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Chapter 16: Problem 571

A certain brand of cigarettes is advertised by the manufacturer as having a mean nicotine content of 15 milligrams per cigarette. A sample of 200 cigarettes is tested by an independent research laboratory and found to have an average of \(16.2\) milligrams of nicotine content and a standard deviation of \(3.6\). Using a \(.01\) level of significance, can we conclude based on this sample that the actual mean nicotine content of this brand of cigarettes is greater than 15 milligrams?

Short Answer

Expert verified
Based on the one-tailed hypothesis test with a 0.01 level of significance, we reject the null hypothesis (H鈧�: 渭 = 15) in favor of the alternative hypothesis (H鈧�: 渭 > 15). Thus, we can conclude that the actual mean nicotine content of this brand of cigarettes is greater than 15 milligrams.

Step by step solution

01

State the Null and Alternative Hypotheses

The null hypothesis (H鈧�) states that the actual mean nicotine content is equal to the stated mean of 15 milligrams. The alternative hypothesis (H鈧�) states that the actual mean nicotine content is greater than 15 milligrams. H鈧�: 渭 = 15 H鈧�: 渭 > 15
02

Calculate the Test Statistic

To calculate the test statistic, we will use the following formula: \(t = \frac{\bar{x} - 渭}{\frac{s}{\sqrt{n}}}\) Where: - t is the test statistic - \(\bar{x}\) is the sample mean (16.2) - 渭 is the population mean (15) - s is the sample standard deviation (3.6) - n is the sample size (200) \(t = \frac{16.2 - 15}{\frac{3.6}{\sqrt{200}}}\)
03

Calculate the Critical Value and Rejection Region

Since we are performing a one-tailed hypothesis test at a 0.01 level of significance, we will find the critical value for a t-distribution using a t-table or a statistical calculator, with 199 degrees of freedom (200 - 1). The critical value at 伪 = 0.01 and 199 degrees of freedom is approximately 2.346. The rejection region is defined as: Reject H鈧� if t > 2.346
04

Compare the Test Statistic to the Critical Value

Use the formula determined in step 2 to calculate the value of the test statistic and compare it to the critical value (2.346) found in step 3. \(t = \frac{16.2 - 15}{\frac{3.6}{\sqrt{200}}}\) = 5.377 Since the test statistic (5.377) is greater than the critical value (2.346), we are in the rejection region.
05

Make a Decision

As the test statistic falls in the rejection region, we reject the null hypothesis (H鈧�: 渭 = 15) in favor of the alternative hypothesis (H鈧�: 渭 > 15). Therefore, based on this sample, we can conclude at a 0.01 level of significance that the actual mean nicotine content of this brand of cigarettes is greater than 15 milligrams.

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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 (often symbolized as \(H_0\)) is an initial statement that proposes no significant effect or relationship between variables. It serves as a starting point for testing whether a specific claim is credible. In our example, the null hypothesis asserts that the mean nicotine content in the cigarettes is 15 milligrams. Thus, the statement is \( H_0: \mu = 15 \). The null hypothesis is presumed true until adequate evidence suggests otherwise.
  • This assumption implies no difference between the sample mean and the population mean.
  • It forms the foundation for hypothesis testing.
  • Rejection of the null hypothesis indicates statistical significance.
Alternative Hypothesis
The alternative hypothesis (denoted as \(H_1\)) contradicts the null hypothesis. It suggests that there is a statistically significant effect or relationship present. In our example, the alternative hypothesis contends that the actual mean nicotine content is greater than 15 milligrams. Thus, it's symbolized as \( H_1: \mu > 15 \).
  • This hypothesis represents the researcher's prediction.
  • It is accepted if the null hypothesis is rejected.
  • In one-tailed tests, it focuses on one direction (either greater than or less than).
Level of Significance
The level of significance (often denoted as \(\alpha\)) is a threshold set by the researcher which determines the probability of rejecting the null hypothesis when it is true. Commonly used levels of significance are 0.05, 0.01, and 0.10. In this scenario, a 0.01 level of significance was used, meaning there is only a 1% risk of concluding that the mean nicotine content differs from 15 milligrams when it doesn't.
  • Lower \(\alpha\) levels imply more stringent criteria for rejecting \(H_0\).
  • It dictates the rejection region where the null hypothesis is disproven.
  • A lower significance level can reduce the chance of Type I error (false positive).
Test Statistic
The test statistic is a standardized value calculated from sample data, used to determine the likelihood of observing the data under the null hypothesis. In this situation, the test statistic formula applied was:\[ t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]Here:- \(\bar{x}\) is the sample mean (16.2)- \(\mu\) represents the population mean hypothesized (15)- \(s\) is the sample standard deviation (3.6)- \(n\) signifies the sample size (200)The derived test statistic is \(t = 5.377\), providing a means to further evaluate the hypothesis.
  • It indicates how far the sample mean deviates from the population mean.
  • A higher absolute value often implies stronger evidence against \(H_0\).
  • It is compared against the critical value to draw conclusions.
Critical Value
Critical value refers to the boundary in the test statistic's distribution. It is determined by the level of significance and the degrees of freedom specific to the test. For this one-tailed test involving a t-distribution with degrees of freedom (\(n-1 = 199\)), the critical value at \(\alpha = 0.01\) is approximately 2.346.
  • This value dictates the cutoff for the rejection region.
  • If the test statistic exceeds the critical value, the null hypothesis is rejected.
  • Critical values help to assess whether the evidence is strong enough to refute \(H_0\).
In this case, with a test statistic of 5.377 being greater than the critical value 2.346, \(H_0\) is rejected, indicating the mean nicotine content likely exceeds 15 milligrams.

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Most popular questions from this chapter

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