A study of the differences in cognitive function between...

A study of the differences in cognitive function between normal individuals and patients diagnosed with schizophrenia was published in the American Journal of Psychiatry (Apr. 2010). The total time (in minutes) a subject spent on the Trail Making Test (a standard psychological test) was used as a measure of cognitive function. The researchers theorize that the mean time on the Trail Making Test for schizophrenics will be larger than the corresponding mean for normal subjects. The data for independent random samples of 41 schizophrenics and 49 normal individuals yielded the following results: $$ \begin{array}{lcc} \hline & \text { Schizophrenia } & \text { Normal } \\ \hline \text { Sample size } & 41 & 49 \\ \text { Mean time } & 104.23 & 62.24 \\ \text { Standard deviation } & 45.45 & 16.34 \\ \hline \end{array} $$ a. Define the parameter of interest to the researchers. b. Set up the null and alternative hypothesis for testing the researchers' theory. c. The researchers conducted the test, part $\mathbf{b},$ and reported a $p$ -value of .001. What conclusions can you draw from this result? (Use $\alpha=.01 .)$ d. Find a $99 \%$ confidence interval for the target parameter. Interpret the result. Does your conclusion agree with that of part $\mathbf{c}$ ?

A study of the differences in cognitive function between normal individuals and patients diagnosed with schizophrenia was published in the American Journal of Psychiatry (Apr. 2010). The total time (in minutes) a subject spent on the Trail Making Test (a standard psychological test) was used as a measure of cognitive function. The researchers theorize that the mean time on the Trail Making Test for schizophrenics will be larger than the corresponding mean for normal subjects. The data for independent random samples of 41 schizophrenics and 49 normal individuals yielded the following results:
$$
\begin{array}{lcc}
\hline & \text { Schizophrenia } & \text { Normal } \\
\hline \text { Sample size } & 41 & 49 \\
\text { Mean time } & 104.23 & 62.24 \\
\text { Standard deviation } & 45.45 & 16.34 \\
\hline
\end{array}
$$
a. Define the parameter of interest to the researchers.
b. Set up the null and alternative hypothesis for testing the researchers' theory.
c. The researchers conducted the test, part $\mathbf{b},$ and reported a $p$ -value of .001. What conclusions can you draw from this result? (Use $\alpha=.01 .)$
d. Find a $99 \%$ confidence interval for the target parameter. Interpret the result. Does your conclusion agree with that of part $\mathbf{c}$ ?

Key Concepts

Confidence Interval

A confidence interval provides a range of plausible values for the population parameter, in this case, the difference in means between the two groups. A confidence interval is typically constructed such that, under repeated sampling, a certain percentage of these intervals will contain the true parameter value. The interpretation of the interval is linked to understanding the precision and uncertainty of the estimated effect, and if it does not include zero, it aligns with rejecting the null hypothesis in hypothesis testing.

Significance Level

The significance level, denoted by alpha, sets the threshold for making a decision about the null hypothesis. It represents the probability of committing a Type I error, which is rejecting the null hypothesis when it is actually true. By comparing the p-value to this predetermined alpha level, researchers decide whether the evidence is strong enough to conclude that the observed effect is statistically significant.

Null and Alternative Hypotheses

The formulation of the null and alternative hypotheses is a foundational aspect of hypothesis testing. The null hypothesis usually asserts that there is no difference between the population means (or that the difference is zero), serving as a default claim to be tested against. The alternative hypothesis reflects the researchers' theory or expectation, indicating that one group’s population mean is larger than the other's. This framework is how statistical testing determines if observed sample differences are likely to be due to chance or reflect a real effect.

Population Parameter

The parameter of interest in such studies is typically the difference between the population means of two groups. In this context, it represents the difference in the mean time taken on the cognitive test between the two populations. This concept is central in comparing groups because it quantifies the effect of a condition (such as a medical or psychological disorder) on a measurable outcome.

p-Value Interpretation

The p-value is a probability measure that helps determine the significance of the observed sample data under the assumption that the null hypothesis is true. A very small p-value, compared to the chosen significance level, suggests that obtaining the observed data (or more extreme) is highly unlikely if the null hypothesis were true. Thus, a low p-value is used to reject the null hypothesis in favor of the alternative.

Transcript

00:01 Here's a solution problem number 916.

00:04 So there was a study on the differences in cognitive function between normal individuals and patients with schizophrenia.

00:10 And the test was used to measure cognitive function.

00:14 And the researchers believed that are theorized that the meantime on the trailmaking test for schizophrenic patients will be larger than the corresponding meantime for normal subjects.

00:30 So the data was given as this, so some sample sizes that are, you know, big enough.

00:39 And we're supposed to have the parameter of interest.

00:42 And the parameter of interest here is the difference in mean time for the cognitive test, and whatever it was called.

01:01 It was called the trail making test.

01:06 All right.

01:06 And then part b, it says to set up the null and alternative hypothesis for testing the researchers ' theory.

01:14 So the mean difference would be zero for the null.

01:18 It's always equal to zero, or it's not always, but usually equal to zero.

01:22 And then the alternative would be mu one minus mu two is greater than zero, because we're theorizing that schizophrenic patients take longer.

01:33 All right.

01:34 And now we're given that the p value is 0 .001.

01:37 So what conclusions can you draw? well, that's going to be less than our alpha because our alpha is 0 .01.

01:42 So that means we're going to reject the null hypothesis, meaning that there is enough evidence to suggest that mu1 is greater than mu 2.

01:51 So there is enough evidence to suggest mu 1 is greater than mu 2.

02:05 So the mean response time or the mean cognitive time for schizophrenic patients is larger than for normal patients...

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