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The researcher from Chapter 17 ?, Exercise 39 ? tests whether the mean cholesterol level among those who eat frozen pizza exceeds the value considered to indicate a health risk. She gets a P-value of 0.07. Explain in this context what the " \(7 \%\) " represents.

Short Answer

Expert verified
The '7%' (P-value of 0.07) represents the probability of observing a result as extreme or more extreme than the one obtained in the study, assuming that there is no health risk from eating frozen pizza (null hypothesis is true). It suggests that the result could be due to random chance 7% of the time, indicating that the researcher does not have strong evidence to conclude that eating frozen pizza leads to dangerous cholesterol levels.

Step by step solution

01

Step 1

Understand the scenario and identify the P-value: In the mentioned research, the researcher is testing whether mean cholesterol level among those who eat frozen pizza exceeds the health risk value. The P-value from this hypothesis test is 0.07, or 7%.
02

Step 2

Interpret the P-value: A P-value of 0.07 means that, assuming the null hypothesis is true, there is a 7% chance of observing a statistical result as extreme as, or more extreme than, the one observed. In other words, if there is really no health risk from eating frozen pizza (null hypothesis), we would expect to see results at least as extreme as these 7% of the time purely due to random chance.
03

Step 3

Discuss the significance: As the P-value is greater than the typical significance level of 0.05, the researcher would not reject the null hypothesis. This means that the researcher does not have strong evidence to conclude that the mean cholesterol level of those who eat frozen pizza exceeds the health risk value.

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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 a fundamental concept in hypothesis testing. It is a statement that indicates no effect or no difference in the context of your experimental or observational data. In this case, the null hypothesis is that the mean cholesterol level among frozen pizza consumers does not exceed the health risk threshold.

Statisticians and researchers use the null hypothesis to test against their actual findings. They assume the null hypothesis is true and look for evidence to refute it, rather than proving an alternative hypothesis outright. This can often involve determining whether the data collected provides enough evidence to reject the null hypothesis.

When performing any statistical test, understanding and correctly positing the null hypothesis is crucial, as it forms the basis on which the test is conducted. The goal is often to determine if there is enough statistical evidence to reject the null hypothesis and introduce alternative explanations.
Significance Level
The significance level is another critical part of hypothesis testing. It is a threshold set by the researcher, which determines how extreme the data must be before we can reject the null hypothesis. This threshold is usually denoted by \( \alpha \), and in many scientific studies, it is set at 0.05 or 5%.

The significance level represents the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. A lower significance level means stricter criteria for rejecting the null hypothesis, reducing the risk of this type of error. In the given exercise, the P-value is 0.07, which is above the common significance level of 0.05. Hence, the null hypothesis is not rejected.

Choosing an appropriate significance level is important as it balances the risk of making Type I and Type II errors, ensuring the results of a study are both reliable and practical.
Hypothesis Testing
Hypothesis testing is a systematic method for evaluating ideas about populations based on samples. It involves making an initial assumption, called the null hypothesis, and obtaining statistical evidence to test this assumption.

Here is a step-by-step process to understand hypothesis testing:
  • Formulate the null and alternative hypotheses. The null hypothesis often suggests no effect or no difference, while the alternative proposes an effect or difference exists.
  • Select a significance level (e.g., \( \alpha = 0.05 \)). This helps define the threshold for determining statistical significance.
  • Conduct the experiment and calculate the test statistic and P-value. The P-value helps quantify evidence against the null hypothesis.
  • Compare the P-value to the significance level to make a decision. If the P-value is less than the significance level, reject the null hypothesis; otherwise, do not reject it.

Hypothesis testing helps researchers make informed decisions using sample data, systematically and objectively. Through this method, they can quantify uncertainty and draw valid conclusions.

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Most popular questions from this chapter

In Chapter 17 ?, Exercise 57 ? we saw that Yvon Hopps ran an experiment to determine optimum power and time settings for microwave popcorn. His goal was to find a combination of power and time that would deliver high-quality popcorn with less than \(10 \%\) of the kernels left unpopped, on average. After experimenting with several bags, he determined that power 9 at 4 minutes was the best combination. To be sure that the method was successful, he popped 8 more bags of popcorn (selected at random) at this setting. All were of high quality, with the following percentages of uncooked popcorn: \(7,13.2,10,6,7.8,2.8,2.2,5.2 .\) Use a test of hypothesis to decide if Yvon has met his goal.

Absentees The National Center for Education Statistics monitors many aspects of elementary and secondary education nationwide. Their 1996 numbers are often used as a baseline to assess changes. In \(1996,34 \%\) of students had not been absent from school even once during the previous month. In a 2000 survey, responses from 8302 students showed that this figure had slipped to \(33 \%\). Officials would, of course, be concerned if student attendance were declining. Do these figures give evidence of a change in student attendance? a. Write appropriate hypotheses. b. Check the assumptions and conditions. c. Perform the test and find the P-value. d. State your conclusion. e. Do you think this difference is meaningful? Explain.

Candy Someone hands you a box of a dozen chocolatecovered candies, telling you that half are vanilla creams and the other half peanut butter. You pick candies at random and discover the first three you eat are all vanilla. a. If there really were 6 vanilla and 6 peanut butter candies in the box, what is the probability that you would have picked three vanillas in a row? b. Do you think there really might have been 6 of each? Explain. c. Would you continue to believe that half are vanilla if the fourth one you try is also vanilla? Explain.

Marriage In \(1960,\) census results indicated that the age at which American men first married had a mean of 23.3 years. It is widely suspected that young people today are waiting longer to get married. We want to find out if the mean age of first marriage has increased since then. a. Write appropriate hypotheses. b. We plan to test our hypothesis by selecting a random sample of 40 men who married for the first time last year. Do you think the necessary assumptions for inference are satisfied? Explain. c. Describe the approximate sampling distribution model for the mean age in such samples. d. The men in our sample married at an average age of 24.2 years, with a standard deviation of 5.3 years.

Obesity 2016 In \(2016,\) the Centers for Disease Control and Prevention reported that \(36.5 \%\) of adults in the United States are obese. A county health service planning a new awareness campaign polls a random sample of 750 adults living there. In this sample, 228 people were found to be obese based on their answers to a health questionnaire. Do these responses provide strong evidence that the \(36.5 \%\) figure is not accurate for this region? Correct the mistakes you find in a student's attempt to test an appropriate hypothesis. $$ \begin{array}{l} \mathrm{H}_{0}: \hat{p}=0.365 \\ \mathrm{H}_{\mathrm{A}}: \hat{p}<0.365 \\ \mathrm{SRS}, 750 \geq 10 \\ \frac{228}{750}=0.304 ; S D(\hat{p})=\sqrt{\frac{(0.304)(0.696)}{750}}=0.017 \end{array} $$ \(z=\frac{0.304-0.365}{0.017}=-3.588\) $$ \mathrm{P}=P(z>-3.588)=0.9998 $$ There is more than a \(99.98 \%\) chance that the stated percentage is correct for this region.

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