/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Choice blindness is the term tha... [FREE SOLUTION] | 91影视

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Choice blindness is the term that psychologists use to describe a situation in which a person expresses a preference and then doesn't notice when they receive something different than what they asked for. The authors of the paper "Can Chocolate Cure Blindness? Investigating the Effect of Preference Strength and Incentives on the Incidence of Choice Blindness" Uournal of Behavioral and Experimental Economics [2016]: 1-11) wondered if choice blindness would occur more often if people made their initial selection by looking at pictures of different kinds of chocolate compared with if they made their initial selection by looking as the actual different chocolate candies. Suppose that 200 people were randomly assigned to one of two groups. The 100 people in the first group are shown a picture of eight different kinds of chocolate candy and asked which one they would like to have. After they selected, the picture is removed and they are given a chocolate candy, but not the one they actually selected. The 100 people in the second group are shown a tray with the eight different kinds of candy and asked which one they would like to receive. Then the tray is removed and they are given a chocolate candy, but not the one they selected. If 20 of the people in the picture group and 12 of the people in the actual candy group failed to detect the switch, would you conclude that there is convincing evidence that the proportion who experience choice blindness is different for the two treatments (choice based on a picture and choice based on seeing the actual candy)? Test the relevant hypotheses using a 0.01 significance level.

Short Answer

Expert verified
In summary, based on a hypothesis test at a 0.01 significance level, there is not enough evidence to conclude that there is a difference in choice blindness between the group who made their initial selection from a picture and the group who made their selection from actual candies. The calculated p-value of 0.050 is greater than the significance level, so we fail to reject the null hypothesis.

Step by step solution

01

Define the null and alternative hypotheses

The null hypothesis (H鈧) states that there is no difference in choice blindness between the two groups, while the alternative hypothesis (H鈧) states that there is a difference between the two groups. Mathematically, let p鈧 be the proportion of people experiencing choice blindness in the picture group, and p鈧 be the proportion in the actual candy group. The hypotheses can be defined as follows: H鈧: p鈧 = p鈧 H鈧: p鈧 鈮 p鈧
02

Calculate the sample proportions

For each group, we will calculate the proportion of people experiencing choice blindness (failed to detect the switch). Picture group (n鈧 = 100): 20 people failed to detect the switch. Actual candy group (n鈧 = 100): 12 people failed to detect the switch. Therefore, \( \bar{p}_1 = \frac{20}{100} = 0.20 \) \( \bar{p}_2 = \frac{12}{100} = 0.12 \)
03

Calculate the combined proportion

For the hypothesis test, calculate the combined proportion of choice blindness in both the groups, which is denoted as \( \bar{p} \): \( \bar{p} = \frac{(20 + 12)}{(100 + 100)} = \frac{32}{200} = 0.16 \)
04

Calculate the test statistic

The test statistic for comparing two proportions is: \( z = \frac{(\bar{p}_1 - \bar{p}_2) - 0}{\sqrt{\bar{p}(1 - \bar{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \) Plug in the calculated proportions and sample sizes: \( z = \frac{(0.20 - 0.12) - 0}{\sqrt{0.16(1 - 0.16)(\frac{1}{100} + \frac{1}{100})}} \) \( z = \frac{0.08}{\sqrt{0.16 \cdot 0.84 \cdot (\frac{1}{50})}} \) \( z \approx 1.96 \)
05

Find the p-value and make a conclusion

Since this is a two-tailed test, we will multiply the p-value corresponding to the z-score by 2: \( p = 2P(Z > 1.96) = 0.050 \) Our calculated p-value is 0.050, which is larger than the significance level of 0.01. Since the calculated p-value is greater than the significance level (0.050 > 0.01), we fail to reject the null hypothesis. There is not enough evidence to conclude that there is a difference in choice blindness between the picture group and the actual candy group at the 0.01 significance level.

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Key Concepts

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

Choice Blindness
Choice blindness is a fascinating psychological phenomenon where people do not notice a change in the choice they made. Imagine you expressed a preference and later received something different, yet you didn't notice the swap. That's choice blindness. In experiments, participants might choose an item expecting one result, but receive another, and often fail to realize or object to the change. This concept challenges our awareness and sheds light on how we might sometimes unconsciously accept outcomes. Researchers study choice blindness to understand the limits of our conscious decision-making and how external factors can influence our awareness.
Proportions
In statistics, proportions refer to the fraction of a total that exhibits a certain characteristic. When talking about proportions in hypothesis testing, it involves calculating the ratio of specific occurrences to the total number of possible cases. Such calculations help in understanding how likely it is for a particular event to happen. For evaluating choice blindness, proportions show us how many people did not detect the switch from their total group. We calculate this by dividing the number of those who experienced choice blindness by the total number of participants in each group. Understanding these proportions is key for comparing different samples.
Significance Level
The significance level, often denoted as alpha (\( \alpha \)), is a threshold set by researchers to determine how confidently they can reject a null hypothesis. It represents the probability of making a Type I error, which is rejecting a true null hypothesis. In most studies, common significance levels are 0.05, 0.01, and 0.10. In our case involving choice blindness, the significance level is 0.01, indicating a stringent criteria for rejecting the null hypothesis. This means we need strong evidence to claim there is a significant difference in the incidence of choice blindness between the two groups. Understanding and setting a proper significance level helps researchers control for error and make valid conclusions.
Test Statistic
A test statistic is a standardized value calculated from sample data during a hypothesis test. It's used to compare your data against what the null hypothesis predicts. The calculated value helps determine how extreme the sample result is, relative to the null hypothesis. For comparing two proportions, as in the choice blindness experiment, the test statistic follows a z-distribution. In our scenario, we calculate the z-value to determine if there is a significant difference between two groups of participants. The larger the test statistic, the more evidence against the null hypothesis. But, if it isn鈥檛 large enough as compared to critical values derived from a significance level, the null hypothesis cannot be rejected. Thus, test statistics are crucial for interpreting results and deciding on the validity of your hypotheses.

