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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?

Short Answer

Expert verified
To summarise, since the calculated test statistic was -1.65, refer to the t-distribution table for a one-tailed test with degrees of freedom of 296. If the associated p-value is less than 0.05 (assuming a significance level of 0.05), the null hypothesis is rejected and it can be concluded that there is a significant difference in the means. If the p-value is more than 0.05, then it can be concluded there is not enough evidence to reject the null hypothesis and the GPAs of working and non-working students do not significantly differ.

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.

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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.

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Most popular questions from this chapter

Some people believe that talking on a cell phone while driving slows reaction time, increasing the risk of accidents. The study described in the paper "A Comparison of the Cell Phone Driver and the Drunk Driver" (Human Factors [2006]: \(381-391\) ) investigated the braking reaction time of people driving in a driving simulator. Drivers followed a pace car in the simulator, and when the pace car's brake lights came on, the drivers were supposed to step on the brake. The time between the pace car brake lights coming on and the driver stepping on the brake was measured. Two samples of 40 drivers participated in the study. The 40 people in one sample used a cell phone while driving. The 40 people in the second sample drank a mixture of orange juice and alcohol in an amount calculated to achieve a blood alcohol level of \(0.08 \%\) (a value considered legally drunk in most states). For the cell phone sample, the mean braking reaction time was 779 milliseconds and the standard deviation was 209 milliseconds. For the alcohol sample, the mean breaking reaction time was 849 milliseconds and the standard deviation was \(228 .\) Is there convincing evidence that the mean braking reaction time is different for the population of drivers talking on a cell phone and the population of drivers who have a blood alcohol level of \(0.08 \%\) ? For purposes of this exercise, you can assume that the two samples are representative of the two populations of interest.

In a study of malpractice claims where a settlement had been reached, two random samples were selected: a random sample of 515 closed malpractice claims that were found not to involve medical errors and a random sample of 889 claims that were found to involve errors (New England Journal of Medicine [2006]: \(2024-2033\) ). The following statement appeared in the paper: "When claims not involving errors were compensated, payments were significantly lower on average than were payments for claims involving errors \((\$ 313,205\) vs. \(\$ 521,560, P=0.004)\) a. What hypotheses did the researchers test to reach the stated conclusion? b. Which of the following could have been the value of the test statistic for the hypothesis test? Explain your reasoning. i. \(\quad t=5.00\) iii. \(t=2.33\) ii. \(t=2.65\) iv. \(t=1.47\)

The Oregon Department of Health web site provides information on cost-to- charge ratio (the percentage of billed charges that are actual costs to the hospital). The following table gives cost-to-charge ratios for both inpatient and outpatient care in 2002 for a random sample of six hospitals in Oregon. $$ \begin{array}{ccc} & \begin{array}{c} 2002 \\ \text { Inpatient } \end{array} & \begin{array}{c} 2002 \\ \text { Hospital } \end{array} & \begin{array}{c} \text { Ratio } \\ \text { Ratient } \end{array} \\ \hline 1 & 68 & \text { Ratio } \\ 2 & 100 & 54 \\ 3 & 71 & 75 \\ 4 & 74 & 53 \\ 5 & 100 & 56 \\ 6 & 83 & 74 \\ & 88 \end{array} $$ Is there evidence that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care? Use a significance level of \(0.05 .\)

Do male college students spend more time studying than female college students? This was one of the questions investigated by the authors of the paper "An Ecological Momentary Assessment of the Physical Activity and Sedentary Behaviour Patterns of University Students" (Heath Education Journal \([2010]: 116-125)\). Each student in a random sample of 46 male students at a university in England and each student in a random sample of 38 female students from the same university kept a diary of how he or she spent time over a 3 -week period. For the sample of males, the mean time spent studying per day was 280.0 minutes, and the standard deviation was 160.4 minutes. For the sample of females, the mean time spent studying per day was 184.8 minutes, and the standard deviation was 166.4 minutes. Is there convincing evidence that the mean time male students at this university spend studying is greater than the mean time for female students? Test the appropriate hypotheses using \(\alpha=0.05\).

For each of the following hypothesis testing scenarios, indicate whether or not the appropriate hypothesis test would be about a difference in population means. If not, explain why not. Scenario 1: The international polling organization Ipsos reported data from a survey of 2,000 randomly selected Canadians who carry debit cards (Canadian Account Habits Survey, July 24,2006 ). Participants in this survey were asked what they considered the minimum purchase amount for which it would be acceptable to use a debit card. You would like to determine if there is convincing evidence that the mean minimum purchase amount for which Canadians consider the use of a debit card to be acceptable is less than \(\$ 10\). Scenario 2: Each person in a random sample of 247 male working adults and a random sample of 253 female working adults living in Calgary, Canada, was asked how long, in minutes, his or her typical daily commute was ("Calgary Herald Traffic Study," Ipsos, September 17,2005 ). You would like to determine if there is convincing evidence that the mean commute times differ for male workers and female workers. Scenario 3: A hotel chain is interested in evaluating reservation processes. Guests can reserve a room using either a telephone system or an online system. Independent random samples of 80 guests who reserved a room by phone and 60 guests who reserved a room online were selected. Of those who reserved by phone, 57 reported that they were satisfied with the reservation process. Of those who reserved online, 50 reported that they were satisfied. You would like to determine if it reasonable to conclude that the proportion who are satisfied is higher for those who reserve a room online.

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