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Chapter 8: Problem 11

The average household size in a certain region several years ago was 3.14 persons. A sociologist wishes to test, at the \(5 \%\) level of significance, whether it is different now. Perform the test using the information collected by the sociologist: in a random sample of 75 households, the average size was 2.98 persons, with sample standard deviation 0.82 person.

Short Answer

Expert verified
Fail to reject the null hypothesis; no significant difference detected.

Step by step solution

01

Identify the Hypotheses

To perform this test, we first set up the null and alternative hypotheses. The null hypothesis \( H_0 \) assumes that the average household size is still 3.14 (i.e., \( \mu = 3.14 \)). The alternative hypothesis \( H_a \) assumes that the average household size is different from 3.14 (i.e., \( \mu eq 3.14 \)).
02

Determine the Test Statistic

Since the sample size is large (\( n = 75 \)), we will use a z-test for the mean. The test statistic is calculated using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]where \( \bar{x} = 2.98 \), \( \mu = 3.14 \), \( s = 0.82 \), and \( n = 75 \). Filling in the values, we get:\[ z = \frac{2.98 - 3.14}{\frac{0.82}{\sqrt{75}}} \]
03

Perform the Calculations

Calculate the test statistic using the formula from Step 2:\[ z = \frac{2.98 - 3.14}{\frac{0.82}{\sqrt{75}}} = \frac{-0.16}{\frac{0.82}{8.66}} \approx \frac{-0.16}{0.09473} \approx -1.688 \]
04

Determine the Critical Values and Decision Rule

Since we are testing at a \(5\%\) level of significance for a two-tailed test, we find the critical z-values from standard normal distribution tables. The critical values for \( \alpha = 0.05 \) are \( z = -1.96 \) and \( z = 1.96 \). If the calculated z-value is less than -1.96 or greater than 1.96, we reject the null hypothesis.
05

Make a Decision

Compare the calculated z-value of approximately \( -1.688 \) with the critical values \( -1.96 \) and \( 1.96 \). Since \( -1.688 \) is not less than \( -1.96 \) and not greater than \( 1.96 \), the null hypothesis is not rejected at the \(5\%\) level of significance.

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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, the null hypothesis, denoted as \( H_0 \), is a statement that there is no effect or no difference. It serves as the starting point for statistical testing.
In our example, the null hypothesis claims that the average household size has not changed and still is 3.14 persons. This means that any observed changes are purely due to chance or sampling variation.
  • It is a specific statement often reflecting the status quo or what is believed to be true before testing.
  • Usually includes an equality sign ( \( = \) ), indicating no difference.
For this exercise, the null hypothesis is: \( H_0: \mu = 3.14 \), where \( \mu \) is the true average household size.
Alternative Hypothesis
The alternative hypothesis, denoted by \( H_a \) or \( H_1 \), proposes that there is a change or a difference. It is what the researcher wants to prove or test for.
It represents a new condition or situation contrary to the null hypothesis. In our sociologist's case, the alternative hypothesis suggests that the average household size is now different from what it was before, namely, not 3.14 persons.
  • Contains inequalities such as \( eq \), \( > \), or \( < \).
  • Challenging the status quo makes it key for validating research claims.
For this particular hypothesis test, the alternative hypothesis formulated is: \( H_a: \mu eq 3.14 \), implying a shift in average household size.
Z-test
A Z-test is a statistical test used to determine if there is a significant difference between sample and population means when the population variance is known and the sample size is large (over 30 is the rule of thumb).
In our situation, since the sample size is 75, it is appropriate to use the Z-test.
  • It uses the standard normal distribution (Z-distribution) for calculation.

  • The Z-test formula is: \[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean under \( H_0 \), \( s \) is the sample standard deviation, and \( n \) is the sample size.
Our Z-test calculated value here was approximately \(-1.688\). This value tells us how many standard deviations the sample mean is away from the null hypothesis mean.
Level of Significance
The level of significance, denoted by \( \alpha \), is the probability of rejecting the null hypothesis when it is actually true.
It defines our threshold for making decisions, usually set at common levels such as 0.01, 0.05, or 0.10.
  • A 5% level of significance means there's a 5% risk of making a Type I error, which is incorrectly rejecting a true null hypothesis.

  • This level influences the critical regions of our test, affecting how extreme our test statistic must be to reject \( H_0 \).
In our test, with \( \alpha = 0.05 \), the critical z-values were \( -1.96 \) and \( 1.96 \). Since our calculated \( z \)-value (\(-1.688\)) did not fall beyond these limits, we do not reject the null hypothesis.

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Most popular questions from this chapter

A random sample of size 8 drawn from a normal population yielded the following results: \(\bar{x}-280, s=46\). a. Test \(H 0: \mu=250\) vs. Ha:\mu \(>250 @ \alpha=0.05\). b. Estimate the observed significance of the test in part (a) and state a decision based on the \(p\) -value approach to hypothesis testing.

State the null and alternative hypotheses for each of the following situations. (That is, identify the correct number \(\mu_{0}\) and write \(H_{0} \mu=\mu_{0}\) and the appropriate analogous expression for \(H_{a}\).) a. The average time workers spent commuting to work in Verona five years ago was 38.2 minutes. The Verona Chamber of Commerce asserts that the average is less now. b. The mean salary for all men in a certain profession is \(\$ 58,291\). A special interest group thinks that the mean salary for women in the same profession is different. c. The accepted figure for the caffeine content of an 8 -ounce cup of coffee is \(133 \mathrm{mg}\). A dietitian believes that the average for coffee served in a local restaurants is higher. d. The average yield per acre for all types of corn in a recent year was 161.9 bushels. An economist believes that the average yield per acre is different this year. e. An industry association asserts that the average age of all self-described fly fishermen is 42.8 years. A sociologist suspects that it is higher.

Researchers wish to test the efficacy of a program intended to reduce the length of labor in childbirth. The accepted mean labor time in the birth of a first child is 15.3 hours. The mean length of the labors of 13 first- time mothers in a pilot program was 8.8 hours with standard deviation 3.1 hours. Assuming a normal distribution of times of labor, test at the \(10 \%\) level of significance test whether the mean labor time for all women following this program is less than 15.3 hours.

A rule of thumb is that for working individuals one-quarter of household income should be spent on housing. A financial advisor believes that the average proportion of income spent on housing is more than \(0.25 .\) In a sample of 30 households, the mean proportion of household income spent on housing was 0.285 with a standard deviation of \(0.063 .\) Perform the relevant test of hypotheses at the \(1 \%\) level of significance.

Five years ago \(3.9 \%\) of children in a certain region lived with someone other than a parent. A sociologist wishes to test whether the current proportion is different. Perform the relevant test at the \(5 \%\) level of significance using the following data: in a random sample of 2,759 children, 119 lived with someone other than a parent.
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