91Ó°ÊÓ

Equip male and female students with a small device that secretly records sound for a random 30 seconds during each \(12.5\) minute period over two days. Count the words each subject speaks during each recording period, and from this, estimate how many words per day each subject speaks. The published report includes a table summarizing six such studies. \({ }^{12}\) Here are two of the six: $$ \begin{array}{ccccc} \hline & \text { Sample Size } & & \text { Estimated Average Number (SD) of Words Spoken per Day } \\ \text { Study } & \text { Women Men } & \text { Women } & \text { Men } \\ \hline 1 & 56 & 56 & 16,177(7520) & 16,569(9108) \\ \hline 2 & 27 & 20 & 16,496(7914) & 12,867(8343) \\ \hline \end{array} $$ Readers are supposed to understand that, for example, the 56 women in the first study had \(\mathrm{x}^{-} \bar{x}=16,177\) and \(s=7520\). It is commonly thought that women talk more than men. Does either of the two samples support this idea? For each study: (a) state hypotheses in terms of the population means for men \(\left(\mu_{M}\right)\) and women \(\left(\mu_{F}\right)\). (b) find the two-sample \(t\) statistic. (c) what degrees of freedom does Option 2 use to get a conservative \(P\)-value? (d) compare your value of \(t\) with the critical values in Table C. What can you say about the \(P\)-value of the test? (e) what do you conclude from the results of these two studies?

Step by step solution

01

State Hypotheses

We want to test if women speak more than men. So our hypotheses for each study is:
- Null Hypothesis (H_0): \( \mu_F = \mu_M \) (The mean number of words spoken per day by women is equal to that of men)
- Alternative Hypothesis (H_a): \( \mu_F > \mu_M \) (Women speak more words per day than men)

02

Compute the Two-Sample t-Statistic for Study 1

The formula for the two-sample t-statistic when testing the difference between two means is:
\[t = \frac{\bar{x}_F - \bar{x}_M}{\sqrt{\frac{s_F^2}{n_F} + \frac{s_M^2}{n_M}}}\]
Substitute the given values for Study 1:
\( \bar{x}_F = 16,177 \), \( \bar{x}_M = 16,569 \), \( s_F = 7520 \), \( s_M = 9108 \), \( n_F = 56 \), \( n_M = 56 \).
\[t = \frac{16177 - 16569}{\sqrt{\frac{7520^2}{56} + \frac{9108^2}{56}}} = \frac{-392}{\sqrt{1008800.571}} = \frac{-392}{1004.38} \approx -0.39\]

03

Compute the Two-Sample t-Statistic for Study 2

Substitute the given values for Study 2:
\( \bar{x}_F = 16,496 \), \( \bar{x}_M = 12,867 \), \( s_F = 7914 \), \( s_M = 8343 \), \( n_F = 27 \), \( n_M = 20 \).
\[t = \frac{16496 - 12867}{\sqrt{\frac{7914^2}{27} + \frac{8343^2}{20}}} = \frac{3629}{\sqrt{2317909.556}} = \frac{3629}{1522.45} \approx 2.38\]

04

Determine Degrees of Freedom for Conservative t-test (Option 2)

For a conservative approach, use the smaller of \( n_1 - 1 \) or \( n_2 - 1 \) for degrees of freedom. For Study 1, \( df = 55 \) and for Study 2, \( df = 19 \).

05

Compare with Critical Values from t-Table

Using a t-table for a one-tailed test with a significance level (\alpha) of 0.05:
- For Study 1: \( t = -0.39 \) is less than the critical value at \( df = 55 \) (about 1.67), indicating \( P > 0.05 \), so not significant.
- For Study 2: \( t = 2.38 \) exceeds the critical value at \( df = 19 \) (about 1.73), indicating \( P < 0.05 \), so significant.

06

Conclusion

In Study 1, there is not enough evidence to claim that women speak more than men. However, Study 2 provides evidence that supports the claim that women speak more words per day than men at a 5% significance level.

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Key Concepts

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

two-sample t-test

When we need to compare the means of two independent groups, the two-sample t-test is the right tool for the job. This statistical method helps to determine if there is a significant difference between the means of the two groups.
For instance, in the exercise, we have male and female students, and we aim to compare their average word count per day.
The two-sample t-test comes into play as we have two independent samples (men and women) and wish to test a hypothesis about their means.Here's how it works:

• We calculate the means and standard deviations for both groups.
• We use these values to compute the t-statistic, using the formula:
  \[t = \frac{\bar{x}_F - \bar{x}_M}{\sqrt{\frac{s_F^2}{n_F} + \frac{s_M^2}{n_M}}}\] where \(\bar{x}_F\) and \(\bar{x}_M\) are the means of the two groups, \(s_F\) and \(s_M\) their standard deviations, and \(n_F\) and \(n_M\) their sample sizes.
• This t-statistic tells us how many standard deviations our observed difference is from the null hypothesis that there is no difference.

degrees of freedom

Degrees of freedom (df) might sound complex, but they are simply a statistical concept that indicates the number of independent values that can vary in an analysis without breaking any constraints.
In the context of the two-sample t-test, degrees of freedom are essential for determining the significance of our t-statistic.Here's how to think about it:

• When comparing two groups, the degrees of freedom are related to the sample size of each group. Specifically, for a conservative approach in the exercise, we take the smaller of \(n_1 - 1\) or \(n_2 - 1\).
• In Study 1, the degrees of freedom are calculated as 55 (since both male and female groups have 56 participants).
• For Study 2, since the smaller group has 20 participants, we use 19 degrees of freedom.
• Degrees of freedom help us decide which critical value to use from the t-table, affecting decision-making in hypothesis testing.

critical value

A critical value is a cutoff point that helps us determine the statistical significance of our test results.
In hypothesis testing, it's crucial to understand where this threshold lies in order to make informed conclusions. Here's how critical values come into play:

• They are derived from the t-distribution table based on the degrees of freedom and the chosen significance level (like 0.05 in our exercise).
• The critical value determines the borderline between significance and non-significance of results.
• In hypothesis testing, if our calculated t-statistic is more extreme than the critical value, we reject the null hypothesis.
• For example, in Study 1, the critical value is about 1.67 at 55 degrees of freedom; since the t-statistic is -0.39, it shows a non-significant result.
• Conversely, Study 2, with a t-statistic of 2.38, exceeds the critical value of 1.73 at 19 degrees of freedom, indicating statistical significance.

significance level

The significance level, denoted by \(\alpha\), is the threshold for determining whether a test result is statistically significant.
It represents the probability of rejecting the null hypothesis when it is actually true, often set at 5% (0.05).Understanding significance level is key:

• It's the chance of making a Type I error, which is falsely finding a significant effect.
• A common choice for \(\alpha\) is 0.05, meaning we are willing to accept a 5% risk of error.
• When the p-value of a test is less than the significance level, we consider the result statistically significant.
• In the original exercise, Study 2 shows a p-value less than 0.05, supporting the hypothesis that women speak more than men.
• On the other hand, Study 1 does not meet this threshold, showing us that we do not have enough evidence to assert a difference in speech between genders.