Questions: Positive events are great, but recent research suggests that unexpected positive outcomes (e.g., an unseasonably sunny day) predict greater-than-normal amounts of risk-taking and gambling (Otto, Fleming, Glimcher, 2016). Researchers demonstrated this by comparing lottery sales-indicative of risk-taking-on normal days with lottery sales on days when some unexpected positive event occurred in the city. They observed increased sales after unexpected positive outcomes. Suppose that a researcher extends this observation to the laboratory and randomly assigns participants to two groups. Group 1 receives an unexpectedly large payment for participating and Group 2 receives the expected amount of compensation. The researcher then measures how much money the participants are willing to gamble in a game of chance. Unexpected Positive Outcome Expected Outcome --- --- n=16 n=16 M=5.75 M=5.00 SS=6.5 SS=10.0 Test the one-tailed hypothesis that an unexpected positive outcome increased the amount of money that participants were willing to gamble. Use a α=.01.

Positive events are great, but recent research suggests that unexpected positive outcomes (e.g., an unseasonably sunny day) predict greater-than-normal amounts of risk-taking and gambling (Otto, Fleming, Glimcher, 2016). Researchers demonstrated this by comparing lottery sales-indicative of risk-taking-on normal days with lottery sales on days when some unexpected positive event occurred in the city. They observed increased sales after unexpected positive outcomes. Suppose that a researcher extends this observation to the laboratory and randomly assigns participants to two groups. Group 1 receives an unexpectedly large payment for participating and Group 2 receives the expected amount of compensation. The researcher then measures how much money the participants are willing to gamble in a game of chance. Unexpected Positive Outcome Expected Outcome --- --- n=16 n=16 M=5.75 M=5.00 SS=6.5 SS=10.0 Test the one-tailed hypothesis that an unexpected positive outcome increased the amount of money that participants were willing to gamble. Use a α=.01.
Transcript text: Positive events are great, but recent research suggests that unexpected positive outcomes (e.g., an unseasonably sunny day) predict greater-than-normal amounts of risk-taking and gambling (Otto, Fleming, & Glimcher, 2016). Researchers demonstrated this by comparing lottery sales-indicative of risk-taking-on normal days with lottery sales on days when some unexpected positive event occurred in the city. They observed increased sales after unexpected positive outcomes. Suppose that a researcher extends this observation to the laboratory and randomly assigns participants to two groups. Group 1 receives an unexpectedly large payment for participating and Group 2 receives the expected amount of compensation. The researcher then measures how much money the participants are willing to gamble in a game of chance. \begin{tabular}{cc} \hline \begin{tabular}{c} Unexpected \\ Positive \\ Outcome \end{tabular} & \begin{tabular}{c} Expected \\ Outcome \end{tabular} \\ \hline$n=16$ & $n=16$ \\ $M=5.75$ & $M=5.00$ \\ $S S=6.5$ & $S S=10.0$ \\ \hline \end{tabular} Test the one-tailed hypothesis that an unexpected positive outcome increased the amount of money that participants were willing to gamble. Use a $\alpha=.01$.
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Solution

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Solution Steps

Step 1: Standard Error Calculation

The standard error \( SE \) is calculated using the formula:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]

Given that both sample variances \( s_1^2 \) and \( s_2^2 \) are \( 0 \) (since all values in each group are the same), we have:

\[ SE = \sqrt{\frac{0.0}{16} + \frac{0.0}{16}} = 0.0 \]

Step 2: Test Statistic Calculation

The test statistic \( t \) is calculated using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} \]

Substituting the sample means and the standard error:

\[ t = \frac{5.75 - 5.0}{0.0} = \infty \]

Step 3: Degrees of Freedom Calculation

The degrees of freedom \( df \) for Welch's t-test is calculated using the formula:

\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]

Since both variances are \( 0 \), we find:

\[ df = \frac{0.0}{0.0} = \text{nan} \]

Step 4: p-value Calculation

The p-value is calculated as:

\[ P = 1 - T(t) \]

Since \( t = \infty \):

\[ P = 1 - T(\infty) = \text{nan} \]

Step 5: Summary of Results

The results of the Welch's t-test are as follows:

  • \( t \)-statistic: \( \infty \)
  • p-value: \( \text{nan} \)
  • Degrees of freedom: \( \text{nan} \)
  • Critical value: \( \text{nan} \)

Final Answer

Due to the calculations resulting in undefined values (nan), we cannot draw a conclusion from the test. Therefore, the results are inconclusive.

\(\boxed{\text{Inconclusive results due to undefined values}}\)

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