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Chapter 11: Problem 46

Many fundraisers ask for donations using e-mail and text messages. 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 whether the proportion of people who make a donation when asked for a donation by e-mail is different from the proportion of people who make a donation when asked for a donation in a text message. In this experiment, \(1.32 \%\) of those who received and opened an e-mail request for a donation and \(7.77 \%\) of those who received a text message asking for a donation actually made a donation. Assume that the people who received these requests were randomly assigned to one of the two groups (e-mail or text message) and suppose that the given percentages are based on sample sizes of 2000 (the actual sample sizes in the experiment were much larger). a. The study described is an experiment with two treatments. What are the two treatments? b. Is there convincing evidence that the proportion who make a donation is not the same for the two different methods? Carry out a hypothesis test using a significance level of 0.05 . c. Use a \(90 \%\) confidence interval to estimate the difference in the proportions who donate for the two different treatments.

Short Answer

Expert verified
a. The two treatments in the experiment are: 1. People who receive and open an e-mail request for a donation. 2. People who receive a text message asking for a donation. b. Results of the hypothesis test: Null hypothesis (H0): \(p_1 = p_2\) Alternative hypothesis (Ha): \(p_1 \neq p_2\) Significance level: 伪 = 0.05 Sample proportions: \(p_1 = 0.0132\), \(p_2 = 0.0777\) Sample sizes: \(n_1 = 2000\), \(n_2 = 2000\) Pooled sample proportion: \(\hat{p} = 0.04545\) Test statistic (z-score): \(z = -32.086\) P-value: P < 伪, so we reject the null hypothesis in favor of the alternative hypothesis. There is convincing evidence that the proportion of people who make a donation is different for the two methods (email and text message). c. 90% confidence interval for the difference in proportions: \((\hat{p_1} - \hat{p_2}) \pm z_{\alpha/2} \sqrt{\frac{\hat{p_1} (1 - \hat{p_1})}{n_1} + \frac{\hat{p_2} (1 - \hat{p_2})}{n_2}}\) Confidence interval: \((-0.0674, -0.0586)\) The 90% confidence interval for the difference in proportions for the two treatments (e-mail and text message) is \((-0.0674, -0.0586)\). This indicates that the proportion of people who make a donation through text message is significantly higher than through email, and the difference in proportions is estimated to be between 5.86% and 6.74% higher for the text message group.

Step by step solution

01

a. Identify the treatments

In this experiment, there are two treatments. They are: 1. People who receive and open an e-mail request for a donation. 2. People who receive a text message asking for a donation.
02

b. Perform a hypothesis test

To carry out the hypothesis test, we need to set up our null and alternative hypotheses. Null hypothesis (H0): \(p_1 = p_2\) (The proportion of people who make a donation when asked through email is the same as those who received a text message request.) Alternative hypothesis (Ha): \(p_1 \neq p_2\) (The proportion of people who make a donation when asked through email is different than those who received a text message request.) Significance level: 伪 = 0.05 The sample proportions are: \(p_1 = 0.0132\) (1.32% of people who received and opened an email request for a donation) \(p_2 = 0.0777\) (7.77% of people who received a text message asking for a donation) The sample sizes are: \(n_1 = 2000\) (email group) \(n_2 = 2000\) (text message group) The pooled sample proportion is: \[\hat{p} = \frac{n_1 p_1 + n_2 p_2}{n_1 + n_2} = \frac{2000(0.0132) + 2000(0.0777)}{2000 + 2000}\] Now calculate the test statistic (z-score): \[z = \frac{(\hat{p_1} - \hat{p_2}) - (p_1 - p_2)}{\sqrt{\hat{p}(1 - \hat{p})(\frac{1}{n_1} + \frac{1}{n_2})}} \] Calculate the p-value by finding the probability of the test statistic under the null hypothesis. We use the standard normal distribution to find the area under the curve that corresponds to the calculated z-score. If the p-value < 伪, we reject the null hypothesis.
03

c. Calculate a 90% confidence interval

To calculate the 90% confidence interval for the difference in the proportions, we will use the following formula: \(\hat{p_1} - \hat{p_2} \pm z_{\alpha/2} \sqrt{\frac{\hat{p_1} (1 - \hat{p_1})}{n_1} + \frac{\hat{p_2} (1 - \hat{p_2})}{n_2}}\) where \(\hat{p_1}\) and \(\hat{p_2}\) are the sample proportions, \(n_1\) and \(n_2\) are the sample sizes, and \(z_{\alpha/2}\) is the z-score corresponding to 90% confidence level. For a 90% confidence level, the critical z-score in a two-tailed test is 1.645. Plug in the sample proportions and sample sizes into the formula to obtain the confidence interval. This will give an estimate of the difference in donation proportions for the two treatments (e-mail and text message).

