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Chapter 13: Problem 3

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

Expert verified
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

01

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.
02

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 \(t\).
03

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.
04

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.

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Key Concepts

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

Mean Comparison
Mean comparison is an essential concept in statistics. It allows us to determine if there is a significant difference between the means of two groups. When tackling a hypothesis test concerning means, we define two hypotheses: the null hypothesis (\(H_0\)) that assumes no difference in the means of the groups, and the alternative hypothesis (\(H_1\)) which suggests that there is a difference.

In this exercise, the focus is on comparing the mean computer usage times between male and female college students. Here, the null hypothesis is that males and females spend the same average time using computers. Meanwhile, the alternative hypothesis suggests that males spend more time on computers compared to females.

This comparison is performed using statistical tools, allowing us to objectively determine if the observed differences in means hold any significance. This step sets the stage for further analysis using statistical methods like the calculation of a test statistic and interpreting results with reference to distributions.
T-Distribution
The t-distribution is a probability distribution used in hypothesis testing, especially when dealing with small sample sizes or unknown population variances. It is similar to the normal distribution but has heavier tails, which means it is more prone to producing values that fall far from its mean. This characteristic makes it ideal for smaller sample sizes.

In hypothesis testing, the test statistic is calculated and then compared against critical values from the t-distribution. These critical values depend on the chosen significance level, denoted by \(\alpha\), and the degrees of freedom, which in this exercise, is calculated as the sum of the individuals in both groups minus two (\(df = n_M + n_F - 2\)).

For the exercise, the degrees of freedom were calculated to be 82, and with \(\alpha = 0.05\), the critical value can be obtained from a t-distribution table. The test statistic is then compared to this critical value to decide whether to accept or reject the null hypothesis. The t-distribution is a critical part of ensuring the reliability of such hypothesis tests.
Statistical Significance
Statistical significance helps determine whether the observed effect in data is genuine or if it's likely due to chance. It is denoted by a significance level, \(\alpha\), which is commonly set at 0.05 in many statistical tests. This threshold indicates a 5% risk of concluding that a difference exists when there is none.

To assess statistical significance, the test statistic calculated in the hypothesis testing process is compared to a critical value derived from the t-distribution. If the test statistic exceeds this critical value, the null hypothesis is rejected, indicating that the result is statistically significant.

In the given exercise, if the calculated t-value is higher than the critical t-value, this suggests statistical significance. This would then imply convincing evidence that the average computer usage time by males is indeed greater than that by females. Statistical significance indicates that the results are not simply by random variation but are likely reflective of a true difference in the means.

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Most popular questions from this chapter

Do children diagnosed with attention deficit/hyperactivity disorder (ADHD) have smaller brains than children without this condition? This question was the topic of a research study described in the paper "Developmental Trajectories of Brain Volume Abnormalities in Children and Adolescents with Attention Deficit/Hyperactivity Disorder" ( Journal of the American Medical Association [2002]: \(1740-1747\) ). Brain scans were completed for a representative sample of 152 children with ADHD and a representative sample of 139 children without ADHD. Summary values for total cerebral volume (in cubic milliliters) are given in the following table:$$ \begin{array}{lccc} & n & \bar{x} & s \\ \hline \text { Children with ADHD } & 152 & 1,059.4 & 117.5 \\ \text { Children without ADHD } & 139 & 1,104.5 & 111.3 \\ \hline \end{array} $$ Is there convincing evidence that the mean brain volume for children with ADHD is smaller than the mean for children without ADHD? Test the relevant hypotheses using a 0.05 level of significance.

Two proposed computer mouse designs were compared by recording wrist extension in degrees for 24 people who each used both mouse designs ("Comparative Study of Two Computer Mouse Designs," Cornell Human Factors Laboratory Technical Report RP7992). The difference in wrist extension was calculated by subtracting extension for mouse type \(\mathrm{B}\) from the wrist extension for mouse type \(\mathrm{A}\) for each person. The mean difference was reported to be 8.82 degrees. Assume that this sample of 24 people is representative of the population of computer users. a. Suppose that the standard deviation of the differences was 10 degrees. Is there convincing evidence that the mean wrist extension for mouse type \(A\) is greater than for mouse type \(\mathrm{B}\) ? Use a 0.05 significance level. b. Suppose that the standard deviation of the differences was 26 degrees. Is there convincing evidence that the mean wrist extension for mouse type \(A\) is greater than for mouse type \(\mathrm{B}\) ? Use a 0.05 significance level. c. Briefly explain why different conclusions were reached in the hypothesis tests of Parts (a) and (b).

The paper referenced in the Preview Example of this chapter ("Mood Food: Chocolate and Depressive Symptoms in a Cross-Sectional Analysis," Archives of Internal Medicine [2010]: \(699-703\) ) describes a study that investigated the relationship between depression and chocolate consumption. Participants in the study were 931 adults who were not currently taking medication for depression. These participants were screened for depression using a widely used screening test. The participants were then divided into two samples based on their test score. One sample consisted of people who screened positive for depression, and the other sample consisted of people who did not screen positive for depression. Each of the study participants also completed a food frequency survey. The researchers believed that the two samples were representative of the two populations of interest-adults who would screen positive for depression and adults who would not screen positive. The paper reported that the mean number of servings per month of chocolate for the sample of people that screened positive for depression was 8.39 , and the sample standard deviation was \(14.83 .\) For the sample of people who did not screen positive for depression, the mean was \(5.39,\) and the standard deviation was \(8.76 .\) The paper did not say how many individuals were in each sample, but for the purposes of this exercise, you can assume that the 931 study participants included 311 who screened positive for depression and 620 who did not screen positive. Carry out a hypothesis test to confirm the researchers' conclusion that the mean number of servings of chocolate per month for people who would screen positive for depression is higher than the mean number of chocolate servings per month for people who would not screen positive.

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.

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 \(\alpha=0.05\).
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