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For Exercises 5 through \(20,\) 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. After-School Snacks 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?

Short Answer

Expert verified
The hypothesis test did not find significant evidence to conclude that the proportion differs from 54%. Parents shouldn't be overly concerned based on this test.

Step by step solution

01

State the Hypotheses

The null hypothesis \( H_0 \) states that the proportion of kids who have a snack after school is equal to the reported proportion, \( p = 0.54 \). The alternative hypothesis \( H_1 \) states that the proportion is different, \( p eq 0.54 \). Thus, our hypotheses are: \( H_0: p = 0.54 \) and \( H_1: p eq 0.54 \). The claim is represented by the null hypothesis here.
02

Find the Critical Value(s)

Since the test is a two-tailed test with \( \alpha = 0.01 \), look up the critical value for a standard normal distribution. A two-tailed test with \( \alpha = 0.01 \) gives critical values at \( z = \pm 2.576 \). These are the cutoff values in the z-table.
03

Compute the Test Value

To find the test statistic \( z \), use the formula for the sample proportion:\[ \hat{p} = \frac{x}{n} = \frac{36}{60} = 0.6 \]The standard error (SE) is calculated as:\[ SE = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.54 \times 0.46}{60}} \approx 0.06452 \]Then, compute the test statistic:\[ z = \frac{\hat{p} - p}{SE} = \frac{0.6 - 0.54}{0.06452} \approx 0.929 \]
04

Make the Decision

The test statistic \( z \approx 0.929 \) does not fall into the critical region since \( |0.929| < 2.576 \). Therefore, we do not reject the null hypothesis.
05

Summarize the Results

At \( \alpha = 0.01 \), there is not enough evidence to reject the claim that 54% of kids have a snack after school. Parents may not need to be concerned that this proportion has changed significantly regarding healthy snacks.

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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 (\( H_0 \)) serves as a starting point. It is a statement that there's no effect or no difference, and it is what we seek to test against. In our exercise, the null hypothesis asserts that 54% of kids have a snack after school. Essentially, it is saying the current condition or belief is true, and there's no significant change.

The null hypothesis is often paired with an alternative hypothesis (\( H_1 \)), which claims the contrary—for instance, the proportion differs from 54%. By comparing sample data against these propositions, we evaluate if the null hypothesis holds. If results show that the data significantly deviates from the expectation under the null hypothesis, we may reject it in favor of the alternative hypothesis.
critical value
The critical value acts like a threshold in hypothesis testing. It helps determine if our test statistic falls within the region where we can reject the null hypothesis. You could think of it as a border between acceptance and rejection of \( H_0 \).

For a two-tailed test at a significance level (\( \alpha \)) of 0.01, as in the exercise, the critical values are \( z = \pm 2.576 \). These values mark the extreme areas in a standard normal distribution where results would be considered statistically significant. If our test statistic moves into these areas, we have grounds to reject the null hypothesis.

Think of critical values as guardrails guiding the decision process in hypothesis testing. Without them, determining significance would lack structure and consistency.
test statistic
The test statistic is at the core of hypothesis testing.
It quantifies the degree to which the observed data diverges from what we would expect under the null hypothesis. In our scenario, it is computed using the sample data—specifically, the proportion of kids who reported having a snack after school.

We calculate the sample proportion (\( \hat{p}\)) first, then the standard error (SE). Using these, we determine the test statistic (\( z \)). The test statistic tells us where our sample statistic lies in a standard normal distribution.

If this value is high enough, it implies our observed results are unlikely under \( H_0 \). Conversely, if it is within the normal range, it suggests that our observations are not unusual enough to reject the null hypothesis.
P-value method
The P-value method is a streamlined way to make decisions in hypothesis testing.
It relies on calculating the probability that the observed data—or something more extreme—could occur under the null hypothesis. By comparing this P-value against a preset significance level (\( \alpha \)), we decide whether or not to reject \( H_0 \).

If the P-value is less than \( \alpha \), we have statistical grounds to reject the null hypothesis, suggesting the observed effect is significant. On the contrary, a higher P-value means the data doesn't provide enough evidence, and we do not reject \( H_0 \).

This method shines for its flexible interpretation and clearer understanding—P-values offer a direct measure of how compatible our results are with \( H_0 \). Thus, it acts as a powerful tool in hypothesis testing for discerning significance.

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Most popular questions from this chapter

Define null and alternative hypotheses, and give an example of each.

For Exercises 5 through \(20,\) 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. Television Set Ownership According to Nielsen Media Research, of all the U.S. households that owned at least one television set, \(83 \%\) had two or more sets. A local cable company canvassing the town to promote a new cable service found that of the 300 randomly selected households visited, 240 had two or more television sets. At \(\alpha=0.05,\) is there sufficient evidence to conclude that the proportion is less than the one in the report?

For Exercises I through 25, 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 diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Dress Shirts In a previous study conducted several years ago, a man owned on average 15 dress shirts. The standard deviation of the population is \(3 .\) A researcher wishes to see if that average has changed. He selected a random sample of 42 men and found that the average number of dress shirts that they owned was \(13.8 .\) At \(\alpha=0.05,\) is there enough evidence to support the claim that the average has changed?

For Exercises I through 25, 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 diagrams to show the critical region (or regions), and use the traditional method of hypothesis testing unless otherwise specified. Telephone Calls A researcher knew that before cell phones, a person made on average 2.8 calls per day. He believes that the number of calls made per day today is higher. He selects a random sample of 30 individuals who use a cell phone and asks them to keep track of the number of calls that they made on a certain day. The mean was \(3.1 .\) At \(\alpha=0.01\) is there enough evidence to support the researcher's claim? The standard deviation for the population found by a previous study is \(0.8 .\) Would the null hypothesis be rejected at \(\alpha=0.05 ?\)

For Exercises 5 through \(20,\) 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. Automobiles Purchased An automobile owner found that 20 years ago, \(76 \%\) of Americans said that they would prefer to purchase an American automobile. He believes that the number is much less than \(76 \%\) today. He selected a random sample of 56 Americans and found that 38 said that they would prefer an American automobile. Can it be concluded that the percentage today is less than 76 \(\% ?\) At \(\alpha=0.01,\) is he correct?

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