91Ó°ÊÓ

For Exercises 7 through \(23,\) perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Find the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Assume that the population is approximately normally distributed. Cigarette Smoking A researcher found that a cigarette smoker smokes on average 31 cigarettes a day. She feels that this average is too high. She selected a random sample of 10 smokers and found that the mean number of cigarettes they smoked per day was \(28 .\) The sample standard deviation was \(2.7 .\) At \(\alpha=0.05\) is there enough evidence to support her claim?

Step by step solution

01

State the Hypotheses

First, we establish the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)). For this scenario, the null hypothesis is that the average number of cigarettes smoked per day is 31 (\(H_0: \mu = 31\)). The researcher's claim is that the average is less than 31, which forms our alternative hypothesis (\(H_1: \mu < 31\)). The claim is identified as the alternative hypothesis.

02

Find the Critical Value

We use the significance level \(\alpha = 0.05\) and since the test is a left-tailed test, we find the critical value using the t-distribution table for \(n-1 = 9\) degrees of freedom. At \(\alpha = 0.05\), the critical value \(t_c\) is approximately \(-1.833\).

03

Calculate the Test Statistic

The test statistic \(t\) is calculated using the formula: \[t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\]where \(\bar{x} = 28\) is the sample mean, \(\mu = 31\) is the population mean, \(s = 2.7\) is the sample standard deviation, and \(n = 10\) is the sample size. Substituting these values, we get: \[t = \frac{28 - 31}{\frac{2.7}{\sqrt{10}}} \approx -3.51\]Thus, the test value is approximately \(-3.51\).

04

Make the Decision

Compare the test statistic \(t\) with the critical value \(t_c\). Since \(t \approx -3.51\) is less than \(-1.833\), we reject the null hypothesis \(H_0\).

05

Summarize the Results

Rejecting the null hypothesis suggests that there is significant evidence at the \(\alpha = 0.05\) level to support the researcher's claim that cigarette smokers smoke less than 31 cigarettes a day on average.

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Key Concepts

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

Null Hypothesis

The null hypothesis is the starting assumption in hypothesis testing. It represents the default position that there is no effect or no difference. In this exercise, the null hypothesis is that the population mean of cigarettes smoked per day is 31. It is stated as \(H_0: \mu = 31\).

• Purpose: The null hypothesis serves as a baseline to test against the research claim.
• Interpretation: Rejecting the null hypothesis suggests that there is enough evidence to support the claim, whereas failing to reject it indicates insufficient evidence.

It is essential to clearly define the null hypothesis, as it frames the statistical test and decision-making process.

Alternative Hypothesis

The alternative hypothesis presents what the researcher believes to be true. It opposes the null hypothesis. In the context of this exercise, the researcher's claim is that smokers, on average, smoke fewer than 31 cigarettes a day. This is the alternative hypothesis, denoted as \(H_1: \mu < 31\).

• Directionality: Since the claim is about smoking less, it is a left-tailed test.
• Significance: Acceptance of the alternative hypothesis implies sufficient evidence supporting the claim.

Understanding the alternative hypothesis is crucial, as it guides the inquiry into whether there is enough evidence to support the researcher's assertion.

Critical Value

The critical value marks the boundary between the region where the null hypothesis is not rejected and the region where it is rejected. For this test, knowing that it's left-tailed and that the significance level \(\alpha = 0.05\), we use the t-distribution to find the critical value.

• The critical value is \(t_c = -1.833\) for 9 degrees of freedom.
• It provides a cutoff point: If the test statistic is more extreme, in this case less than -1.833, you reject the null hypothesis.

The critical value plays a pivotal role in the decision-making process, as it delineates when the evidence is strong enough to reject the null hypothesis.

Test Statistic

The test statistic helps to decide whether to reject the null hypothesis. It is calculated using sample data in conjunction with the parameters described in the null hypothesis. For this exercise, the test statistic is derived using the formula for the t-test:\[t = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}\]where \(\bar{x} = 28\) is the sample mean, \(\mu = 31\) is the population mean, \(s = 2.7\) is the sample standard deviation, and \(n = 10\) is the sample size. This calculates to \(t \approx -3.51\).

• Comparison: If \(t\) is less than \(t_c\), the null hypothesis is rejected.
• Meaning: Here, since \(t \approx -3.51\) is less than \(-1.833\), evidence suggests the average number smoked is less than 31.

The test statistic gives quantitative evidence concerning the likelihood of the null hypothesis, guiding the final decision.