91Ó°ÊÓ

In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were reported for a sample of students who worked and for a sample of students who did not work (University of Central Florida Undergraduate Research Journal, Spring 2005\()\) : $$ \begin{array}{cccc} & \begin{array}{c} \text { Sample } \\ \text { Size } \end{array} & \begin{array}{c} \text { Mean } \\ \text { GPA } \end{array} & \begin{array}{c} \text { Sandard } \\ \text { Deviation } \end{array} \\ \begin{array}{c} \text { Students Who Are } \\ \text { Employed } \end{array} & 184 & 3.12 & 0.485 \\ \begin{array}{c} \text { Students Who Are } \\ \text { Not Employed } \end{array} & 114 & 3.23 & 0.524 \\ \hline \end{array} $$ The samples were selected at random from working and nonworking students at the University of Central Florida. Does this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed?

Step by step solution

01

State the hypothesis

The null hypothesis \(H_0\) is that the two population means are equal or that the mean GPA of employed students is higher. The alternate hypothesis \(H_1\) is that the population mean GPA of unemployeed students is higher. So, \(H_0: \mu_1 - \mu_2 \leq 0\) and \(H_1: \mu_1 - \mu_2 > 0\).

02

Calculate pooled variance

The pooled variance, denoted as \(s_p^2\), is the weighted average of the sample variances, so it's given by: \(s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}\). Given that \(n_1 = 184, s_1 = 0.485, n_2 = 114\) and \(s_2 = 0.524\), the pooled variance \(s_p^2 = \frac{(184-1)(0.485)^2 + (114-1)(0.524)^2}{184+114-2} = 0.250\).

03

Calculate the test statistic

The test statistic for the hypothesis test is given by: \(t = \frac{\overline{x}_1 - \overline{x}_2}{s_p \sqrt{1/n_1 + 1/n_2}}\). Given that \(\overline{x}_1 = 3.12, \overline{x}_2 = 3.23\), and you already calculated \(s_p = \sqrt{s_p^2} = \sqrt{0.250} = 0.5\), the test statistic is \(t = \frac{3.12 - 3.23}{0.5 \sqrt{1/184 + 1/114}} = -1.65\).

04

Find the p-value and draw a conclusion

You should use a t-distribution table to find the p-value associated with \(t = -1.65\). Degrees of freedom in this case is \(df = n_1 + n_2 - 2 = 184 + 114 - 2 = 296\). Since this is a one-sided test, you can conclude that the test is significant at the 0.05 level (or any level above the p-value found) and reject the null hypothesis if the p-value is less than 0.05. If p-value is about 0.05 or more, then you fail to reject the null hypothesis.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Hypothesis Testing

Hypothesis testing is a statistical method used to make inferences about population parameters based on sample data. It starts with an initial assumption called the null hypothesis, which is a statement of 'no effect' or 'no difference'. For instance, in evaluating academic performance, we might hypothesize that being employed does not affect a student's GPA. To test such hypotheses, we use sample data to calculate a test statistic, which can help us decide whether to reject the null hypothesis or not. If the test statistic falls in the critical region of the distribution associated with a specified significance level, we reject the null hypothesis in favor of the alternative hypothesis, which states that there is indeed an effect or a difference.

The importance of choosing the right significance level and understanding the power of the test cannot be overstated. It's crucial to note that a failure to reject the null hypothesis does not necessarily confirm its truth; it simply means there is not enough evidence against it, based on the sample data provided.

T-test

The t-test is a statistical test used to compare the means of two groups to see if they are significantly different from each other. It's especially useful when dealing with small sample sizes or when the population standard deviation is unknown. There are different types of t-tests designed for specific situations, such as the independent samples t-test, paired samples t-test, and one-sample t-test. In our context, the independent samples t-test is applicable as we are comparing GPA means of two different groups of students - those who are employed and those who are not. The calculated t-statistic allows us to measure the size of the difference relative to the variation in the sample data. Significance is determined by comparing the t-statistic to a t-distribution, after which a p-value is obtained to make a conclusion about the hypothesis.

Pooled Variance

Pooled variance is a technique used to estimate the overall variance of two or more groups. When conducting a t-test between two independent samples of different sizes or variances, it's essential to calculate a combined estimate of variance, which provides a weighted average of the variances from each sample. This combined estimate is known as the pooled variance. Calculating pooled variance involves using the sample sizes and variances to find a middle ground between the group variances, taking into account the degrees of freedom from each group. It serves as a crucial component in determining the standard error of the difference between two sample means, which subsequently is used to calculate the t-statistic. Consistent estimation of variance is vital for accurate hypothesis testing.

Null Hypothesis

The null hypothesis, symbolized as \( H_0 \), is a fundamental aspect of hypothesis testing. It represents the statement being tested and typically suggests that there is no effect or no significant difference between groups. In our study of student GPAs, the null hypothesis might assert that the mean GPA for students who are employed is equal to or greater than that of those who are not employed. The null hypothesis acts as the default or starting point for statistical tests. To challenge this hypothesis, evidence from the sample data must be strong enough to warrant its rejection, otherwise, we fail to reject it. This concept is the cornerstone of hypothesis testing; it establishes the basis for determining whether sample data provides sufficient evidence to support a specific claim about the population.