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The following data are based on information from the Regis University Psychology Department. In an effort to determine if rats perform certain tasks more quickly if offered larger rewards, the following experiment was performed. On day 1 , a group of three rats was given a reward of one food pellet each time they ran a maze. A second group of three rats was given a reward of five food pellets each time they ran the maze. On day 2, the groups were reversed, so the first group now got five food pellets for running the maze and the second group got only one pellet for running the same maze. The average times in seconds for each rat to run the maze 30 times are shown in the following table. \(\begin{array}{l|cccccc} \hline \text { Rat } & A & B & C & D & E & F \\ \hline \text { Time with one food pellet } & 3.6 & 4.2 & 2.9 & 3.1 & 3.5 & 3.9 \\\ \hline \text { Time with five food pellets } & 3.0 & 3.7 & 3.0 & 3.3 & 2.8 & 3.0 \\ \hline \end{array}\) Do these data indicate that rats receiving larger rewards tend to run the maze in less time? Use a \(5 \%\) level of significance.

Step by step solution

01

Define the Hypotheses

We want to test if larger rewards reduce the time taken by rats to complete the maze. Set up the null hypothesis (H_0) such that there is no difference in maze completion times. The alternative hypothesis (H_1) will state that rats with larger rewards complete the maze in less time.\[\begin{align*}H_0 &: \mu_1 = \mu_5 \H_1 &: \mu_1 > \mu_5\end{align*}\]where \mu_1 is the mean time for one pellet and \mu_5 is the mean time for five pellets.

02

Collect and Summarize Data

Calculate the mean and standard deviation of the times for both conditions (one pellet and five pellets) from the given data. Mean time with one pellet is calculated as:\[\mu_1 = \frac{3.6 + 4.2 + 2.9 + 3.1 + 3.5 + 3.9}{6} = 3.53\, \text{seconds} \]Mean time with five pellets:\[\mu_5 = \frac{3.0 + 3.7 + 3.0 + 3.3 + 2.8 + 3.0}{6} = 3.13\, \text{seconds}\]Calculate the standard deviations for both sets.

03

Conduct a t-Test

Perform a paired t-test because the same rats are used for both conditions, making the samples dependent. First, find the differences between paired observations and calculate the mean and standard deviation of these differences.Difference array: \([-0.6, -0.5, 0.1, 0.2, -0.7, -0.9]\)Calculate the mean and standard deviation of differences. Mean of differences: \(-0.4\), Standard Deviation (s_d): use appropriate formulas.

04

Calculate the t-Statistic

Calculate the t-statistic using the formula for dependent samples:\[t = \frac{\text{Mean difference} - 0}{s_d / \sqrt{n}}\]where \(n = 6\) is the number of paired observations, and s_d is the standard deviation of the differences. Substitute the values to get:\[t = \frac{-0.4}{s_d / \sqrt{6}}\]Compute this value.

05

Determine the Critical Value and Conclusion

Use a t-table to find the critical value for a 5% significance level with 5 degrees of freedom (n-1) for a one-tailed test. Compare the calculated t-statistic to the critical value. If the t-statistic is greater than the critical value, reject H_0 . If it is less, fail to reject H_0 . Interpret the results accordingly.

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Key Concepts

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

t-test

In hypothesis testing, a **t-test** is used to determine if there is a significant difference between the means of two groups. In the given problem, a paired t-test is appropriate due to the use of **dependent samples**. This means the same group of rats was tested under different conditions (one food pellet vs. five food pellets).

A t-test evaluates whether the means of two paired sets are statistically different from each other. It takes into account the mean of the differences between paired observations, the standard deviation of these differences, and the number of samples.

The test calculation results in a t-statistic, which can be compared to a critical value from a t-distribution table. The decision to accept or reject the null hypothesis (usually denoting no effect) is based on this comparison. When conducting a t-test, it's crucial to decide whether the test will be one-tailed or two-tailed, which affects the determination of the critical value from the t-table.

significance level

The **significance level**, often denoted by \(\alpha\), is a critical concept in hypothesis testing. It represents the probability of rejecting the null hypothesis when it is true — also known as a Type I error. In simpler terms, it's the threshold for determining the strength of the evidence against the null hypothesis.

For this exercise, a significance level of 5% was used (\(\alpha = 0.05\)). A 5% significance level means there is a 5% risk of concluding that a difference exists when there actually is none. Selecting a significance level involves a trade-off: lower values (e.g., 1%) reduce the chance of making a Type I error but might increase Type II errors (incorrectly failing to reject a false null hypothesis).

In many scientific studies, a 5% significance level is standard, balancing sensitivity and specificity, and is often considered the threshold for "statistical significance" in hypothesis testing.

dependent samples

**Dependent samples** refer to samples in which the observations are not independent of one another. This is the case in paired sample tests, where each observation in one sample can be linked directly to an observation in the other sample.

In the given problem, the same group of rats experienced both conditions — running the maze with one food pellet first and then with five pellets. Each rat's time in both conditions can be seen as a pair, making the data dependent.

Using dependent samples affects how the t-test is conducted, especially in calculating the differences between pairs. It helps control for individual variability because each subject serves as their own control. This design reduces the variability and often has more power to detect differences than independent samples.

null hypothesis

The **null hypothesis** (**null hypothesis** or \(H_0\)) is a fundamental part of hypothesis testing, proposing that there is no effect or no difference. It acts as a default or starting assumption that any observed effect is due to chance.

In our scenario, the null hypothesis was that the rats' maze completion times are the same regardless of receiving one or five food pellets. This is mathematically stated as \(\mu_1 = \mu_5\), where \(\mu_1\) is the mean time with one pellet and \(\mu_5\) is the mean time with five pellets.

Testing the null hypothesis involves calculating the t-statistic and comparing it against a critical value from a t-distribution. If data provides sufficient evidence (beyond the set significance level) to reject the null hypothesis, it suggests that there's a statistically significant effect, prompting acceptance of the alternative hypothesis.