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Gender and Frequency of "Liking" Content on Facebook Exercise 7.48 introduces a study about users of social networking sites such as Facebook. Table 7.37 shows the frequency of users "liking" content on Facebook, with the data shown by gender. Does the frequency of "liking" depend on the gender of the user? Show all details of the test. $$ \begin{array}{l|rr|r} \hline \downarrow \text { Liking/Gender } \rightarrow & \text { Male } & \text { Female } & \text { Total } \\ \hline \text { Every day } & 77 & 142 & 219 \\ \text { 3-5 days/week } & 39 & 54 & 93 \\ \text { 1-2 days/week } & 62 & 69 & 131 \\ \text { Every few weeks } & 42 & 44 & 86 \\ \text { Less often } & 166 & 182 & 348 \\ \hline \text { Total } & 386 & 491 & 877 \end{array} $$

Short Answer

Expert verified
The complete answer requires calculation and depends on the results of those calculations. After calculating expected frequencies, chi-square test statistic, degrees of freedom and comparing the p-value to the significance level, one can determine whether the liking behavior on Facebook depends on gender.

Step by step solution

01

Calculations of Expected Frequencies

Under the null hypothesis of gender and liking behavior being independent, the expected frequency of a particular combination is calculated as (Row Total * Column Total) / Grand Total. Do this for every cell in the table to get the expected frequencies.
02

Chi-Square Test Statistic

The test statistic for the chi-square test is given by the sum over all cells: \((Observed-Expected)^2/Expected\). Compute this value.
03

Degrees of Freedom and P-value

The degrees of freedom for this test is given by (Number of Rows - 1) * (Number of Columns - 1). Use this degree of freedom and the test statistic computed in Step 2 to find the p-value, referencing the chi-square distribution table.
04

Conclusion

If the p-value is less than the significance level (generally .05), reject the null hypothesis and conclude that the frequency of 'liking' does indeed depend on the gender. If not, there is not enough evidence to say that liking behaviors on Facebook depend on gender.

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Key Concepts

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

Expected Frequencies
When performing a chi-square test, a crucial step is calculating the expected frequencies. These figures represent the expected counts in each category if the null hypothesis were true, meaning there is no association between the variables.

In the context of our exercise, the expected frequency for each cell is found by multiplying the total number of observations in that row by the total number of observations in that column, and then dividing by the grand total of all observations. For example, to calculate the expected frequency of males 'liking' content every day, one would use the formula: \( \frac{\text{Total Every day} \times \text{Total Male}}{\text{Grand Total}} \).

It's essential to compute these expected values accurately as they are used as a benchmark to measure how far off the observed frequencies are from what we would expect to see if the variables were truly independent.
Test Statistic
The test statistic in a chi-square test quantifies how much the observed frequencies deviate from the expected frequencies. It’s a measure of the disparity between what was actually observed in the data and what we would anticipate under the null hypothesis of no association.

To calculate the chi-square test statistic, each cell's observed value is compared to its expected value using the formula: \( (\text{Observed} - \text{Expected})^2 / \text{Expected} \).

After summing these values for all cells, you get a single number that indicates the overall difference. The larger the test statistic, the more evidence there is of discrepancy between the observed and expected data, suggesting a possible association between the variables in question.
Degrees of Freedom
Degrees of freedom in statistics can be thought of as the number of values that can vary without violating given constraints. They play a key role in determining the critical value against which the test statistic is compared in a chi-square test.

The formula for calculating degrees of freedom in a contingency table like our exercise is: \( (\text{Number of Rows} - 1) \times (\text{Number of Columns} - 1) \). In this scenario, with a 2x5 contingency table (2 genders and 5 frequency categories), the chi-square test has \( (2-1) \times (5-1) = 4 \) degrees of freedom.

This informs us about the chi-square distribution that we will reference to interpret the test statistic and ultimately reach a conclusion about our hypothesis.
P-value
The p-value is a probability that provides a measure of the evidence against the null hypothesis provided by the data. Specifically, it's the probability of observing a test statistic at least as extreme as the one computed, assuming the null hypothesis is true.

In our sample exercise, once the chi-square statistic is calculated, the corresponding p-value can be found by looking up this statistic in the chi-square distribution table, with the degrees of freedom we calculated.

A smaller p-value means stronger evidence against the null hypothesis. Typically, if the p-value is below the predefined level of significance, which is often 0.05, we would reject the null hypothesis, indicating that there is a statistically significant association between the variables.
Statistical Significance
Statistical significance is used to determine whether the observed results are due to chance or if they reflect a true effect. In hypothesis testing, statistical significance is assessed by comparing the p-value to a predetermined significance level, often set at 0.05.

