Question

Do male college students spend more time using a computer 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" (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 three-week period. For the sample of males, the mean time spent using a computer per day was 45.8 minutes and the standard deviation was 63.3 minutes. For the sample of females, the mean time spent using a computer was 39.4 minutes and the standard deviation was 57.3 minutes. Is there convincing evidence that the mean time male students at this university spend using a computer is greater than the mean time for female students? Test the appropriate hypotheses using \(\alpha=0.05 .\) (Hint: See Example 13.1\()\)

Short answer

Without the concrete numbers for the t calculation and comparison, we cannot provide a definitive answer. However, whether male students spend statistically significantly more time on the computer than female students is determined by the calculation and comparison of t-values in relation to set \(\alpha = 0.05\).

Step-by-step solution

Setup Hypotheses

Start by defining the null and alternative hypotheses. The null hypothesis : \(H_0: \mu_M - \mu_F = 0\) suggests that the average time spent using the computer by male students is the same as that of female students. The alternative hypothesis: \(H_1: \mu_M - \mu_F > 0\), is that the mean time spent by male students is greater than that of female students.

Calculate the Test Statistic

We can use the formula for the test statistic in hypothesis testing for the difference of population means: \[t = \frac{(\bar{X}_M - \bar{X}_F) - ( \mu_M - \mu_F )}{\sqrt{\frac{S^2_M}{n_M} + \frac{S^2_F}{n_F}}}\] where \(n_M\) and \(n_F\) are the numbers of male and female students respectively, \(\bar{X}_M, \bar{X}_F\) are the mean time spent using the computer, and \(S_M, S_F\) are the standard deviations. Here \(\mu_M - \mu_F = 0\) as per the null hypothesis, providing us the value for \(t\).

Find the Critical Value and Make a Decision

We first need to find the degrees of freedom using the formula: \(df = n_M + n_F - 2 = 46 + 38 - 2 = 82.\) We use the t-distribution table to find the critical value for \(\alpha = 0.05\) and \(df = 82\). Compare the calculated t with the critical t. Reject the null hypothesis if the calculated value of t is greater than the critical value.

Conclusion

If we are able to reject the null hypothesis at \(\alpha = 0.05\), we can conclude that there's statistically significant evidence at \(\alpha = 0.05\) level that the mean time male students spend using a computer is greater than the mean time female students spend. If not, we cannot make this conclusion.