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Chapter 10: Problem 44

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: Time Reaction \begin{tabular}{crr} & \multicolumn{2}{c} { Reaction Time } \\\ \cline { 2 - 3 } Subject & Before & After \\ \hline 1 & 7 & 13 \\ 2 & 2 & 3 \\\ 3 & 12 & 18 \\ 4 & 12 & 13 \end{tabular} Test at the \(5 \%\) level of significance to determine whether the drug significantly increases reaction time.

Short Answer

Expert verified
Answer: Yes, the drug significantly increases the reaction time to the given stimulus at the 5% level of significance.

Step by step solution

01

State the hypotheses

The null hypothesis, \(H_0\), states that there is no significant difference in reaction times after taking the drug: \(H_0 : \mu_{D} = 0\) The alternative hypothesis, \(H_1\), states that the reaction time increases after taking the drug: \(H_1 : \mu_{D} > 0\)
02

Compute the differences

Calculate the differences between the after and before reaction times for each subject: \(d_1 = 13 - 7 = 6\) \(d_2 = 3 - 2 = 1\) \(d_3 = 18 - 12 = 6\) \(d_4 = 13 - 12 = 1\)
03

Calculate the mean and standard deviation of the differences

Mean of differences, \(\bar{d}=\frac{6+1+6+1}{4}=\frac{14}{4}=3.5\) Standard deviation, \(s_d=\sqrt{\frac{(6-3.5)^2+(1-3.5)^2+(6-3.5)^2+(1-3.5)^2}{4-1}}=\sqrt{\frac{18.75}{3}}=\sqrt{6.25}=2.5\)
04

Find the critical value from t-distribution

For a \(5 \%\) level of significance and a one-tailed test, the critical value (\(t_{crit}\)) with \(3\) degrees of freedom (sample size \(- 1 = 4 - 1 = 3\)) can be found using a t-table or online calculator. The critical value is \(t_{crit} = 2.353\)
05

Calculate the test statistic (t-score)

Calculate the t-score using the formula: \(t = \frac{\bar{d}-\mu_D}{\frac{s_d}{\sqrt{n}}} = \frac{3.5-0}{\frac{2.5}{\sqrt{4}}} =\frac{3.5}{1.25} = 2.8\)
06

Compare the t-score to the critical value

Since the t-score \(2.8\) is greater than the critical value \(2.353\), we reject the null hypothesis.
07

Make conclusion based on the comparison

We reject the null hypothesis, which means that the drug significantly increases the reaction time to the given stimulus at the \(5 \%\) level of significance.

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Key Concepts

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

Understanding t-distribution
In statistical analysis, the t-distribution plays a crucial role, especially when dealing with small sample sizes. This distribution is similar to the normal distribution but has thicker tails, which means there's more probability in the extremes. It's especially useful when the population standard deviation is unknown, and you have a small sample size, typically less than 30. In this exercise, since we have just four subjects, the t-distribution is the appropriate choice for our hypothesis test. It allows us to make inferences about the mean differences in reaction times using the calculated t-score, which we compare against a critical value from the t-table.
Significance level in hypothesis testing
The significance level is a key component in hypothesis testing. It reflects the probability of rejecting the null hypothesis when it is actually true, often denoted by 伪. Commonly, a 5% significance level is used, meaning there's a 5% risk of concluding that a difference exists when there isn't one. In our context, we're testing whether the drug significantly increases reaction time. Using the 5% significance level ensures that we are confident in our results. If our calculated t-score surpasses the critical value associated with this level, we reject the null hypothesis and accept that the drug has a statistically significant effect.
The concept of reaction time
Reaction time is the duration it takes for someone to respond to a stimulus. It's a crucial measure in various fields, including psychology and neuroscience. Changes in reaction time can indicate alterations in cognitive processing or the effects of substances like drugs. In this exercise, the psychologist measures reaction times before and after administering a drug to determine its impact. An increase in reaction time could suggest that the drug slows down cognitive response, which is what we analyzed through our hypothesis testing.
Solving a statistics problem
Tackling a statistics problem involves several systematic steps. First, you define clear hypotheses: a null hypothesis that suggests no effect, and an alternative hypothesis that indicates a difference. Next, gather data and compute relevant statistics, such as means and standard deviations. Calculating the t-score helps quantify the effect size in comparison to the standard error. The critical step is comparing the t-score against a critical value from the t-distribution to decide whether to reject the null hypothesis. This method ensures that your conclusion about the significance of effects, like the drug's impact on reaction time, is statistically valid.

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Most popular questions from this chapter

Cholesterol The serum cholesterol Elevels of 50 subjects randomly selected from the L.A. Heart Data, data from an epidemiological heart disease study on Los Angeles County employees, follow. \(\begin{array}{llllllllll}148 & 304 & 300 & 240 & 368 & 139 & 203 & 249 & 265 & 229 \\ 303 & 315 & 174 & 209 & 253 & 169 & 170 & 254 & 212 & 255 \\ 262 & 284 & 275 & 229 & 261 & 239 & 254 & 222 & 273 & 299 \\ 278 & 227 & 220 & 260 & 221 & 247 & 178 & 204 & 250 & 256 \\ 305 & 225 & 306 & 184 & 242 & 282 & 311 & 271 & 276 & 248\end{array}\) a. Construct a histogram for the data. Are the data approximately mound- shaped? b. Use a \(t\) -distribution to construct a \(95 \%\) confidence interval for the average serum cholesterol levels for L.A. County employees.

A producer of machine parts claimed that the diameters of the connector rods produced by his plant had a variance of at most .03 inch \(^{2}\). A random sample of 15 connector rods produced by his plant produced a sample mean and variance of .55 inch and .053 inch \(^{2},\) respectively. a. Is there sufficient evidence to reject his claim at the \(\alpha=.05\) level of significance? b. Find a \(95 \%\) confidence interval for the variance of the rod diameters.

At a time when energy conservation is so important, some scientists think closer scrutiny should be given to the cost (in energy) of producing various forms of food. Suppose you wish to compare the mean amount of oil required to produce 1 acre of corn versus 1 acre of cauliflower. The readings (in barrels of oil per acre), based on 20-acre plots, seven for each crop, are shown in the table. Use these data to find a \(90 \%\) confidence interval for the difference between the mean amounts of oil required to produce these two crops. Corn Cauliflower \(\begin{array}{ll}5.6 & 15.9\end{array}\) 7.1 13.4 \(\begin{array}{ll}4.5 & 17.6\end{array}\) 6.0 16.8 7.9 15.8 \(\begin{array}{ll}4.8 & 16.3\end{array}\) 57 \(17 .\)

The calcium (Ca) content of a powdered mineral substance was analyzed 10 times with the following percent compositions recorded: \(\begin{array}{lllll}.0271 & .0282 & .0279 & .0281 & .0268\end{array}\) \(\begin{array}{llll}.0271 & .0281 & .0269 & .0275 & .0276\end{array}\) a. Find a \(99 \%\) confidence interval for the true calcium content of this substance. b. What does the phrase " \(99 \%\) confident" mean? c. What assumptions must you make about the sampling procedure so that this confidence interval will be valid? What does this mean to the chemist who is performing the analysis?

Wishing to demonstrate that the variability of fills is less for her model than for her competitor's, a sales representative for company A acquired a sample of 30 fills from her company's model and a sample of 10 fills from her competitor's model. The sample variances were \(s_{A}^{2}=.027\) and \(s_{B}^{2}=.065,\) respectively. Does this result provide statistical support at the .05 level of significance for the sales representative's claim?
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