Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

In the paper "Happiness for Sale: Do Experiential Purchases Make Consumers Happier than Material Purchases?" (Journal of Consumer Research [2009]: \(188-197\) ), the authors distinguish between spending money on experiences (such as travel) and spending money on material possessions (such as a car). In an experiment to determine if the type of purchase affects how happy people are after the purchase has been made, 185 college students were randomly assigned to one of two groups. The students in the "experiential" group were asked to recall a time when they spent about \(\$ 300\) on an experience. They rated this purchase on three different happiness scales that were then combined into an overall measure of happiness. The students assigned to the "material" group recalled a time that they spent about \(\$ 300\) on an object and rated this purchase in the same manner. The mean happiness score was 5.75 for the experiential group and 5.27 for the material group. Standard deviations and sample sizes were not given in the paper, but for purposes of this exercise, suppose that they were as follows: \begin{tabular}{|ll|} \hline Experiential & Material \\ \hline\(n_{1}=92\) & \(n_{2}=93\) \\ \(s_{1}=1.2\) & \(s_{2}=1.5\) \\ \hline \end{tabular} Using the following Minitab output, carry out a hypothesis test to determine if these data support the authors' conclusion that, on average, "experiential purchases induced more reported happiness." Use \(\alpha=0.05\) Two-Sample T-Test and Cl Sample \(\begin{array}{rrrrr}\text { ple } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 1 & 92 & 5.75 & 1.20 & 0.13 \\ 2 & 93 & 5.27 & 1.50 & 0.16\end{array}\) Difference \(=\operatorname{mu}(1)-\operatorname{mu}(2)\) Estimate for difference: 0.480000 \(95 \%\) lower bound for difference: 0.149917 T-Test of difference \(=0(\mathrm{vs}>): \mathrm{T}\) -Value \(=2.40 \mathrm{P}\) -Value \(=\) \(0.009 \mathrm{DF}=175\)

Short Answer

Expert verified
The statistical test supports the claim that, on average, 'experiential purchases induced more reported happiness.' The test provides enough evidence to reject the null hypothesis and conclude that the mean of reported happiness is greater for the experiential group compared to the material one.

Step by step solution

01

Setup Hypotheses

Firstly, it's important to define the null and alternative hypotheses. The null hypothesis (\(H_0\)) is that there is no difference in the average happiness level between the experiential purchase group and the material purchase group. The alternative hypothesis (\(H_a\)) is that there is a difference. These can be written as follows:\n\nNull hypothesis (\(H_0\)): \(\mu_1 = \mu_2\)\n, Alternate hypothesis (\(H_a\)): \(\mu_1 > \mu_2\)
02

Test calculation

The calculated test statistic from the Minitab output is 2.40 and p-value is 0.009.
03

Decision making

Now compare the p-value with the level of significance (Alpha). If the p-value is less than the significance level (\(\alpha = 0.05\)), reject the null hypothesis. Since p-value (0.009) < significance level (0.05) We reject \(H_0\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Null and Alternative Hypotheses
Understanding the null and alternative hypotheses is foundational to any hypothesis testing procedure. In the context of our exercise, the null hypothesis (\(H_0\)) posits that there is no statistical difference in happiness levels between groups making experiential purchases and those making material purchases. Formally, it states that \(\mu_1 = \mu_2\), where \(\mu_1\) and \(\mu_2\) represent the mean happiness scores for the experiential and material groups, respectively.

The alternative hypothesis (\(H_a\) or \(H_1\)), on the other hand, challenges the null by proposing that a specific difference exists; in our case, that experiential purchases lead to higher happiness, formally stated as \(\mu_1 > \mu_2\). It guides the direction of the statistical test, indicating that we're specifically looking for evidence of greater happiness from experiential spending.
P-value
The p-value plays a critical role in hypothesis testing. It quantifies the probability of observing a test statistic as extreme as, or more extreme than, the value observed in your study, assuming that the null hypothesis is true. In simpler terms, it's a measure of how surprising your data is under the assumption that there are no real effects or differences ('surprising' generally means how well your data supports the null hypothesis).

For our example, the p-value is 0.009. This number is a probability, falling between 0 and 1, and it tells us how compatible our data are with the null hypothesis. A very small p-value, such as ours, indicates that if the null hypothesis were true, it would be very unlikely to observe the data we have. This leads to stronger evidence against the null hypothesis, suggesting that our alternative hypothesis might be the correct one.
Statistical Significance
Statistical significance is the determination of whether or not the results of a statistical test, such as our t-test, supports the rejection of the null hypothesis. This decision is made by comparing the p-value to the chosen significance level, often denoted by \(\alpha\). The significance level represents the probability of committing a Type I error — rejecting a true null hypothesis.

