllll} \hline 31 & 27 & 24 & 30 & 24 \\ \hline 27 & 32 & 24 & 23 & 22 \\ \hline 25 & 23 & 28 & 22 & 26 \\ \hline 21 & 30 & 25 & 24 & 27 \\ \hline \end{array} $$ Do the data support the sociologist's feelings at the \(\alpha=0.05\) level of significance?

Short Answer

Expert verified
Reject the null hypothesis. The median age is less than 26.9 years.

Step by step solution

01

Define Hypotheses

Begin by stating the null and alternative hypotheses. The null hypothesis states that the median age at which women marry in Cook County is equal to 26.9 years. The alternative hypothesis states that the median age is less than 26.9 years. Formally, we express them as: \( H_0: \text{Median age} \ = 26.9 \) \( H_a: \text{Median age} < 26.9 \)
02

Collect Sample Data

Utilize the given sample data of ages: \[ 31, 27, 24, 30, 24 \ 27, 32, 24, 23, 22 \ 25, 23, 28, 22, 26 \ 21, 30, 25, 24, 27 \]
03

Calculate the Test Statistic

Since the population median is assumed to be 26.9, subtract 26.9 from each value to get the deviations: \[ 31-26.9, 27-26.9, 24-26.9, \text{...} \ 4.1, 0.1, -2.9, \text{...} \] Count the number of positive and negative deviations. Here, there are 8 positive deviations and 12 negative deviations. Using the sign test, the test statistic would be the number of negative deviations.
04

Determine the Critical Value and P-Value

At \( \alpha=0.05 \) significance level for a one-tailed test with a sample size of 20, the critical value from the sign test table is approximately 7 negative deviations. If we find the actual number of negative deviations, which is 12, we compare this count against the critical value.
05

Make Decision

Since our calculated value of 12 negative deviations is greater than the critical value of 7, we reject the null hypothesis at the \( \alpha=0.05 \) significance level.
06

Conclusion

We have enough evidence to support the sociologist's feeling that the median age at which women marry in Cook County, Illinois, is less than the national median age of 26.9 years.

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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, we start with the null hypothesis. This is a statement asserting that there is no difference or effect in the population from what is stated. In our example, the null hypothesis (denoted as \(H_0\)) suggests that the median age at which women marry in Cook County is the same as the national median age, 26.9 years. By assuming there is no effect, we aim to test if the sample data provides enough evidence to contradict this assumption.
alternative hypothesis
The alternative hypothesis (denoted as \(H_a\)) is what you want to prove. It is a statement that suggests there is a difference or effect. In our case, it states that the median age at which women marry in Cook County is less than the national median age, 26.9 years. This hypothesis is typically what the researcher aims to support with evidence from the data.
sign test
The sign test is a non-parametric test used to determine if there is a median difference in matched pairs or between the sample and the population. It is ideal for small sample sizes that do not require any assumptions about the distribution of the data. To perform the sign test, we calculate the difference between each sample data point and the hypothesized median. We then count how many differences are positive and how many are negative. In our example, we calculated the differences from 26.9 and found 8 positive and 12 negative deviations. The number of negative deviations is used as the test statistic.
significance level
The significance level, denoted by \(\alpha\), is the threshold for deciding whether to reject the null hypothesis. It represents the probability of rejecting the null hypothesis when it is actually true. Common significance levels are 0.05, 0.01, and 0.10. In this problem, \(\alpha = 0.05\), meaning we accept a 5% risk of concluding that there is a difference when there is none.
critical value
The critical value is a point on the test distribution that is compared to the test statistic to determine whether to reject the null hypothesis. Using a sign test table for \(\alpha = 0.05\) and a sample size of 20, the critical value is 7 negative deviations. Because our calculated test statistic (12 negative deviations) exceeds this critical value, we reject the null hypothesis.

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Most popular questions from this chapter

Use the sign test to test the given alternative hypothesis at the \(\alpha=0.05\) level of significance. The median is more than 50. An analysis of the data reveals that there are 7 minus signs and 12 plus signs

Write a paragraph that describes the logic of the test statistic in a two- tailed sign test.

Problems 17 and 18 illustrate the use of the sign test to test hypotheses regarding a population proportion. The only requirement for the sign test is that our sample be obtained randomly. When dealing with nominal data, we can identify a characteristic of interest and then determine whether each individual in the sample possesses this characteristic. Under the null hypothesis in the sign test, we expect that half of the data will result in minus signs and half in plus signs. If we let a plus sign indicate the presence of the characteristic (and a minus sign indicate the absence), we expect half of our sample to possess the characteristic while the other half will not. Letting \(p\) represent the proportion of the population that possesses the characteristic, our null hypothesis will be \(H_{0}: p=0.5 .\) Use the sign test for Problems 17 and 18 , following the sign convention indicated previously. Trusting the Press In a study of 2302 U.S. adults surveyed online by Harris Interactive 1243 respondents indicated that they tend to not trust the press. Using an \(\alpha=0.05\) level of significance, does this indicate that more than half of U.S. adults tend to not trust the press?

You Explain It! Dietary Habits A student in an exercise science program wishes to study dietary habits of married couples. She surveys married couples from her local gym and asks them (individually) what percent of their daily calories are from fat. She analyzes the results using the Wilcoxon signedranks test. Explain why her results are questionable.

Every year Money magazine publishes its list of top places to live. The following data represent a list of top places to live for a recent year, along with the median family income and median commute time. $$ \begin{array}{lcc} \text { City } & \begin{array}{c} \text { Family Income } \\ (\$ \mathbf{1 0 0 0 s}) \end{array} & \begin{array}{c} \text { Commute Time } \\ \text { (minutes) } \end{array} \\ \hline \text { Woodridge, Illinois } & 83 & 27.1 \\ \hline \text { Urbandale, Idaho } & 82 & 17.0 \\ \hline \text { La Palma, California } & 86 & 26.9 \\ \hline \text { Friendswood, Texas } & 90 & 26.0 \\ \hline \text { Suwanee, Georgia } & 101 & 32.1 \\ \hline \text { Somers, Connecticut } & 83 & 22.6 \\ \hline \end{array} $$ (a) Does a positive association exist between income and commute time at the \(\alpha=0.10\) level of significance? (b) Draw a scatter diagram to support your conclusion.
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