/*! 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

The report "Young People Living on the Edge" (Greenberg Quinlan Rosner Research, 2008) summarizes a survey of people in two independent random samples. One sample consisted of 600 young adults (age 19 to 35 ), and the other sample consisted of 300 parents of young adults age 19 to \(35 .\) The young adults were presented with a variety of situations (such as getting married or buying a house) and were asked if they thought that their parents were likely to provide financial support in that situation. The parents of young adults were presented with the same situations and asked if they would be likely to provide financial support in that situation. When asked about getting married, \(41 \%\) of the young adults said they thought parents would provide financial support and \(43 \%\) of the parents said they would provide support. Carry out a hypothesis test to determine if there is convincing evidence that the proportion of young adults who think their parents would provide financial support and the proportion of parents who say they would provide support are different.

The Interactive Advertising Bureau surveyed a representative sample of 1000 adult Americans and a representative sample of 1000 adults in China (鈥淢ajority of Digital Users in U.S. and China Regularly Shop and Purchase via E-Commerce," November \(10,2016,\) www.iab.com, retrieved December 15,2016 ). They reported that American shoppers are much more likely to use a credit or a debit card to make an online purchase. This conclusion was based on finding that \(63 \%\) of the people in the United States sample said they pay with a credit or a debit card, while only \(34 \%\) of those in the China sample said that they used a credit card or a debit card to pay for online purchases. To determine if the stated conclusion is justified, you want to carry out a test of hypotheses to determine if there is convincing evidence that the proportion who pay with a credit card or a debit card is greater for adult Americans than it is for adult Chinese. a. What hypotheses should be tested to answer the question of interest? b. Are the two samples large enough for the large-sample test for a difference in population proportions to be appropriate? Explain. c. Based on the following Minitab output, what is the value of the test statistic and what is the value of the associated \(P\) -value? If a significance level of 0.01 is selected for the test, will you reject or fail to reject the null hypothesis? d. Interpret the result of the hypothesis test in the context of this problem.

In a test of hypotheses about a difference in treatment proportions, what does it mean when the null hypothesis is not rejected?

Many people believe that they experience "information overload" in today's digital world. The report "Information Overload" (Pew Research Center, December 7, 2016) describes a survey in which people were asked if they feel overloaded by information. In a representative sample of 634 college graduates, 101 indicated that they suffered from information overload, while 119 people in an independent representative sample of 496 people who had never attended college said that they suffered from information overload. a. Construct and interpret a \(95 \%\) large-sample confidence interval for the proportion of college graduates who experience information overload. (Hint: This is a onesample confidence interval.) b. Construct and interpret a \(95 \%\) large-sample confidence interval for the proportion of people who have never attended college who experience information overload. c. Do the confidence intervals from Parts (a) and (b) overlap? What does this suggest about the two population proportions? d. Construct and interpret a \(95 \%\) large-sample confidence interval for the difference in the proportions who have experienced information overload for college graduates and for people who have never attended college. e. Is the interval in Part (d) consistent with your answer in Part (c)? Explain.

Some fundraisers believe that people are more likely to make a donation if there is a relatively quick deadline given for making the donation. The paper "Now or Never! The Effect of Deadlines on Charitable Giving: Evidence from Two Natural Field Experiments" (Journal of Behavioral and Experimental Economics [2016]: \(1-10\) ) describes an experiment to investigate the influence of deadlines. In this experiment, \(1.2 \%\) of those who received an e-mail request for a donation that had a three-day deadline to make a donation and \(0.8 \%\) of those who received the same e-mail request but without a deadline made a donation. The people who received the e-mail request were randomly assigned to one of the two groups (e-mail with deadline and e-mail without deadline). Suppose that the given percentages are based on sample sizes of 2000 (the actual sample sizes in the experiment were much larger). Use a \(90 \%\) confidence interval to estimate the difference in the proportion who donate for the two different treatments.

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