In hypothesis testing, if the p-value is less than the chosen \(\alpha\) level (usually 0.05), the results are deemed statistically significant. This implies that the evidence in the sample is strong enough to conclude that there is a statistically detectable effect. In the given exercise, with a p-value of 0.009 and an \(\alpha\) of 0.05, the result is statistically significant, and we would reject the null hypothesis in favor of the alternative.
Two-Sample t-Test
The two-sample t-test is a statistical test used to compare the means of two independent groups. It offers a way to evaluate whether the difference in means is statistically significant or if it likely occurred by random chance. The t-test takes into account both the sample sizes and variability within the sample data of the two groups, which is often reflected in the standard deviation values.

Applying this to our situation, the two-sample t-test is used to contrast the happiness scores of the experiential and material purchase groups. We are provided Minitab output data showing a calculated test statistic (also called t-value) and an associated p-value. With a t-value of 2.40 and our previously examined low p-value, the test indicates that we have sufficient evidence to believe that there is a significant difference in happiness scores, favoring the experiential group.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article "A "White' Name Found to Help in Job Search" (Associated Press, January 15,2003 ) described an experiment to investigate if it helps to have a "white-sounding" first name when looking for a job. Researchers sent resumes in response to 5,000 ads that appeared in the Boston Globe and Chicago Tribune. The resumes were identical except that 2,500 used "white-sounding" first names, such as Brett and Emily, whereas the other 2,500 used "black- sounding" names such as Tamika and Rasheed. The 5,000 job ads were assigned at random to either the white-sounding name group or the black-sounding name group. Resumes with whitesounding names received 250 responses while resumes with black sounding names received only 167 responses. a. What are the two treatments in this experiment? b. Use the data from this experiment to estimate the difference in response proportions for the two treatments.

The article "An Alternative Vote: Applying Science to the Teaching of Science" (The Economist, May 12,2011 ) describes an experiment conducted at the University of British Columbia. A total of 850 engineering students enrolled in a physics course participated in the experiment. Students were randomly assigned to one of two experimental groups. Both groups attended the same lectures for the first 11 weeks of the semester. In the twelfth week, one of the groups was switched to a style of teaching where students were expected to do reading assignments prior to class, and then class time was used to focus on problem solving, discussion, and group work. The second group continued with the traditional lecture approach. At the end of the twelfth week, students were given a test over the course material from that week. The mean test score for students in the new teaching method group was \(74,\) and the mean test score for students in the traditional lecture group was \(41 .\) Suppose that the two groups each consisted of 425 students. Also suppose that the standard deviations of test scores for the new teaching method group and the traditional lecture method group were 20 and 24 , respectively. Estimate the difference in mean test score for the two teaching methods using a \(95 \%\) confidence interval. Be sure to give an interpretation of the interval.

(C1) The paper "Effects of Caffeine on Repeated Sprint Ability, Reactive Agility Time, Sleep and Next Day Performance" (Journal of Sports Medicine and Physical Fitness [2010]: 455 - 464) describes an experiment in which male athlete volunteers who were considered low caffeine consumers were assigned at random to one of two experimental groups. Those assigned to the caffeine group drank a beverage which contained caffeine 1 hour before an exercise session. Those in the no- caffeine group drank a beverage that did not contain caffeine 1 hour before an exercise session. That night, participants wore a device that measures sleep activity. The researchers reported that there was no significant difference in mean sleep duration for the two experimental groups. In the context of this experiment, explain what it means to say that there is no significant difference in the group means. In particular, explain if this means that the mean sleep durations for the two groups are equal.

The article "Why We Fall for This" (AARP Magazine, May/June 2011 ) describes an experiment investigating the effect of money on emotions. In this experiment, students at University of Minnesota were randomly assigned to one of two groups. One group counted a stack of dollar bills. The other group counted a stack of blank pieces of paper. After counting, each student placed a finger in very hot water and then reported a discomfort level. It was reported that the mean discomfort level was significantly lower for the group that had counted money. In the context of this experiment, explain what it means to say that the money group mean was significantly lower than the blank- paper group mean.

14.32 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.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free