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Chapter 8: Problem 13

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?

Short Answer

Expert verified
There is not enough evidence to conclude the proportion is different from 54%, so parents need not worry based on this data.

Step by step solution

01

State the Hypotheses and Identify the Claim

In this step, we identify our null hypothesis and alternative hypothesis. The null hypothesis, denoted by \(H_0\), states that the proportion of kids who have a snack after school is 54%, i.e., \( p = 0.54 \). The alternative hypothesis, \(H_1\), is that the proportion of kids who have a snack is not 54%, i.e., \( p eq 0.54 \). This indicates a two-tailed test. The claim we are testing here is the one made in the research, \( p = 0.54 \).
02

Find the Critical Value(s)

Since we are conducting a two-tailed test at \( \alpha = 0.01 \), we need to find the critical values for a standard normal distribution (Z-distribution) that correspond to \( \alpha/2 = 0.005 \) in each tail. For \( \alpha = 0.01 \), the critical Z-values are approximately \( \pm 2.576 \).
03

Compute the Test Value

To compute the test statistic, we use the formula: \[Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \]where \(\hat{p}\) is the sample proportion, \(p\) is the population proportion, and \(n\) is the sample size. Given \(\hat{p} = \frac{36}{60} = 0.6\), \(p = 0.54\), and \(n = 60\), the test statistic is \[Z = \frac{0.6 - 0.54}{\sqrt{\frac{0.54 \times 0.46}{60}}} \approx 0.941\]
04

Make the Decision

Given that the computed test value \(Z = 0.941\) does not exceed the critical value of \( \pm 2.576 \), we fail to reject the null hypothesis at the 0.01 significance level.
05

Summarize the Results

At the 0.01 level of significance, there is not enough evidence to support the claim that the proportion of kids who have a snack after school is different from 54%. Thus, parents need not be particularly concerned based on this data that the proportion of kids snacking after school deviates from what was previously reported.

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Key Concepts

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

Understanding Critical Values
Critical values play a crucial role in hypothesis testing and help determine the threshold for making decisions about the hypothesis. In simple terms, critical values are the points on the Z-distribution that separate the regions where the null hypothesis is either rejected or not rejected.
  • In a two-tailed test, like in our exercise, there are two critical value thresholds. One in each tail (left and right) of the distribution.

  • The significance level, often denoted by \( \alpha \), helps us find these critical values. For instance, with \( \alpha = 0.01 \), as in the given example, we divide this significance level by two for a two-tailed test. This results in \( \alpha/2 = 0.005 \) for each tail.
Critical values for a standard normal distribution (Z-distribution) can be found using Z-tables or computational tools. In this exercise, these values are \( \pm 2.576 \). If our test value lies beyond these boundaries, we reject the null hypothesis.
The Z-Distribution Simplified
The Z-distribution, also known as the standard normal distribution, is essential in the world of hypothesis testing. It is a continuous probability distribution that is symmetric around the mean, which is zero. The standard deviation of the Z-distribution is one.
  • This distribution allows us to understand the probability of observing data points under the null hypothesis.
  • It is characterized by its bell-shaped curve and is used to find probabilities and critical values when the population standard deviation is known.
In hypothesis testing, we often transform our test statistic to a Z-score. This conversion simplifies decision-making because we can compare this Z-score against critical values to decide whether to reject or fail to reject the null hypothesis. Such was the case in our exercise where we calculated a Z-score of 0.941.
Decoding the Null Hypothesis
The null hypothesis, symbolized as \( H_0 \), is a cornerstone of hypothesis testing. It is a statement of no effect or no difference and serves as a neutral claim that we aim to test.
  • In our example, the null hypothesis claims that the proportion of kids who have a snack after school is 54% (\( p = 0.54 \)).
  • The null hypothesis is what our statistical tests put to the test; if the observed data is significantly different from what the null hypothesis predicts, we may consider rejecting it.
We use the null hypothesis as a default or starting assumption, and our goal is to determine if the data provides enough evidence to render this assumption unlikely. If our computed test value does not fall in the critical region, we "fail to reject" the null hypothesis, remaining with our initial claim as potentially true.
Exploring the Alternative Hypothesis
The alternative hypothesis, denoted as \( H_1 \) or sometimes \( H_a \), complements the null hypothesis. It presents the scenario that there is an effect or a difference, contrary to \( H_0 \).
  • In the context of our exercise, \( H_1 \) posits that the actual proportion of kids who enjoy a snack after school is not 54% (\( p eq 0.54 \)). This suggests variability from the reported statistic and introduces the possibility of change or deviation.
  • If the test result is statistically significant, indicating the observed data is very unlikely under the null hypothesis, we adopt the alternative hypothesis.
The alternative hypothesis can guide further research or considerations, such as reassessing diets in our case study. It is crucial when the null hypothesis faces rejection, implying evidence for the scenario the alternative hypothesis presents.

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Most popular questions from this chapter

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. The U.S. Bureau of Labor and Statistics reported that a person between the ages of 18 and 34 has had an average of 9.2 jobs. To see if this average is correct, a researcher selected a random sample of 8 workers between the ages of 18 and 34 and asked how many different places they had worked. The results were as follows: $$ \begin{array}{llllllll} 8 & 12 & 15 & 6 & 1 & 9 & 13 & 2 \end{array} $$ At \(\alpha=0.05,\) can it be concluded that the mean is \(9.2 ?\) Use the \(P\) -value method. Give one reason why the respondents might not have given the exact number of jobs that they have worked.

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. The Energy Information Administration reported that \(51.7 \%\) of homes in the United States were heated by natural gas. A random sample of 200 homes found that 115 were heated by natural gas. Does the evidence support the claim, or has the percentage changed? Use \(\alpha=0.05\) and the \(P\) -value method. What could be different if the sample were taken in a different geographic area?

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. A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute in his city is less and wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At \(\alpha=0.10\), is he correct?

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

A manager states that in his factory, the average number of days per year missed by the employees due to illness is less than the national average of \(10 .\) The following data show the number of days missed by 40 randomly selected employees last year. Is there sufficient evidence to believe the manager's statement at \(\alpha=0.05 ? \sigma=3.63 .\) Use the \(P\) -value method. \(\begin{array}{rrrrrrrr}0 & 6 & 12 & 3 & 3 & 5 & 4 & 1 \\ 3 & 9 & 6 & 0 & 7 & 6 & 3 & 4 \\ 7 & 4 & 7 & 1 & 0 & 8 & 12 & 3 \\ 2 & 5 & 10 & 5 & 15 & 3 & 2 & 5 \\ 3 & 11 & 8 & 2 & 2 & 4 & 1 & 9\end{array}\)
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