Question

Age Distribution of Social Networking Site Users The Pew Research Center conducted a survey of randomly sampled American adults in 2008 and in 2010 , asking them about their use of social networking sites such as Facebook. \(^{21}\) Table 7.34 shows age groups of social networking site users in 2008 and in 2010 . Did the age distribution change significantly in the two-year time span? Show all details of the test. $$ \begin{array}{l|rr|r} \downarrow \text { Age/Year } \rightarrow & 2008 & 2010 & \text { Total } \\ \hline 18-22 & 138 & 152 & 290 \\ 23-35 & 197 & 303 & 500 \\ 36-49 & 108 & 246 & 354 \\ 50+ & 52 & 246 & 298 \\ \hline \text { Total } & 495 & 947 & 1442 \\ \hline \end{array} $$

Short answer

The step-by-step test procedure of the Chi-Square test for independence allows us to conclude whether the age distribution of social networking site users changed significantly from 2008 to 2010. Conclusion will be based on p-value comparison with the significance level.

Step-by-step solution

State the Null and Alternative Hypotheses

The first step in any statistical test is to state the null and alternative hypotheses.\n\n\(H_0\): The year of usage and age group are independent. There is no significant association between the age distribution of users and the year.\n\n\(H_A\): The year of usage and age group are not independent. There is a significant association between the age distribution of users and the year.

State the Level of Significance

Commonly, a level of significance (alpha) is chosen to be 0.05. This means that if the calculated Chi-square value will lead to a probability (p-value) less than 0.05, then the null hypothesis will be rejected.

Calculate Observed and Expected Frequencies

The next step would be to calculate the Expected frequency for every group in the table. The expected frequency can be calculated using the formula:\n\n\(E = (row total * column total) / overall total\)\n\nAfter, we calculate the observed frequency which is the actual collected data given in the problem.

Calculate the Chi-Square Test Statistic

The Chi-Square Test Statistic can be calculated using the formula:\n\n\(\chi^2 = ∑\[((O - E)^2 / E)\]\n\nwhere\nO = Observed frequency\nE = Expected frequency

Calculate the p-value and draw the conclusion

Compare the p-value to the level of significance and reject or fail to reject the null hypothesis. If the p-value is less than the significance level (0.05), reject the null hypothesis. Which means that there is a significant association between the age distribution of social network users and the year.

Key concepts

Null and Alternative Hypotheses

Understanding the null and alternative hypotheses is essential in any statistical testing, including the Chi-Square Test for Independence. The null hypothesis ((H_0)) represents a statement of no effect or no difference. It suggests that any observed variation is purely due to chance, and that variables are independent. In the context of our exercise, the null hypothesis posits that the age distribution of social networking site users in 2008 and 2010 are independent.

The alternative hypothesis ((H_A)), on the other hand, is the statement which a researcher is trying to prove. It proposes that there is an effect or a difference, and that variables are not independent. For our scenario, the alternative hypothesis suggests that there is a significant association between the age distribution of social networking site users and the year.

Essentially, the null hypothesis serves as the default or starting assumption that stands until evidence in the form of data provides a strong enough reason to reject it in favor of the alternative hypothesis.

Level of Significance

The level of significance, often denoted as alpha ((α)), is a critical concept in hypothesis testing. It represents the threshold for rejecting the null hypothesis and is pre-set by the researcher before any data is observed. This threshold is the likelihood of observing a test statistic at least as extreme as the one calculated, assuming the null hypothesis is true.

In many studies, a common level of significance used is 0.05, meaning there is a 5% chance of rejecting the null hypothesis when it is actually true (Type I error). If the p-value obtained from the test is less than the chosen significance level, we conclude that the data provides sufficient evidence to reject the null hypothesis. This is a pivotal point in making inferences since it directly relates to the risk one is willing to take in making a conclusion about the statistical evidence.

Observed and Expected Frequencies

In Chi-Square tests, the observed frequencies represent the actual data collected, as shown in the table of the exercise. These are the counts of individuals or items divided among the categories of the variable being analyzed.

The expected frequencies are theoretical frequencies calculated under the assumption that the null hypothesis is true. They represent what we would expect to observe if the categories of the variable were distributed purely by chance according to certain ratios. To calculate these frequencies, we use the formula (E = (row total * column total) / overall total). Comparing the observed frequencies to the expected frequencies is key to determining if there's a statistically significant difference, or if any observed variation could reasonably occur by random chance alone.

Statistical Significance

Statistical significance plays a vital role in determining the validity of our hypotheses. It addresses the question: 'Is the observed effect real, or could it be due to random chance?' In the context of the Chi-Square Test for Independence, statistical significance is assessed by comparing the p-value to the level of significance.

If the p-value is less than or equal to the level of significance, the results are considered statistically significant, leading us to reject the null hypothesis. This finding would indicate that the variables we are studying are likely associated with one another in the population. For our exercise, if the calculated p-value is below 0.05, we would conclude that the age distribution of social networking site users changed significantly between 2008 and 2010. It's this concept of statistical significance that helps to guard against attributing patterns to random variations in the data.