If the p-value is less than the chosen significance level, we reject the null hypothesis, suggesting that the results are unlikely to have occurred by random chance, thus inferring a statistically significant relationship between the variables.

In the context of our exercise, if the p-value is below 0.05, we would conclude that the frequency of 'liking' content on Facebook is dependent on gender. If the p-value is above 0.05, we lack sufficient evidence to claim a relationship, and the null hypothesis that 'liking' behavior is independent of gender would not be rejected.

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Most popular questions from this chapter

Gender and Award Preference Example 2.6 on page 53 contains a two-way table showing preferences for an award (Academy Award, Nobel Prize, Olympic gold medal) by gender for the students sampled in StudentSurvey. The data are reproduced in Table \(7.28 .\) Test whether the data indicate there is some association between gender and preferred award. $$ \begin{array}{l|ccc|c} \hline & \text { Academy } & \text { Nobel } & \text { Olympic } & \text { Total } \\ \hline \text { Female } & 20 & 76 & 73 & 169 \\ \text { Male } & 11 & 73 & 109 & 193 \\ \hline \text { Total } & 31 & 149 & 182 & 362 \\ \hline \end{array} $$

Favorite Skittles Flavor? Exercise 7.13 on page 518 discusses a sample of people choosing their favorite Skittles flavor by color (green, orange, purple, red, or yellow). A separate poll sampled 91 people, again asking them their favorite Skittles flavor, but rather than by color they asked by the actual flavor (lime, orange, grape, strawberry, and lemon, respectively). \(^{19}\) Table 7.32 shows the results from both polls. Does the way people choose their favorite Skittles type, by color or flavor, appear to be related to which type is chosen? (a) State the null and alternative hypotheses. (b) Give a table with the expected counts for each of the 10 cells. (c) Are the expected counts large enough for a chisquare test? (d) How many degrees of freedom do we have for this test? (e) Calculate the chi-square test statistic. (f) Determine the p-value. Do we find evidence that method of choice affects which is chosen? $$ \begin{array}{lcrccc} \hline & \begin{array}{l} \text { Green } \\ \text { (Lime) } \end{array} & \begin{array}{c} \text { Purple } \\ \text { Orange } \end{array} & \begin{array}{c} \text { Red } \\ \text { (Grape) } \end{array} & \begin{array}{c} \text { Yellow } \\ \text { (Strawberry) } \end{array} & \text { (Lemon) } \\ \hline \text { Color } & 18 & 9 & 15 & 13 & 11 \\ \text { Flavor } & 13 & 16 & 19 & 34 & 9 \end{array} $$

Can People Delay Death? A study indicates that elderly people are able to postpone death for a short time to reach an important occasion. The researchers \({ }^{10}\) studied deaths from natural causes among 1200 elderly people of Chinese descent in California during six months before and after the Harbor Moon Festival. Thirty-three deaths occurred in the week before the Chinese festival, compared with an estimated 50.82 deaths expected in that period. In the week following the festival, 70 deaths occurred, compared with an estimated 52. "The numbers are so significant that it would be unlikely to occur by chance," said one of the researchers. (a) Given the information in the problem, is the \(\chi^{2}\) statistic likely to be relatively large or relatively small? (b) Is the p-value likely to be relatively large or relatively small? (c) In the week before the festival, which is higher: the observed count or the expected count? What does this tell us about the ability of elderly people to delay death? (d) What is the contribution to the \(\chi^{2}\) -statistic for the week before the festival? (e) In the week after the festival, which is higher: the observed count or the expected count? What does this tell us about the ability of elderly people to delay death? (f) What is the contribution to the \(\chi^{2}\) -statistic for the week after the festival? (g) The researchers tell us that in a control group of elderly people in California who are not of Chinese descent, the same effect was not seen. Why did the researchers also include a control group?

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. $$ \begin{array}{l|rr|r} \downarrow \text { Age/Year } \rightarrow & 2008 & 2010 & \text { Total } \\ \hline 18-22 & 138 & 152 & 290 \\ 23-35 & 197 & 303 & 500 \\ 36-49 & 108 & 246 & 354 \\ 50+ & 52 & 246 & 298 \\ \hline \text { Total } & 495 & 947 & 1442 \\ \hline \end{array} $$

In Exercises 7.1 to \(7.4,\) find the expected counts in each category using the given sample size and null hypothesis. \(H_{0}:\) All three categories \(A, B, C\) are equally likely; \(\quad n=1200\)

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