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91Ó°ÊÓ

Let \(p\) denote the proportion of grocery store customers who use the store's club card. For a largesample \(z\) test of \(H_{0}: p=.5\) versus \(H_{a}: p>.5\), find the \(P\) -value associated with each of the given values of the test statistic: a. \(\quad 1.40\) d. \(2.45\) b. \(0.93\) e. \(-0.17\) c. \(1.96\)

Short Answer

Expert verified
The associated P-values with the given test statistic values are: for z = 1.40, P-value = 0.0808; for z = 0.93, P-value = 0.1762; for z = -0.17, P-value = 0.5675; for z = 1.96, P-value = 0.025; for z = 2.45, P-value = 0.0071.

Step by step solution

01

Understanding the meaning of P-value

The P-value measures the strength of evidence in support of the null hypothesis. Here, since we're testing an alternative hypothesis of \(p > 0.5\), the P-value is the probability of obtaining a value greater than the observed test statistic value under the null hypothesis.
02

Calculate P-value for z = 1.40

For a test statistic value of 1.40, the P-value can be determined by looking up this value in the standard normal distribution table. The given value corresponds to the probability of 0.9192. Since this is a one-tailed test (greater than), the P-value is calculated as 1 - 0.9192 = 0.0808.
03

Continue for other z-values

Applying the same method for other values, we get:\n- For z = 0.93, P-value = 1 - 0.8238 = 0.1762\n- For z = -0.17, P-value = 1 - 0.4325 = 0.5675 (Since this is a negative z value, it's on the left of the mean. The normal lookup table value is less than 0.5, subtract it from 0.5, not from 1)\n- For z = 1.96, P-value = 1 - 0.9750 = 0.025\n- For z = 2.45, P-value = 1 - 0.9929 = 0.0071.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about data. It involves setting up two opposing hypotheses—the null hypothesis and the alternative hypothesis—and then using data to determine which one is more likely to be true.
In this exercise, we're dealing with the proportion of grocery store customers who use a club card. Our null hypothesis, denoted as \(H_0\), is that \(p=0.5\), meaning that 50% of customers use the card. The alternative hypothesis, \(H_a\), is that more than 50% of customers use the card (\(p>0.5\)).
  • The null hypothesis (\(H_0\)) represents a statement of no effect or no difference. It's what we seek to test against.
  • The alternative hypothesis (\(H_a\)) represents a statement that there's an effect or a difference.
The purpose of the test is to compute the probability of observing our sample data if the null hypothesis is true. This probability is known as the P-value.
Z-test
The z-test is a type of hypothesis test used when the sample size is large, and the population variance is known or assumed to be the standard normal distribution. It is perfect for evaluating the mean or proportion in hypothesis testing when data meets these conditions.
In this particular problem, the z-test is applied to determine whether the proportion of club card users exceeds 0.5.
  • The z-test helps determine how far, in standard deviations, the observed statistic is from the null hypothesis claim.
  • We use z-test by converting a sample statistic into a z-score, based on the standard normal distribution.
A z-score tells us where a value fits within the standard normal distribution and is essential in calculating the P-value.
Standard Normal Distribution
The standard normal distribution is a way of describing data that shows a bell-shaped, symmetric curve with a mean of 0 and a standard deviation of 1. It is used heavily in statistical techniques, including hypothesis testing with z-tests.
  • It forms the basis for z-tests, as z-scores are the number of standard deviations away from the mean a data point is.
  • Standard normal distribution allows us to calculate probabilities for z-scores, which are crucial in determining P-values.
In this exercise, the z-scores for each test statistic are looked up in the standard normal distribution table. This table tells us the probability or area to the left of any given z-score, which we then use to calculate the P-value.
Null Hypothesis
The null hypothesis is a foundational concept in hypothesis testing. It is the hypothesis that there is no effect or difference. It's the assumption that any kind of difference or significance you see in a set of data is purely by chance.
  • In our context, \(H_0: p=0.5\) suggests no preference or increase in proportion for club card users.
  • Null hypothesis is usually the default or starting assumption for any statistical test.
The idea is to either reject the null hypothesis in favor of the alternative hypothesis if our data provides strong evidence against it, or fail to reject it if the evidence is not strong enough. The P-value is critical in this decision-making process—it quantifies the probability of observing the data when the null hypothesis is true. Lower P-values indicate stronger evidence against the null hypothesis, altering our belief concerning it.

