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Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Researchers suspect that \(18 \%\) of all high school students smoke at least one pack of cigarettes a day. At Wilson High School, a randomly selected sample of 300 students found that 50 students smoked at least one pack of cigarettes a day. At \(\alpha=0.05,\) test the claim that less than \(18 \%\) of all high school students smoke at least one pack of cigarettes a day. Use the \(P\) -value method.

Short Answer

Expert verified
The claim that less than 18% of students smoke is not supported; we do not reject the null hypothesis.

Step by step solution

01

Stating the Hypotheses and Identifying the Claim

Let's formulate our null and alternative hypotheses. The claim is that less than 18% of all high school students smoke. Therefore:- Null hypothesis (\(H_0\)): \(p = 0.18\)- Alternative hypothesis (\(H_1\)): \(p < 0.18\)The claim is represented by the alternative hypothesis \(H_1\).
02

Finding the Critical Value

For the given significance level \(\alpha = 0.05\), and because this is a left-tailed test, we need to find the critical value corresponding to 0.05 in the standard normal distribution table (Z-table). The critical value for 0.05 is approximately \(-1.645\).
03

Computing the Test Value

To find the test value, we use the formula for the test statistic for a proportion:\[ z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]where \(\hat{p}\) is the sample proportion, \(p\) is the population proportion (0.18), and \(n\) is the sample size (300).First, calculate the sample proportion, \(\hat{p}:\)\[ \hat{p} = \frac{50}{300} = 0.1667 \]Then, plug the values into the formula:\[ z = \frac{0.1667 - 0.18}{\sqrt{\frac{0.18(1-0.18)}{300}}} \approx \frac{-0.0133}{\sqrt{\frac{0.1476}{300}}} \approx \frac{-0.0133}{0.0218} \approx -0.610 \]
04

Making the Decision

Compare the test statistic \(z\) value to the critical value. Our calculated test value is \(-0.610\), and the critical value at \(\alpha = 0.05\) is \(-1.645\).Since \(-0.610 > -1.645\), we do not reject the null hypothesis \(H_0\).
05

Summarizing the Results

At the 0.05 significance level, there is not enough evidence to support the claim that less than 18% of all high school students smoke at least one pack of cigarettes a day. The sample proportion is not significantly lower than 18%.

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Key Concepts

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

Null Hypothesis
The concept of the null hypothesis is a fundamental part of statistical testing. It serves as the starting point for most hypothesis tests. The null hypothesis, often denoted as \(H_0\), is a statement we aim to test. It typically states that there is no effect or no difference, and in the mathematical context, it expresses that a parameter is equal to a certain value.

In the context of our exercise, the null hypothesis is: \(H_0: p = 0.18\). This means we assume at the start that 18% of all high school students smoke at least one pack of cigarettes daily. The goal of the test is then to determine whether we have sufficient evidence to reject this assumption.

The null hypothesis is essential because it provides a baseline to compare against. It helps determine whether the evidence from the sample data is strong enough to conclude that the alternative hypothesis might be true.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_1\) or \(H_a\), is what you want to prove when conducting a statistical test. It is usually the opposite of the null hypothesis and suggests that there is an effect or a difference.

For our example, the alternative hypothesis is: \(H_1: p < 0.18\). This suggests that less than 18% of all high school students smoke at least one pack of cigarettes daily. The alternative hypothesis reflects the claim made by researchers and is what we test against the null hypothesis.

The formulation of the alternative hypothesis is crucial as it defines the direction of the test. In this case, it's a left-tailed test, meaning we are looking for evidence that the population proportion is less than 18%.

It's important to note that while the null hypothesis is assumed true until evidence suggests otherwise, the alternative hypothesis is only accepted if enough statistical evidence is gathered to reject the null hypothesis.
Critical Value
Understanding the critical value is key to making decisions in hypothesis testing. The critical value is a point on the test distribution that is compared to the test statistic to determine whether to reject the null hypothesis.

In our example, since we have a significance level of \(\alpha = 0.05\) and the test is left-tailed, we find the critical value from the standard normal distribution or Z-table. For a left-tailed test with \(\alpha = 0.05\), the critical value is approximately \(-1.645\).

The critical value separates the region where we would reject the null hypothesis from the region where we would not. If the test statistic falls below \(-1.645\), it indicates enough evidence to reject the null hypothesis in favor of the alternative.

Thus, by comparing the test statistic to this critical value, we can make a statistical decision about the hypotheses. In our scenario, since the test value \(-0.610\) is not less than \(-1.645\), we do not reject the null hypothesis.
P-value Method
The P-value method is another way to make decisions in hypothesis testing. It involves calculating a P-value, which is the probability of observing the test statistic, or something more extreme, assuming the null hypothesis is true.

A smaller P-value indicates stronger evidence against the null hypothesis. In general, if the P-value is less than the significance level \(\alpha\), you reject the null hypothesis. If it is greater, you fail to reject the null hypothesis.

For the exercise involving smoking students at Wilson High School, we are tasked to use the P-value method with \(\alpha = 0.05\). The calculated test statistic is \(-0.610\). Checking this against a Z-table will give us the P-value. If this value is less than 0.05, we would reject \(H_0\), but in this test, the comparison with the critical value already suggested not rejecting \(H_0\).

The P-value method provides an additional perspective on the hypothesis test, giving a measure of the result's strength. It's a pragmatic approach for making decisions based on statistical evidence.

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Most popular questions from this chapter

Explain the difference between a one-tailed and a two-tailed test.

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. 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.

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. The manager of a large company claims that the standard deviation of the time (in minutes) that it takes a telephone call to be transferred to the correct office in her company is 1.2 minutes or less. A random sample of 15 calls is selected, and the calls are timed. The standard deviation of the sample is 1.8 minutes. At \(\alpha=0.01\), test the claim that the standard deviation is less than or equal to 1.2 minutes. Use the \(P\) -value method.

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In the Journal of the American Dietetic Association, it was reported that \(54 \%\) of kids said that they had a snack after school. A random sample of 60 kids was selected, and 36 said that they had a snack after school. Use \(\alpha=0.01\) and the \(P\) -value method to test the claim. On the basis of the results, should parents be concerned about their children eating a healthy snack?

The average one-way airfare from Pittsburgh to Washington, D.C., is \(\$ 236 .\) A random sample of 20 one-way fares during a particular month had a mean of \(\$ 210\) with a standard deviation of \(\$ 43 .\) At \(\alpha=0.02\), is there sufficient evidence to conclude a difference from the stated mean? Use the sample statistics to construct a \(98 \%\) confidence interval for the true mean one-way airfare from Pittsburgh to Washington, D.C. and compare your interval to the results of the test. Do they support or contradict one another?

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