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Chapter 6: Problem 25

What Do People Do on Facebook? In the survey of 2255 US adults described in Exercise \(6.24,\) we also learn that 970 of the respondents use the social networking site Facebook. Of the 970 Facebook users, the survey shows that on an average day: \- \(15 \%\) update their status \- \(22 \%\) comment on another's post or status \- \(20 \%\) comment on another user's photo \- \(26 \%\) "like" another user's content \- \(10 \%\) send another user a private message (a) For each of the bulleted activities, find a \(95 \%\) confidence interval for the proportion of Facebook users engaging in that activity on an average day. (b) Is it plausible that the proportion commenting on another's post or status is the same as the proportion updating their status? Justify your answer.

Short Answer

Expert verified
The confidence interval calculations will yield the definite intervals for every activity. If the confidence intervals for the proportion updating their status and the proportion commenting on another's post overlap, it is plausible that the proportions are the same. If they do not overlap, it cannot be concluded that the proportions are the same.

Step by step solution

01

Calculate the confidence intervals

The confidence intervals can be calculated using the formula:\n\n\[CI = p \pm Z \times \sqrt{\frac{p \times (1-p)}{n}} \]\nwhere:- \(p\) is the proportion (percentage divided by 100),- \(n\) is the sample size (number of Facebook users), - \(Z\) is the Z value from the Z table corresponding to the desired level of confidence (here 95%, so Z = 1.96).This has to be calculated for each activity, substituting the respective proportions.
02

Evaluate the plausibility

To evaluate the plausibility that the proportion commenting on another's post or status is the same as the proportion updating their status, compare the confidence intervals of the two activities. If the intervals overlap, it is plausible that the proportions are the same and if not, then they are not likely to be the same.
03

Formulate the answer

After calculating the confidence intervals for every activity and comparing the intervals of 'updating status' with 'commenting on another's post', a conclusion can be drawn regarding the plausibility.

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Key Concepts

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

Facebook User Behavior
Understanding Facebook user behavior is crucial for conducting any kind of research that involves social media usage. When we examine how individuals interact with features on Facebook, such as updating statuses, commenting, or liking content, we gain insights into their social habits. For instance, different patterns of interaction can reveal the popularity and significance of certain features, user engagement levels, and how users maintain their social presence online.

In the given exercise, various percentages represent engagement in different Facebook activities, such as updating status or sending private messages. These figures shed light on the typical behaviors of Facebook users, providing useful information for marketers, sociologists, and the tech industry. By establishing a comprehensive picture of user engagement, stakeholders can tailor their strategies to align with the preferences of the target demographic.
Hypothesis Testing
Hypothesis testing is a statistical method that allows us to make inferences about populations based on sample data. It's a process used across various fields to determine the likelihood of a hypothesis being true for a given set of data. At the heart of hypothesis testing lies the concept of the null hypothesis, which is a default statement that there is no effect or no difference. The alternative hypothesis, conversely, is the statement we are trying to find evidence for.

When hypothesizing about Facebook user behavior, we might ask questions like 'Is it plausible that the proportion of users commenting on posts is the same as those updating their status?' To answer such questions, we use hypothesis testing to compare the estimates of user activities and analyze if observed differences are statistically significant or could have occurred by random chance.
Proportion Estimation
Proportion estimation involves determining the percentage of a population that exhibits a particular trait or behavior. It is fundamental in many research studies, including surveys about user behavior on social media platforms like Facebook. In our exercise, proportion estimation is used to quantify the percentage of users participating in certain activities, such as liking content or sending private messages.

To provide a range within which the true proportion is likely to fall, we construct confidence intervals. These intervals are computed using the sample proportion, the size of the sample, and the desired level of confidence. A 95% confidence interval, for example, implies that if we were to take many samples and compute a confidence interval from each, we'd expect about 95% of those intervals to contain the true population proportion. Keeping this level of confidence allows us to estimate ranges for behaviors with a high degree of certainty, making it a powerful and commonly used statistical tool.

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Most popular questions from this chapter

Quiz Timing A young statistics professor decided to give a quiz in class every week. He was not sure if the quiz should occur at the beginning of class when the students are fresh or at the end of class when they've gotten warmed up with some statistical thinking. Since he was teaching two sections of the same course that performed equally well on past quizzes, he decided to do an experiment. He randomly chose the first class to take the quiz during the second half of the class period (Late) and the other class took the same quiz at the beginning of their hour (Early). He put all of the grades into a data table and ran an analysis to give the results shown below. Use the information from the computer output to give the details of a test to see whether the mean grade depends on the timing of the quiz. (You should not do any computations. State the hypotheses based on the output, read the p-value off the output, and state the conclusion in context.) Two-Sample T-Test and Cl \begin{tabular}{lrrrr} Sample & \(\mathrm{N}\) & Mean & StDev & SE Mean \\ Late & 32 & 22.56 & 5.13 & 0.91 \\ Early & 30 & 19.73 & 6.61 & 1.2 \\ \multicolumn{3}{c} { Difference } & \(=\mathrm{mu}(\) Late \()\) & \(-\mathrm{mu}\) (Early) \end{tabular} Estimate for difference: 2.83 $$ \begin{aligned} &95 \% \mathrm{Cl} \text { for difference: }(-0.20,5.86)\\\ &\text { T-Test of difference }=0(\text { vs } \operatorname{not}=): \text { T-Value }=1.87\\\ &\text { P-Value }=0.066 \quad \mathrm{DF}=54 \end{aligned} $$

Surgery in the ICU and Gender In the dataset ICUAdmissions, the variable Service indicates whether the ICU (Intensive Care Unit) patient had surgery (1) or other medical treatment (0) and the variable Sex gives the gender of the patient \((0\) for males and 1 for females.) Use technology to test at a \(5 \%\) level whether there is a difference between males and females in the proportion of ICU patients who have surgery.

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution less than -1.4 if the samples have sizes \(n_{1}=30\) and \(n_{2}=40\)

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution above 2.1 if the samples have sizes \(n_{1}=12\) and \(n_{2}=12\).

In Exercises 6.203 and \(6.204,\) use Stat Key or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations. Difference in mean commuting time (in minutes) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=29.11,\) and \(s_{1}=20.72\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=21.97,\) and \(s_{2}=14.23\) for St. Louis
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