Question

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 \(\alpha=0.05\).

Short answer

The solution requires setting up and testing hypotheses. The decision to reject or not reject the null hypothesis is based on comparing the test statistic with the critical value.

Step-by-step solution

Formulate the Hypotheses

The Null Hypothesis (\(H_0\)) should be that the mean study time of male students (\(\mu_m\)) equals the mean study time of the female students (\(\mu_f\)). Stated mathematically, \(H_0: \mu_m = \mu_f\). The Alternative Hypothesis(\(H_1\)) states that the mean study time of male students is greater than that of female students. Mathematically, \(H_1: \mu_m > \mu_f\).

Calculate the Test Statistic

The test statistic is calculated as \(Z = \frac{(\bar{x}_m - \bar{x}_f) - (\mu_m - \mu_f)}{\sqrt{\frac{S_m^2}{n_m} + \frac{S_f^2}{n_f}}}\), where \(\bar{x}_m\) and \(\bar{x}_f\) are the sample means, \(S_m\) and \(S_f\) are the sample standard deviations, and \(n_m\) and \(n_f\) are the sample sizes. Here \(\mu_m - \mu_f = 0\) under \(H_0\). Substitute the given values to calculate Z.

Determine the Critical Value

Given that the significance level \(\alpha = 0.05\) and the alternative hypothesis indicates a right-tailed test, the critical value can be determined using a standard Normal (Z) distribution table which yields approximately 1.645.

Make a Decision and Interpret the Result

Compare the test statistic (Z) with the critical value. If \(Z > 1.645\), reject the null hypothesis. This would provide convincing evidence to conclude that the mean study time of male students at the university is greater than that of female students. Otherwise, do not reject the null hypothesis.

Key concepts

Null and Alternative Hypothesis

In hypothesis testing, we begin by stating two contradictory opinions about a population parameter, like a mean or a proportion. These are the null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)).

The null hypothesis is a statement of no difference or no effect; it is what we assume to be true before we collect any data. In our exercise, the null hypothesis is that the mean study time for male and female students is the same, mathematically expressed as \(H_0: \mu_m = \mu_f\).

The alternative hypothesis is what you aim to support. It represents an outcome that contradicts the null hypothesis. In our exercise, the alternative hypothesis claims that male students spend more time studying than female students, denoted as \(H_1: \mu_m > \mu_f\).

Testing these hypotheses involves using sample data to determine whether to support the null hypothesis or to consider the alternative hypothesis more credible.

Test Statistic

The test statistic is a crucial value derived from sample data, which is compared to a threshold value in order to make a decision about the hypotheses. The test statistic in our example involves the difference in sample means, sample standard deviations, and sample sizes of both male and female student groups.

To find this statistic, use the formula \(Z = \frac{(\bar{x}_m - \bar{x}_f) - (\mu_m - \mu_f)}{\sqrt{\frac{S_m^2}{n_m} + \frac{S_f^2}{n_f}}}\), where \(\bar{x}_m\) and \(\bar{x}_f\) are the sample means for males and females, \(S_m\) and \(S_f\) are the sample standard deviations, and \(n_m\) and \(n_f\) are the sample sizes.

Substitute the given values into this formula to find the Z-score. This calculated Z-score is then compared against a critical value to decide if the null hypothesis should be rejected or not.

Standard Deviation

Standard deviation is a measure of how spread out numbers are in a dataset. In other words, it's an indication of variability or diversity within a set of values. A lower standard deviation means that most of the numbers are close to the average (mean), while a higher standard deviation indicates that the numbers are more spread out.

In the context of our exercise, two standard deviations are given: one for male students and one for female students. These reflect the variability in daily study times within each group. When we perform hypothesis testing, these standard deviations help calculate the test statistic and ultimately determine whether the observed difference between group means is statistically significant or due to random chance.

Significance Level

The significance level, usually denoted by \(\alpha\), is a threshold we set to decide whether or not to reject the null hypothesis. It's the probability of rejecting the null hypothesis when it is actually true, known as a Type I error.

In this exercise, the significance level is set at 0.05, which is a common choice in many scientific studies. This means there is a 5% risk of concluding that a difference exists when there is none. To make a decision using this significance level, we compare the test statistic to a critical value from a statistical table based on \(\alpha\). If the test statistic exceeds the critical value, we reject the null hypothesis, suggesting that the results are statistically significant.