91Ó°ÊÓ

Typically, only very brave students are willing to speak out in a college classroom. Student participation may be especially difficult if the individual is from a different culture or country. The article "An Assessment of Class Participation by International Graduate Students" (Journal of College Student Development [1995]: 132- 140) considered a numerical "speaking-up" scale, with possible values from 3 to 15 (a low value means that a student rarely speaks). For a random sample of 64 males from Asian countries where English is not the official language, the sample mean and sample standard deviation were \(8.75\) and \(2.57\), respectively. Suppose that the mean for the population of all males having English as their native language is \(10.0\) (suggested by data in the article). Does it appear that the population mean for males from non-English-speaking Asian countries is smaller than \(10.0 ?\)

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the population mean speaking-up score (\(μ\)) for males from non-English-speaking Asian countries is equal to 10.0: \(H_0: μ = 10.0\). The alternative hypothesis (\(H_1\)) is that the population mean speaking-up score for males from non-English-speaking Asian countries is less than 10.0: \(H_1: μ < 10.0\).

02

Compute the Test Statistic

The test statistic for a sample mean, when the population standard deviation is unknown, follows a t-distribution. It is calculated as: \[ t = \frac{{\overline{x} - μ_0}}{{s / \sqrt{n}}}\], where \(\overline{x}\) is the sample mean, \(μ_0\) is the hypothesized population mean, s is the sample standard deviation, and n is the sample size. Substitute \(\overline{x} = 8.75\), \(μ_0 = 10.0\), s = 2.57, and n = 64 into the formula to get: \[ t = \frac{{8.75 - 10.0}}{{2.57 / \sqrt{64}}} = -4.88 \]

03

Determine the P-value

The P-value is the probability of observing a test statistic as extreme as the one computed, assuming the null hypothesis is true. Using a t-distribution table or a statistical calculator, the P-value associated with a t-score of -4.88, with 63 degrees of freedom, is nearly 0

04

Conclusion

If the P-value is less than the significance level (typically 0.05), reject the null hypothesis in favor of the alternative. Since the P-value is nearly 0, reject the null hypothesis. There is significant evidence to suggest that the mean speaking-up score for males from non-English-speaking Asian countries is less than 10.0

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Key Concepts

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

T-Distribution

In hypothesis testing, the T-distribution is often used when dealing with small sample sizes or when the population standard deviation is unknown. Unlike the normal distribution, which is symmetrical, the T-distribution has heavier tails.
This means it can accommodate more variability, which is common in small samples.
The formula for calculating the T-statistic is:

• \[ t = \frac{{\overline{x} - \mu}}{{s / \sqrt{n}}} \]

Here, \( \overline{x} \) is the sample mean, \( \mu \) is the population mean hypothesized under the null hypothesis, \( s \) is the sample standard deviation, and \( n \) is the sample size.
In our example, the T-distribution is used because the population standard deviation is not known, and it helps in determining how far the sample mean is from the population mean.

Null Hypothesis

The null hypothesis, often denoted as \( H_0 \), is the starting assumption that there is no effect or difference. It posits that any observed effect is due to sampling error.
In the exercise, the null hypothesis states that the mean speaking-up score for males from non-English-speaking Asian countries is equal to 10.0.
This means that any deviation from this mean is assumed to be due to chance unless evidence suggests otherwise.
The null hypothesis is a crucial component of hypothesis testing as it provides a baseline from which to measure changes or differences.

Alternative Hypothesis

The alternative hypothesis, symbolized as \( H_1 \) or \( H_a \), challenges the assumption set by the null hypothesis. It seeks to prove that there is a real effect or difference.
In this analysis, the alternative hypothesis is that the mean speaking-up score is less than 10.0.
The alternative hypothesis carries the implication that if the null hypothesis is rejected based on statistical evidence, we can consider supporting the alternative hypothesis.

• The focus is on demonstrating that the observed data statistically supports \( H_1 \).

P-Value

The P-value is a crucial aspect of hypothesis testing. It represents the probability of obtaining results as extreme as the observed results, assuming that the null hypothesis is true. A small P-value indicates strong evidence against the null hypothesis.
In our case study, the P-value was nearly 0 when the T-score of -4.88 was computed.

• A P-value less than 0.05 typically means that the results are statistically significant, warranting rejection of the null hypothesis.

This resulted in rejecting the null hypothesis, suggesting that the mean speaking-up score for the given group is indeed less than 10.0.

Sample Mean

The sample mean, denoted as \( \overline{x} \), is a measure of central tendency that provides an estimate of the population mean based on the observed sample. In statistics, it is a critical value used in calculations and hypothesis tests.
In this exercise, the sample mean was calculated to be 8.75 for males from non-English-speaking Asian countries.

• The sample mean helps inform whether the sample provides enough evidence to support or reject the null hypothesis.

It is important because it provides the benchmark against which the hypothesized population mean is compared.
Alongside the sample standard deviation and sample size, it is used to compute the T-statistic, helping in assessing the evidence against the null hypothesis.