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Chapter 10: Problem 5

To test \(H_{0}: \sigma=1.8\) versus \(H_{1}: \sigma>1.8,\) a random sample of size \(n=18\) is obtained from a population that is known to be normally distributed. (a) If the sample standard deviation is determined to be \(s=2.4\) compute the test statistic. (b) If the researcher decides to test this hypothesis at the \(\alpha=0.10\) level of significance, determine the critical value. (c) Draw a chi-square distribution and depict the critical region. (d) Will the researcher reject the null hypothesis? Why?

Short Answer

Expert verified
Test statistic: 30.22. Critical value at \(\alpha = 0.10\) is 24.77. Reject the null hypothesis because 30.22 > 24.77.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis is stated as \(H_{0}: \sigma=1.8\). The alternative hypothesis is stated as \(H_{1}: \sigma>1.8\).
02

Identify the Sample Statistics

The sample size is \(n=18\) and the sample standard deviation is \(s=2.4\).
03

Compute the Test Statistic

The test statistic for testing the population standard deviation is given by the chi-square statistic \[ \chi^2 = \frac{(n-1)s^2}{\sigma_0^2} \] where \(s\) is the sample standard deviation and \(\sigma_0\) is the population standard deviation under the null hypothesis. Substituting in the given values: \[ \chi^2 = \frac{(18-1)(2.4)^2}{(1.8)^2} = \frac{17(5.76)}{3.24} = \frac{97.92}{3.24} = 30.22 \]
04

Determine the Critical Value

To find the critical value, identify the degrees of freedom which are \(df = n - 1 = 18 - 1 = 17\). Using a chi-square table, find the critical value at \(\alpha = 0.10\) and \(df = 17\). The critical value is approximately 24.77.
05

Draw the Chi-Square Distribution

Draw a chi-square distribution curve with \(df = 17\). Shade the critical region to the right of the critical value, \(\chi^2 = 24.77\).
06

Make a Decision

Compare the test statistic (\(30.22\)) to the critical value (\(24.77\)). Since \(30.22\) is greater than \(24.77\), it falls in the critical region. Thus, we reject the null hypothesis \(H_{0}\).

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Key Concepts

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

Chi-Square Test
The chi-square test is a statistical method used to determine if there is a significant difference between the expected and observed frequencies in categorical data. It is often applied when comparing observed data with data we would expect to obtain according to a specific hypothesis.
For our exercise, we are testing the population standard deviation (\(\sigma\)) using the chi-square test. The formula for the chi-square statistic is: \(\chi^2 = \frac{(n-1)s^2}{\sigma_0^2}\). Here, \(n\) is the sample size, \(s\) is the sample standard deviation, and \(\sigma_0\) is the hypothesized population standard deviation.
Standard Deviation
Standard deviation measures the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a larger range.
In this problem, the sample standard deviation (\(s\)) is 2.4, and the hypothesized population standard deviation (\(\sigma_0\)) is 1.8. This difference, along with the sample size (\(n = 18\)), helps determine the test statistic.
Critical Value
The critical value is a point on the scale of the test statistic beyond which we reject the null hypothesis. It is determined by the given significance level (\(\alpha\)) and the degrees of freedom.
In our example, the degrees of freedom (\(df\)) are calculated as \(n - 1 = 18 - 1 = 17\). The critical value for \(\alpha = 0.10\) and \(df = 17\) is approximately 24.77. If the test statistic exceeds this value, the null hypothesis is rejected.
Null Hypothesis
The null hypothesis (\(H_{0}\)) is a statement that there is no effect or no difference, and it is assumed to be true until evidence indicates otherwise. In this exercise, the null hypothesis is \(H_{0}: \sigma = 1.8\), which means the population standard deviation is equal to 1.8.
The null hypothesis serves as a starting point for the test, and it is only rejected if the evidence strongly suggests that it is not true.
Alternative Hypothesis
The alternative hypothesis (\(H_{1}\)) is a statement that contradicts the null hypothesis. It represents what we are trying to prove in the test. In our exercise, the alternative hypothesis is \(H_{1}: \sigma > 1.8\), meaning the population standard deviation is greater than 1.8.
When the test statistic falls in the critical region (i.e., it is greater than the critical value), we reject the null hypothesis in favor of the alternative hypothesis.

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Most popular questions from this chapter

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