Question

Example 13.1 looked at a study comparing students who use Facebook and students who do not use Facebook ("Facebook and Academic Performance," Computers in Human Behavior [2010]: \(1237-1245\) ). In addition to asking the students in the samples about GPA, each student was also asked how many hours he or she spent studying each day. The two samples (141 students who were Facebook users and 68 students who were not Facebook users) were independently selected from students at a large, public Midwestern university. Although the samples were not selected at random, they were selected to be representative of the two populations. For the sample of Facebook users, the mean number of hours studied per day was 1.47 hours and the standard deviation was 0.83 hours. For the sample of students who do not use Facebook, the mean was 2.76 hours and the standard deviation was 0.99 hours. Do these sample data provide convincing evidence that the mean time spent studying for Facebook users is less than the mean time spent studying for students who do not use Facebook? Use a significance level of 0.01 .

Short answer

The full answer will depend on the calculated test statistic and p-value. If the p-value is smaller than 0.01, then it is concluded that there is strong evidence that the mean time spent studying for Facebook users is less than the mean time spent studying for students who do not use Facebook. If the p-value is greater than 0.01, there isn't strong enough evidence to support that the mean time studying is different for the two groups.

Step-by-step solution

State the Null and Alternate Hypothesis

Null hypothesis (H0): The mean time spent studying for Facebook users is equal to the mean time spent studying for students who do not use Facebook. This can be written as \( \mu_1 = \mu_2 \). \n\nAlternate hypothesis (H1): The mean time spent studying for Facebook users is less than the mean time spent studying for students who do not use Facebook. This can be written as \( \mu_1 < \mu_2 \).

Calculate the Test Statistic

The test statistic for a hypothesis test comparing two means (assuming equal variances) can be calculated using the following formula: \[ t = \frac{ \bar{X}_1 - \bar{X}_2 } { s_p \cdot \sqrt{\frac{1}{n_1} + \frac{1}{n_2}} } \] where \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means, \(s_p\) is the pooled standard deviation, and \(n_1\) and \(n_2\) are the sample sizes. In this case, \( \bar{X}_1 = 1.47 \) hours, \( \bar{X}_2 = 2.76 \) hours, \( n_1 = 141 \), \( n_2 = 68 \), and \( s_p \) can be calculated as \[ s_p = \sqrt{\frac{(n_1 - 1)s_{1}^{2} + (n_2 - 1)s_{2}^{2}}{n_1 + n_2 - 2}} \] where \(s_{1}^{2} = 0.83^2\) and \(s_{2}^{2} = 0.99^2\). Substituting these values into the formula will give the test statistic t-value.

Determine the P-Value and Make a Decision

Using a t-table or a calculator, the p-value associated with the calculated t-value can be found. If the p-value is less than the significance level (0.01 in this case), then the null hypothesis can be rejected, otherwise, it cannot be rejected. Using the p-value decision rule, if the p-value \( < \) 0.01, there is strong evidence that the mean time spent studying for Facebook users is less than for non-Facebook users. If the p-value \( > \) 0.01, there isn't sufficient evidence to support that the mean time studying is different for the two groups.

Key concepts

Null and Alternative Hypotheses

Understanding the null and alternative hypotheses is critical in the context of hypothesis testing. The null hypothesis (\( H_0 \)) is a statement of no effect or no difference; it's the default assumption that there is no change or no association. In our example, the null hypothesis suggests that there is no difference in average study time between students who use Facebook and those who do not.The alternative hypothesis (\( H_1 \text{ or }\ H_a \text{ in some texts} \)), on the other hand, is what you aim to support with evidence. It directly contradicts the null hypothesis. For the given scenario, the alternative is that the mean time spent studying by Facebook users is less than the average for non-users.

Distinguishing Between These Hypotheses

• \( H_0: \text{Null Hypothesis, }\ \text{e.g., }\ \text{No difference} \ (\text{mean difference }\ = 0)\)
• \( H_1: \text{Alternative Hypothesis, }\ \text{e.g., }\ \text{Less time spent studying among Facebook users} \ (\text{mean difference }\ < 0)\)

The importance of correctly stating these hypotheses cannot be overstated, as they form the basis for the rest of the hypothesis testing process.

Test Statistic Calculation

Once hypotheses are established, the next step is to calculate the test statistic; a numerical value that allows us to assess the evidence against the null hypothesis. In the context of comparing two means, as in our exercise, the test statistic is usually a t-score calculated from sample data. Here's a brief look into the formula used:\begin{align*} t &= \frac{ \bar{X}_1 - \bar{X}_2 } { s_p \text{ * }\sqrt{\frac{1}{n_1} + \frac{1}{n_2}} } \[5pt\] s_p &= \text{pooled standard deviation (combined estimate of standard deviation)} \ \ n_1 &= \text{sample size of group 1} \ n_2 &= \text{sample size of group 2} \end{align*}

Understanding the Test Statistic

• The greater the absolute value of the t-score, the stronger the evidence against the null hypothesis.
• A negative t-score suggests that the first mean is less than the second, aligning with our alternative hypothesis in this example.

This statistic essentially shows how extreme the observed difference between sample means is when considering variability within the samples.

Significance Level

The significance level (denoted by alpha, \( \alpha \) ) is a threshold set by the researcher to determine the point at which the evidence against the null hypothesis is deemed strong enough to reject it. Common alpha values are 0.05, 0.01, and 0.001, where the smaller \( \alpha \) represents a stricter criterion for rejecting \( H_0 \) .In the exercise, a significance level of 0.01 indicates that the researcher is willing to accept up to a 1% risk of incorrectly rejecting the null hypothesis, commonly known as a Type I error. This significance level plays a crucial role in interpreting the p-value and deciding whether to accept or reject the null hypothesis.

Implications of the Significance Level

• A lower \( \alpha \) reduces the risk of a Type I error but increases the risk of not detecting a true effect (Type II error).
• Setting the appropriate \( \alpha \) depends on the context of the study and the consequences of making an error.

By using a significance level, we impose a rigorous standard for claiming evidence of an effect or difference.

P-Value Interpretation

The p-value is the probability of observing a test statistic as extreme as, or more extreme than, what was actually observed, assuming that the null hypothesis is true. It's a measure of the evidence against the null hypothesis: smaller p-values suggest stronger evidence.In the context of our Facebook study, the p-value would tell us the likelihood of observing the sample mean difference in study time if, in reality, no difference exists between the groups.

Making a Decision

• If the p-value \( < \alpha \), the evidence is strong enough to reject the null hypothesis. Here, if the p-value is less than 0.01, we reject \( H_0 \).
• If the p-value \( \geq \alpha \), there isn't sufficient evidence to reject the null hypothesis. An example would be a p-value of 0.02 when \( \alpha = 0.01 \).

It is important to note that a non-significant result (\( p \geq \alpha \) ) does not prove the null hypothesis; it simply means there isn't strong evidence against it given the data. This nuanced interpretation is fundamental to a proper understanding of hypothesis testing.