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Most popular questions from this chapter

A hotel chain is interested in evaluating reservation processes. Guests can reserve a room by using either a telephone system or an online system that is accessed through the hotel's web site. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. Based on these data, is it reasonable to conclude that the proportion who are satisfied is greater for those who reserve a room online? Test the appropriate hypotheses using a significance level of \(0.05 .\) (Hint: See Example \(11.4 .)\)

The article "Spray Flu Vaccine May Work Better than Injections for Tots" (San Luis Obispo Tribune, May 2, 2006) described a study that compared flu vaccine administered by injection and flu vaccine administered as a nasal spray. Each of the 8000 children under the age of 5 who participated in the study received both a nasal spray and an injection, but only one was the real vaccine and the other was salt water. At the end of the flu season, it was determined that of the 4000 children receiving the real vaccine by nasal spray, \(3.9 \%\) got the flu. Of the 4000 children receiving the real vaccine by injection, \(8.6 \%\) got the flu. a. Why would the researchers give every child both a nasal spray and an injection? b. Use a \(99 \%\) confidence interval to estimate the difference in the proportion of children who get the flu after being vaccinated with an injection and the proportion of children who get the flu after being vaccinated with the nasal spray. Based on the confidence interval, would you conclude that the proportion of children who get the flu is different for the two vaccination methods? (Hint: See Example \(11.7 .\) )

Gallup surveyed adult Americans about their consumer debt ("Americans' Big Debt Burden Growing, Not Evenly Distributed," February 4, 2016, www.gallup.com, retrieved December 15,2016 ). They reported that \(47 \%\) of millennials (those born between 1980 and 1996 ) and \(61 \%\) of Gen Xers (those born between 1965 and 1971 ) did not pay off their credit cards each month and therefore carried a balance from month to month. Suppose that these percentages were based on representative samples of 450 millennials and 300 Gen Xers. Is there convincing evidence that the proportion of Gen Xers who do not pay off their credit cards each month is greater than this proportion for millennials? Test the appropriate hypotheses using a significance level of 0.05

The report "The New Food Fights: U.S. Public Divides Over Food Science" (December 1, 2016, www.pewinternet.org, retrieved December 10,2016 ) states that younger adults are more likely to see foods with genetically modified ingredients as being bad for their health than older adults. This statement is based on a representative sample of 178 adult Americans age 18 to 29 and a representative sample of 427 adult Americans age 50 to 64 . Of those in the 18 to 29 age group, \(48 \%\) said they believed these foods were bad for their health, while only \(38 \%\) of those in the 50 to 64 age group believed this. a. Are the sample sizes large enough to use the large-sample confidence interval to estimate the difference in the population proportions? Explain. b. Estimate the difference in the proportion of adult Americans age 18 to 29 who believe the foods made with genetically modified ingredients are bad for their health and the corresponding proportion for adult Americans age 50 to \(64 .\) Use a \(90 \%\) confidence interval. c. Is zero in the confidence interval? What does this suggest about the difference in the two population proportions?

The article "Footwear, Traction, and the Risk of Athletic Injury" (January \(2016,\) www.lermagazine.com/article/footwear -traction-and-the-risk-of-athletic-injury, retrieved December \(15,\) 2016) describes a study in which high school football players were given either a conventional football cleat or a swivel disc shoe. Of 2373 players who wore the conventional cleat, 372 experienced an injury during the study period. Of the 466 players who wore the swivel disc shoe, 24 experienced an injury. The question of interest is whether there is evidence that the injury proportion is smaller for the swivel disc shoe than it is for conventional cleats. a. What are the two treatments in this experiment? b. The article didn't state how the players in the study were assigned to the two groups. Explain why it is important to know if they were assigned to the groups at random. c. For purposes of this example, assume that the players were randomly assigned to the two treatment groups. Carry out a hypothesis test to determine if there is evidence that the injury proportion is smaller for the swivel disc shoe than it is for conventional cleats. Use a significance level of 0.05 .

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