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For Exercises 5 through \(20,\) assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. Nicotine Content of Cigarettes A manufacturer of cigarettes wishes to test the claim that the variance of the nicotine content of the cigarettes the company manufactures is equal to 0.638 milligram. The variance of a random sample of 25 cigarettes is 0.930 milligram. At \(\alpha=0.05,\) test the claim.

Short Answer

Expert verified
Do not reject the null hypothesis.

Step by step solution

01

Identify the Null and Alternative Hypotheses

The null hypothesis (H_0) represents the claim to be tested, which is that the variance of the nicotine content is equal to 0.638 mg: \( H_0: \sigma^2 = 0.638 \). The alternative hypothesis (H_a) is what we suspect might be true if the null is not: \( H_a: \sigma^2 eq 0.638 \). This is a two-tailed test since we are checking for equality.
02

Determine the Test Statistic

Since we are testing for variance, we use the chi-squared statistic, which is calculated using: \( \chi^2 = \frac{(n-1)s^2}{\sigma^2_0} \), where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma^2_0 \) is the variance under the null hypothesis. For this problem, \( n = 25 \), \( s^2 = 0.930 \), and \( \sigma^2_0 = 0.638 \). Substituting these into the formula gives us \( \chi^2 = \frac{(25-1) \times 0.930}{0.638} \).
03

Calculate the Test Statistic

Perform the calculation: \( \chi^2 = \frac{24 \times 0.930}{0.638} = \frac{22.32}{0.638} \approx 34.98 \). So, the test statistic is approximately 34.98.
04

Determine the Critical Value(s)

For a two-tailed test with \( \alpha = 0.05 \) and \( n-1 = 24 \) degrees of freedom, we refer to the chi-square distribution table to find critical values. The critical values for \( \alpha = 0.05 \) are approximately 39.364 (upper critical value) and 12.401 (lower critical value).
05

Make a Decision

Compare the test statistic to the critical values. If the test statistic lies outside the range formed by the critical values, we reject the null hypothesis. Here, \( 34.98 \) lies between 39.364 and 12.401, hence we do not reject \( H_0 \).
06

Conclusion

Since we did not reject the null hypothesis, there is not enough evidence to conclude that the variance of the nicotine content is different from 0.638 mg at the 0.05 level of significance.

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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 a set of observed frequencies differ from expected frequencies. It's commonly used for hypothesis testing on categorical data and variances. In the context of our cigarette manufacturer problem, we use the chi-square test to compare the sample variance to the claimed population variance of nicotine content. This involves calculating the chi-square statistic, \[ \chi^2 = \frac{(n-1)s^2}{\sigma^2_0} \] where \( n \) is the sample size, \( s^2 \) is the sample variance, and \( \sigma^2_0 \) is the population variance under the null hypothesis.
With the sample size, sample variance, and claimed variance known, we can compute the test statistic, which helps assess the validity of the cigarette manufacturer's claim. Understanding this process is crucial when working with data sets that involve analyzing the divergence between observed and expected outcomes.
Variance
Variance is a critical concept in statistics that measures how much data points in a set differ from the mean. It provides insight into the degree of spread and can indicate the reliability of an average. In the cigarette manufacturer scenario, variance reflects the fluctuation in nicotine content among the cigarettes. The formula for variance in a sample is given by \[ s^2 = \frac{\sum{(x_i - \bar{x})^2}}{n-1} \] where \( x_i \) are the individual data points, \( \bar{x} \) is the sample mean, and \( n \) is the sample size.
Variance is essential when working with hypothesis tests on data variability, such as in this exercise where we compare the sample variance to a claimed variance using a chi-square test. Grasping the concept of variance is fundamental to interpreting data accurately and conducting reliable statistical analyses.
Two-Tailed Test
A two-tailed test in hypothesis testing checks for the possibility of a relationship in two directions. It means that the effect or difference can be in either direction from the hypothesized value. For the cigarette manufacturer's variance claim, a two-tailed test is used because the null hypothesis states that the variance is equal to a specific value, not greater or less than it.
Two-tailed tests offer a broader perspective, as they allow for rejection of the null hypothesis if the test statistic is significantly high or low. This is contrasted against a single-tailed test, where concern is on deviation in only one direction. The decision to use a two-tailed test provides more comprehensive insight on whether there is a statistically significant difference from the hypothesized value.
Critical Values
Critical values are essential to hypothesis testing as they define the threshold beyond which the null hypothesis can be rejected. These values are determined based on the significance level, \( \alpha \), and the degrees of freedom in the test. In this chi-square test scenario, the critical values are established from the chi-square distribution table.
For a two-tailed test with \( \alpha = 0.05 \) and 24 degrees of freedom, the critical values are the points where the probability of observing a test statistic as extreme as or more extreme than the critical value is \( \alpha/2 \) for each tail. In our scenario, these critical values are approximately 39.364 (upper critical value) and 12.401 (lower critical value).
  • If the test statistic falls outside this critical range, we reject the null hypothesis.
  • Conversely, if it lies within, the null hypothesis is not rejected.
Understanding how to find and interpret critical values is vital for making informed decisions in hypothesis testing.

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