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Key Concepts

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

Statistical Significance
Understanding the concept of statistical significance is essential when interpreting the results of experiments and surveys in statistics.

Statistical significance helps us determine whether our observations are likely due to chance or reflect true differences in the data. In the context of our exercise, where two methods of soliciting donations are compared, it's important to assess whether the observed difference in donation rates is statistically significant.

This assessment is done by setting a significance level before conducting the hypothesis test. A common choice is the 0.05 level. If the probability of observing the test statistic, known as the p-value, is less than the significance level, we declare the results to be statistically significant. This would suggest that the different donation rates between email and text message recipients are not due to random variation alone, but likely due to the effectiveness of the methods themselves.
Confidence Interval
The confidence interval gives us a range of values within which we believe the true parameter, like a difference in proportions, lies.

In our exercise, a 90% confidence interval is desired for estimating the difference between the two donation proportions. The use of the confidence interval is powerful as it provides a range rather than a single point estimate, offering an interval of plausible values given our data.

The calculation includes the desired level of confidence and utilizes the z-score, which corresponds to the confidence level's critical value. If the confidence interval does not contain zero, this further strengthens the evidence against the null hypothesis, suggesting a significant difference between the proportions.
Proportion Hypothesis Test
In the context of proportion hypothesis testing, we compare the proportions of a categorical outcome between groups.

The exercise involves testing the null hypothesis that the proportion of donors is the same for emails and text messages against the alternative hypothesis that they are different. After setting up our hypotheses and calculating the pooled sample proportion, which combines the information from both groups, we proceed to the hypothesis test.

Using the calculated sample proportions and sizes, we determine the test statistic that measures the degree of deviation from the null hypothesis. This is where the concept of the z-score becomes vital.
Z-Score
The z-score is a fundamental concept in statistics that indicates how many standard deviations a data point is from the mean.

In hypothesis testing, the z-score helps us determine whether the evidence is strong enough to reject the null hypothesis. It is calculated based on the sample data and reflects the significance of the test result. A high magnitude of the z-score suggests that the data point is far from what we would expect under the null hypothesis, possibly indicating a significant difference.

In our exercise, the z-score is used to calculate the p-value and is also crucial in determining the confidence interval, as it provides the critical boundary that our estimate must exceed for the result to be significant at the desired confidence level.

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Most popular questions from this chapter

According to the U.S. Census Bureau (www.census.gov), the percentage of U.S. residents living in poverty in 2015 was \(12.2 \%\) for men and \(14.8 \%\) for women. These percentages were estimates based on data from large representative samples of men and women. Suppose that the sample sizes were 1200 for men and 1000 for women. You would like to use the survey data to estimate the difference in the proportion living in poverty in 2015 for men and women. (Hint: See Example \(11.2 .\).) a. Answer the four key questions (QSTN) for this problem. What method would you consider based on the answers to these questions? b. Use the five-step process for estimation problems \(\left(\mathrm{EMC}^{3}\right)\) to calculate and interpret a \(90 \%\) large- sample confidence interval for the difference in the proportion living in poverty in 2015 for men and women.

In the experiment described in the article "Study Points to Benefits of Knee Replacement Surgery Over Therapy Alone" (The New York Times, October 21,2015 ), adults who were considered candidates for knee replacement were followed for one year. Suppose that 200 patients were randomly assigned to one of two groups. One hundred were assigned to a group that had knee replacement surgery followed by therapy and the other half were assigned to a group that did not have surgery but did receive therapy. After one year, \(86 \%\) of the patients in the surgery group and \(68 \%\) of the patients in the therapy only group reported pain relief. Is there convincing evidence that the proportion experiencing pain relief is greater for the surgery treatment than for the therapy treatment? Use a significance level of 0.05

The article "More Teen Drivers See Marijuana as OK; It's a Dangerous Trend" (USA TODAY, February 23,2012 ) describes two surveys of U.S. high school students. One survey was conducted in 2009 and the other was conducted in \(2011 .\) In \(2009,78 \%\) of the people in a representative sample of 2300 students said marijuana use is very distracting or extremely distracting to their driving. In \(2011,70 \%\) of the people in a representative sample of 2294 students answered this way. Use the five-step process for estimation problems \(\left(\mathrm{EMC}^{3}\right)\) to construct and interpret a \(99 \%\) large-sample confidence interval for the difference in the proportion of high school students who believed marijuana was very distracting or extremely distracting in 2009 and this proportion in 2011 .

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 ). Suppose that this conclusion was based on a \(90 \%\) confidence interval for the difference in treatment proportions. Which of the following three statements is correct? Explain why you chose this statement. Statement 1: Both endpoints of the confidence interval were negative. Statement 2: The confidence interval included \(0 .\) Statement 3: Both endpoints of the confidence interval were positive.

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.
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