Question

Do female college students spend more time watching TV than male 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" (Health 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 watching TV per day was 68.2 minutes, and the standard deviation was 67.5 minutes. For the sample of females, the mean time spent watching TV per day was 93.5 minutes, and the standard deviation was 89.1 minutes. Is there convincing evidence that the mean time female students at this university spend watching \(\mathrm{TV}\) is greater than the mean time for male students? Test the appropriate hypotheses using \(\alpha=0.05\).

Short answer

First compute the t-statistic, then calculate the P-value to test the hypotheses at the given \(\alpha=0.05\) level. Depending on the P-value, decide whether or not to reject the null hypothesis. Please perform the detailed calculations.

Step-by-step solution

Setup the Hypotheses

The null hypothesis (\(H_0\)) is that the mean time for male and female students watching TV is equal. The alternative hypothesis (\(H_A\)) is that female students spend more time watching TV than male students. Mathematically, they can be represented as follows: \(H_0: \mu_f = \mu_m\) and \(H_A: \mu_f > \mu_m\) where \(\mu_f\) and \(\mu_m\) represent the mean time female and male students respectively.

Compute the Test Statistic

The test statistic is calculated using the formula for two-sample t-statistic given by \[ t = \frac{(\bar{x}_f - \bar{x}_m)}{\sqrt{\frac{s^2_f}{n_f} + \frac{s^2_m}{n_m}}} \] where \(\bar{x}_f\) and \(\bar{x}_m\) are sample means, \(s_f\) and \(s_m\) are sample standard deviations, and \(n_f\) and \(n_m\) are sample sizes for female and male students respectively. Plugging the given values: \[ t = \frac{(93.5 - 68.2)}{\sqrt{\frac{89.1^2}{38} + \frac{67.5^2}{46}}} \] Calculate this value.

Calculate the P-value

The P-value is obtained by comparing the calculated t statistic with a t distribution. The degrees of freedom is estimated using the formula \[ \text{df} = \frac{(\frac{s^2_f}{n_f} + \frac{s^2_m}{n_m})^2}{\frac{(\frac{s^2_f}{n_f})^2}{n_f-1} + \frac{(\frac{s^2_m}{n_m})^2}{n_m-1}} \] Gather these values and use software to compute the P-value.

Decision Making

If the calculated p-value from the previous step is less than the level of significance \(\alpha=0.05\), the null hypothesis is rejected, implying there is sufficient evidence that female students spend more time watching TV. If not, do not reject the null hypothesis, indicating insufficient evidence.

Key concepts

Hypothesis Testing

Understanding hypothesis testing is crucial for interpreting results in research and statistical analyses. Essentially, hypothesis testing is a systematic procedure used to evaluate assumptions or guesses we make about a population parameter. In the context of our exercise, we aim to test whether the mean amount of time female students spend watching TV is greater than that of male students at a university.

To start, we establish two opposing statements: the null hypothesis (yopsis{H_0}) and the alternative hypothesis (yopsis{H_A}). The null hypothesis typically represents a default position or a statement of no effect, while the alternative represents what we are attempting to find evidence for. In our case, yopsis{H_0: y_feature{y_mu_f} = y_feature{y_mu_m}} and yopsis{H_A: y_feature{y_mu_f} > y_feature{y_mu_m}}, where yopsis{y_feature{y_mu_f}} and yopsis{y_feature{y_mu_m}} denote the population means for females and males, respectively. By calculating a test statistic and comparing it to a distribution, we can infer whether to reject the null hypothesis or not based on a predefined significance level, yopsis{y_alpha} (usually 0.05).

Standard Deviation

The standard deviation is a measure of how spread out the numbers are in a data set, representing the variation or dispersion of the dataset relative to its mean. A low standard deviation means that the data points tend to be close to the mean, while a high standard deviation indicates that the data are more spread out.

In hypothesis testing, particularly when comparing two means as in our exercise, the standard deviation plays a vital role. It's part of the formula used to calculate the t-statistic: yopsis{s_f} and yopsis{s_m} represent the standard deviations for the female and male student samples, respectively. In this real-world scenario, the large standard deviation relative to the means (67.5 and 89.1 minutes) suggests that there is considerable variability in TV watching times among the students. This variability needs to be accounted for when determining the reliability of our comparison between the two sample means.

P-value

The p-value is a probability that measures the evidence against the null hypothesis provided by the data. Specifically, it's the probability of observing a test statistic as extreme as, or more extreme than, the statistic calculated from the data, assuming the null hypothesis is true. In the context of our exercise, after calculating the t-statistic, we find the corresponding p-value from the t distribution with the appropriate degrees of freedom.

A small p-value (yopsis{< y_alpha}) indicates that the observed data is unlikely under the null hypothesis and thereby provides strong evidence to reject the null hypothesis. Conversely, a large p-value suggests that the evidence isn't strong enough to warrant a rejection of the null hypothesis. Thus, if our calculated p-value is less than 0.05 (our yopsis{y_alpha}-level), we would conclude there is convincing evidence that female students watch more TV than male students.

Degrees of Freedom

Degrees of freedom (df) represent the number of values in a calculation that are free to vary. They play a critical role in the assessment of the t-test because they affect the shape of the t distribution, which we use to find the p-value associated with the test statistic. The degrees of freedom essentially depend on the sample sizes from each group minus the number of parameters estimated.

In a two-sample t-test like ours, calculating the degrees of freedom can be complex as it involves both sample sizes and variances of the groups. The df for a two-sample t-test is determined using the formula provided in the exercise, incorporating both sample sizes and variances, to get a precise value adjusted for the weight of each group's variance. These degrees of freedom are then used to reference the correct t-distribution when determining the p-value and, ultimately, when making decisions about the hypotheses.