/*! 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 27 Some fundraisers believe that pe... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
We are 90% confident that the difference in proportions of donations between those who received an email with a deadline and those who received an email without a deadline is between \(0.0002\) and \(0.0078\). This result suggests that having a deadline has a positive influence on donation rates.

Step by step solution

01

Identify the proportions and sample sizes

From the problem, let's identify the given information: - Proportion who made a donation with a deadline (p1): \(1.2\% = 0.012\) - Sample size for the deadline group (n1): 2000 - Proportion who made a donation without a deadline (p2): \(0.8\% = 0.008\) - Sample size for the no deadline group (n2): 2000
02

Calculate the point estimate

The point estimate for the difference in proportions is the difference between the proportions of donations in the two groups: Point estimate = p1 - p2 = 0.012 - 0.008 = 0.004
03

Calculate the margin of error

To calculate the margin of error, we need to find the standard error (SE). The formula for the standard error for the difference in proportions is: \[SE = \sqrt{ \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} }\] Substitute the given values: \[SE = \sqrt{ \frac{0.012(1-0.012)}{2000} + \frac{0.008(1-0.008)}{2000} }\] Calculate the standard error: \(SE \approx 0.0023\) Now, we need to find the z-score for a 90% confidence interval: Z-score = 1.645 (for 90% confidence interval, use a z-table or calculator) Margin of error = Z-score * SE = 1.645 * 0.0023 ≈ 0.0038
04

Calculate the 90% confidence interval

Now, we have the point estimate of 0.004 and the margin of error is 0.0038. We can calculate the 90% confidence interval for the difference in proportions: Lower limit = Point estimate - Margin of error = 0.004 - 0.0038 ≈ 0.0002 Upper limit = Point estimate + Margin of error = 0.004 + 0.0038 ≈ 0.0078
05

Interpret the results

This means we are 90% confident that the difference in proportions between the group that received an email with a deadline and the group that received an email without a deadline is between 0.0002 and 0.0078. It indicates that there is a positive effect of having a deadline on donation rates, as the confidence interval does not include 0.

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Key Concepts

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

Difference in Proportions
The concept of difference in proportions is crucial when comparing two separate groups and their outcomes in an experiment, like the one described in this exercise. In our particular case, the two groups are those who received a donation request email with a deadline and those who didn't. The proportion refers to the fraction or percentage of individuals in each group that made a donation after receiving the email.

To find the difference in proportions, you simply subtract one group's proportion from the other. Here, the donation proportion for the deadline group, denoted as \(p_1\), was 0.012, or 1.2%. The proportion for the group without a deadline, denoted as \(p_2\), was 0.008, or 0.8%. So, the point estimate of the difference is simply:
  • \( p_1 - p_2 = 0.012 - 0.008 = 0.004 \)
This represents a preliminary estimate of how much more likely individuals were to donate when given a deadline, calculated in a simple and straightforward manner. It's important not to stop here, though, because without considering the variability of this estimate, you're only seeing part of the picture.
Standard Error
The standard error (SE) is a measure of the variability or uncertainty associated with our estimate, specifically, the difference in the proportions across two groups. It accounts for how much our sample-based difference might shift if we were to take many samples similarly.

The formula for standard error with respect to difference in proportions is: \[ SE = \sqrt{ \frac{p_1(1-p_1)}{n_1} + \frac{p_2(1-p_2)}{n_2} } \] Where:
  • \(p_1\) is the proportion with a deadline
  • \(p_2\) is the proportion without a deadline
  • \(n_1\) and \(n_2\) are the sample sizes of the respective groups
Plugging in our numbers from the exercise, we get: \[ SE = \sqrt{ \frac{0.012 \times (1 - 0.012)}{2000} + \frac{0.008 \times (1 - 0.008)}{2000} } \] Which results in a standard error of approximately 0.0023.

This value gives us a sense of spread or dispersion around our estimated differences in proportions and tells us how precise our estimate is likely to be.
Margin of Error
The margin of error expands our understanding of the point estimate by giving a range within which we expect the true population difference in proportions to lie, based on our sample statistics.

To compute the margin of error for a confidence interval, we multiply the standard error by the z-score corresponding to our desired level of confidence. In our scenario, to find the 90% confidence interval, a z-score of 1.645 is appropriate. The margin of error then becomes:
  • Margin of Error = Z-score \( \times \) SE = 1.645 \( \times \) 0.0023 \( \approx 0.0038 \)
This margin indicates the degree of deviation we might expect from our point estimate (difference in proportions) when predicting the true difference in the wider population.

By adding and subtracting the margin of error from our point estimate, we derive a confidence interval. This tells us that with 90% confidence, the true difference between those two email groups' donation rates lies somewhere between 0.0002 and 0.0078. The interval does not contain zero, suggesting a legitimate effect of the deadline on donation behavior.

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Most popular questions from this chapter

Women diagnosed with breast cancer whose tumors have not spread may be faced with a decision between two surgical treatments -mastectomy (removal of the breast) or lumpectomy (only the tumor is removed). In a long-term study of the effectiveness of these two treatments, 701 women with breast cancer were randomly assigned to one of two treatment groups. One group received mastectomies and the other group received lumpectomies and radiation. Both groups were followed for 20 years after surgery. It was reported that there was no statistically significant difference in the proportion surviving for 20 years for the two treatments (Associated Press, October \(17,\) 2002). What hypotheses do you think the researchers tested in order to reach the given conclusion? Did the researchers reject or fail to reject the null hypothesis?

As part of a study described in the report "I Can't Get My Work Done!" (harmon.ie/blog/i-cant-get-my-work \- done-how-collaboration-social-tools-drain-productivity, 2011, retrieved May 6,2017 ), people in a sample of 258 cell phone users ages 20 to 39 were asked if they use their cell phones to stay connected while they are in bed, and 168 said "yes." The same question was also asked of each person in a sample of 129 cell phone users ages 40 to \(49,\) and 61 said "yes." You might expect the proportion who stay connected while in bed to be higher for the 20 to 39 age group than for the 40 to 49 age group, but how much higher? a. Construct and interpret a \(90 \%\) large-sample confidence interval for the difference in the population proportions of cell phone users ages 20 to 39 and those ages 40 to 49 who say that they sleep with their cell phones. Interpret the confidence interval in context. Note that the sample sizes in the two groups - cell phone users ages 20 to 39 and those ages 40 to 49 -are large enough to satisfy the conditions for a large-sample test and a large-sample confidence interval for two population proportions. Even though the sample sizes are large enough, you can still use simulation-based methods. b. Use the output below to identify a \(90 \%\) bootstrap confidence interval for the difference in the population proportions of cell phone users ages 20 to 39 and those ages 40 to 49 who say that they sleep with their cell phones. c. Compare the confidence intervals you computed in Parts (a) and (b). Would your interpretation change using the bootstrap confidence interval compared with the large-sample confidence interval? Explain.

The article referenced in the previous exercise also reported that \(53 \%\) of the Republicans surveyed indicated that they were opposed to making women register for the draft. Would you use the large-sample test for a difference in population proportions to test the hypothesis that a majority of Republicans are opposed to making women register for the draft? Explain why or why not.

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.

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.

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