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91Ó°ÊÓ

A psychologist wishes to verify that a certain drug increases the reaction time to a given stimulus. The following reaction times (in tenths of a second) were recorded before and after injection of the drug for each of four subjects: Test at the \(5 \%\) level of significance to determine whether the drug significantly increases reaction time.

Short Answer

Expert verified
Answer: Based on the paired t-test's results and comparison of the t-statistic with the critical value, [accept/reject] the null hypothesis. Therefore, there is [enough/not enough] evidence to conclude that the drug significantly increases the reaction time at the \(5\%\) level of significance.

Step by step solution

01

State the hypotheses

The null hypothesis, \(H_0\), is that there is no significant difference in the reaction times before and after injecting the drug. The alternative hypothesis, \(H_1\), is that the drug increases the reaction time. \(H_0: \mu_D = 0\) \(H_1: \mu_D > 0\) Here, \(\mu_D\) represents the mean difference in reaction times.
02

Calculate the mean and standard deviation of differences

Calculate the mean \(\bar{D}\) and standard deviation \(s_D\) of the differences between the pre- and post-injection reaction times for each subject. Let \(D_i\) be the difference in reaction times for subject \(i\) (\(D_i =\) post-injection time - pre-injection time).
03

Calculate t-statistic

Calculate the t-statistic by using the following formula: \(t = \frac{\bar{D}}{s_D / \sqrt{n}}\) where \(n\) is the number of subjects.
04

Determine the critical value

Using a t-distribution table, determine the critical value for a one-tailed test at the \(5\%\) level of significance with \(n-1\) degrees of freedom (\(df = n -1\)).
05

Compare the t-statistic and the critical value

Compare the t-statistic calculated in Step 3 with the critical value determined in Step 4. If the t-statistic is greater than the critical value, reject the null hypothesis, otherwise, fail to reject the null hypothesis. If \(t >\) critical value, reject \(H_0\). If not, fail to reject \(H_0\).
06

Draw the conclusion

Based on the comparison in Step 5, draw the conclusion about whether the drug significantly increases the reaction time at the \(5\%\) level of significance. If the null hypothesis is rejected, it means that there is enough evidence to support that the drug increases reaction time, while if we fail to reject the null hypothesis, there is not enough evidence to conclude that the drug significantly increases reaction time at the desired level of significance.

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Key Concepts

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

Null Hypothesis
The null hypothesis, often denoted as \(H_0\), is a fundamental concept in hypothesis testing. It is a statement that there is no effect or difference in a particular situation. In the context of the problem, the null hypothesis asserts that the drug does not significantly alter reaction times. Essentially, it assumes that any observed changes are due to random chance rather than the drug's effect.
For example, in our psychologist's study, the null hypothesis (\(H_0\)) is that the mean difference in reaction times before and after the injection, \(\mu_D\), is equal to 0. This means firstly we presume there is no real change caused by the drug. Always remember, if our test results indicate it's unlikely to happen under the null hypothesis, then we may consider evidence against it.
t-Test
A t-test is a statistical method used to determine whether there is a significant difference between the means of two groups. In this exercise, it helps assess if the drug has an impact on reaction times. The t-test is applicable here because we work with a small sample size and need to compare the before-and-after situation.
The general formula for calculating the t-statistic in our study is:
  • Calculate the mean of the differences between reaction times before and after the drug is administered, denoted as \(\bar{D}\).
  • Calculate the standard deviation of these differences, denoted as \(s_D\).
  • Use the formula \( t = \frac{\bar{D}}{s_D / \sqrt{n}} \) where \( n \) is the sample size.
This calculation helps determine if the mean difference is substantial enough to suggest a true effect, compared to what would typically be expected by random variation alone.
Level of Significance
The level of significance, often represented as \(\alpha\), is a predetermined threshold that guides decision-making in hypothesis testing. In this example, it is set at \(5\%\) or \(0.05\). It indicates the probability of rejecting the null hypothesis if it is actually true (a Type I error).
Choosing a \(5\%\) level of significance means you are willing to accept a 5% chance of incorrectly concluding that the drug affects reaction times.
  • If the p-value (calculated probability of observing the test results under \(H_0\)) is less than \(\alpha\), we reject the null hypothesis.
  • Conversely, if the p-value is greater, we do not reject the null hypothesis.
Thus, the level of significance provides a standard criterion for making decisions based on the t-test results.
Critical Value
A critical value identifies the cutoff point beyond which we will reject the null hypothesis. It is dependent on the chosen level of significance and the degrees of freedom in the data. In practice:
  • The critical value is determined using a t-distribution table corresponding to the \(5\%\) significance level and \(n-1\) degrees of freedom, where \(n\) is the number of subjects.
  • For a one-tailed test, it designates the threshold above which the t-statistic results in rejecting \(H_0\).
When the t-statistic computed from the experiment exceeds the critical value, it demonstrates that the sample data are significantly different enough from what is expected under \(H_0\), providing reason to reject \(H_0\). Conversely, if the t-statistic does not exceed this critical threshold, we do not have sufficient evidence to reject \(H_0\). This role makes the critical value a crucial deciding factor in hypothesis testing.

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Most popular questions from this chapter

To test the effect of alcohol in increasing the reaction time to respond to a given stimulus, the reaction times of seven people were measured. After consuming 3 ounces of \(40 \%\) alcohol, the reaction time for each of the seven people was measured again. Do the following data indicate that the mean reaction time after consuming alcohol was greater than the mean reaction time before consuming alcohol? Use \(\alpha=.05 .\) $$ \begin{array}{llllllll} \text { Person } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text { Before } & 4 & 5 & 5 & 4 & 3 & 6 & 2 \\ \text { After } & 7 & 8 & 3 & 5 & 4 & 5 & 5 \end{array} $$

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