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Most popular questions from this chapter

The paper "Debt Literacy, Financial Experiences and Over-Indebtedness" (Social Science Research Network, Working paper WI4808, 2008 ) included analysis of data from a national sample of 1000 Americans. One question on the survey was: "You owe \(\$ 3000\) on your credit card. You pay a minimum payment of \(\$ 30\) each month. At an Annual Percentage Rate of \(12 \%\) (or \(1 \%\) per month), how many years would it take to eliminate your credit card debt if you made no additional charges?" Answer options for this question were: (a) less than 5 years; (b) between 5 and 10 years; (c) between 10 and 15 years; (d) never-you will continue to be in debt; (e) don't know; and (f) prefer not to answer. a. Only 354 of the 1000 respondents chose the correct answer of never. For purposes of this exercise, you can assume that the sample is representative of adult Americans. Is there convincing evidence that the proportion of adult Americans who can answer this question correctly is less than \(.40(40 \%) ?\) Use \(\alpha=.05\) to test the appropriate hypotheses. b. The paper also reported that \(37.8 \%\) of those in the sample chose one of the wrong answers \((a, b\), and \(c)\) as their response to this question. Is it reasonable to conclude that more than one-third of adult Americans would select a wrong answer to this question? Use \(\alpha=.05\).

The article "Poll Finds Most Oppose Return to Draft, Wouldn't Encourage Children to Enlist" (Associated Press, December 18,2005 ) reports that in a random sample of 1000 American adults, 700 indicated that they oppose the reinstatement of a military draft. Is there convincing evidence that the proportion of American adults who oppose reinstatement of the draft is greater than two-thirds? Use a significance level of \(.05\).

A study of fast-food intake is described in the paper "What People Buy From Fast-Food Restaurants" (Obesity [2009]: \(1369-1374\) ). Adult customers at three hamburger chains (McDonald's, Burger King, and Wendy's) at lunchtime in New York City were approached as they entered the restaurant and asked to provide their receipt when exiting. The receipts were then used to determine what was purchased and the number of calories consumed was determined. In all, 3857 people participated in the study. The sample mean number of calories consumed was 857 and the sample standard deviation was 677 . a. The sample standard deviation is quite large. What does this tell you about number of calories consumed in a hamburger-chain lunchtime fast-food purchase in New York City? b. Given the values of the sample mean and standard deviation and the fact that the number of calories consumed can't be negative, explain why it is not reasonable to assume that the distribution of calories consumed is normal. c. Based on a recommended daily intake of 2000 calories, the online Healthy Dining Finder (www .healthydiningfinder.com) recommends a target of 750 calories for lunch. Assuming that it is reasonable to regard the sample of 3857 fast-food purchases as representative of all hamburger-chain lunchtime purchases in New York City, carry out a hypothesis test to determine if the sample provides convincing evidence that the mean number of calories in a New York City hamburger-chain lunchtime purchase is greater than the lunch recommendation of 750 calories. Use \(\alpha=.01\). d. Would it be reasonable to generalize the conclusion of the test in Part (c) to the lunchtime fast-food purchases of all adult Americans? Explain why or why not.e. Explain why it is better to use the customer receipt to determine what was ordered rather than just asking a customer leaving the restaurant what he or she purchased. f. Do you think that asking a customer to provide his or her receipt before they ordered could have introduced a potential bias? Explain.

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