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Randomly chosen people were observed for about 10 seconds in several public places, such as malls and restaurants, to see whether they smiled during that time. The table shows the results for different age groups. $$ \begin{array}{lccccc} & & & \text { Age Group } \\ & \mathbf{0 - 1 0} & \mathbf{1 1 - 2 0} & \mathbf{2 1 - 4 0} & \mathbf{4 1 - 6 0} & \mathbf{6 1 +} \\ \hline \text { Smile } & 1131 & 1748 & 1608 & 937 & 522 \\ \hline \text { No Smile } & 1187 & 2020 & 3038 & 2124 & 1509 \end{array} $$ a. Find the percentage of each age group that were observed smiling, and compare these percentages. b. Treat this as a single random sample of people, and test whether smiling and age group are associated, using a significance level of \(0.05 .\) Comment on the results.

Short Answer

Expert verified
The percentages of the number of people smiling in each age group were calculated. They were compared to find discernable patterns or differences. Furthermore, to ascertain if there is an association between the age group and smiling, the chi-square test of independence was performed at a 0.05 significance level. The result of this test, whether the p-value is less than or more than 0.05, will determine if there is an association or not.

Step by step solution

01

Calculate the Total Number of Participants in Each Age Group

Add the frequency of 'Smile' and 'No Smile' for each age group. \[Total = Smile + No Smile\] For example, for the 0-10 group, the total would be \(1131 + 1187 = 2318\). Make sure to compute the total for every age group.
02

Calculate the Percentage that Were Observed Smiling

Divide the number of people who were observed smiling ('Smile') in each group by the total participants in that group and multiply by 100. \[Percentage = \frac{Smile}{Total} * 100\] For example, for the 0-10 group, the percentage would be \(\frac{1131}{2318} * 100 ≈ 48.76%\). Compute the percentage for each age group.
03

Compare the Percentages

Analyze the obtained percentages and compare them to each other. Here, we are looking for noticeable differences or patterns between the age groups.
04

Perform Chi-Square Test of Independence

Treat this as a single random sample of people, and perform the chi-square test of independence to determine if there is a significant association between the age group and smiling. Set your significance level at \(alpha = 0.05\). The null hypothesis is that the age group and smiling are not associated (independent), and the alternative hypothesis is that they are associated (dependent).
05

Comment on the Results

Based on the p-value obtained from the chi-square test, if the p-value is less than the significance level (\(alpha = 0.05\)), we reject the null hypothesis and conclude that there is evidence that age group and smiling are associated. Otherwise, we do not reject the null hypothesis and conclude there's no enough evidence for association.

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Key Concepts

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

Probability
Probability in statistics is the measure of how likely an event is to occur. In this exercise, we assess the likelihood of people smiling in different age groups when observed for 10 seconds in public places. The occurrence or non-occurrence of a smile is treated as a random event.

Understanding probability helps us predict outcomes in a given scenario. Here, we're focusing on comparing the percentage of smiles within each age group, such as ages 0-10, 11-20, etc. It gives us a clearer picture of whether certain age groups are more prone to smile during such observations.
  • To calculate the probability or percentage, we divide the number of smiles by the total number of observations in each age group.
  • This allows us to draw a comparison across different groups, highlighting trends or notable differences.
For example, if you find that the 0-10 age group has around 48.76% smiling, it gives a measure of its likelihood against other groups. By looking at these figures, you analyze probabilities to assess whether age impacts smiling behavior.
Age Group Analysis
In this exercise, age group analysis refers to breaking down participants into specific age categories to identify how different age groups behave in terms of smiling during a short observation. This provides insight into behavioral patterns among varying demographics.

Analyzing age groups involves comparing the percentage of smiling individuals across these groups. Each age group - from 0-10 to 61+ - is assessed to see the proportion of people smiling.
  • First, calculate the total number of smiles in each age group.
  • Next, determine what portion these smiles represent out of the total participants in that age group.
  • Finally, compare these proportions to see if younger or older people are smiling more when observed.
The purpose of age group analysis is not just to identify differences but also to understand if these differences suggest that age plays a role in influencing people's likelihood to smile. By examining each group, we uncover patterns that could be tied to age-specific behaviors or social norms.
Hypothesis Testing
Hypothesis testing is a statistical method used to determine if there is enough evidence to conclude if an observation has meaningful insight or if it's just by random chance. In this exercise, we use the Chi-Square Test of Independence to evaluate the relationship between age and smiling.

For hypothesis testing, we start by formulating two hypotheses:
  • The null hypothesis ( H_0 ) claims there is no association between age group and smiling; they are independent.
  • The alternative hypothesis ( H_a ) suggests there is an association; age group does influence smiling.
We set a significance level ( alpha = 0.05 ) and use the Chi-Square test to calculate a p-value. This p-value tells us the probability that the data could occur under the null hypothesis.
  • If the p-value is less than 0.05, we reject the null hypothesis, suggesting an association does exist.
  • If the p-value is greater, we do not reject the null hypothesis, implying no significant association is confirmed.
This testing helps firmly conclude whether age truly impacts smiling probability, allowing us to move beyond guesswork to evidence-driven insights.

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Most popular questions from this chapter

According to a 2017 report, \(64 \%\) of college graduate in Illinois had student loans. Suppose a random sample of 80 college graduates in Illinois is selected and 48 of them had student loans. (Source: Lendedu.com) a. What is the observed frequency of college graduates in the sample who had student loans? b. What is the observed proportion of college graduates in the sample who had student loans? c. What is the expected number of college graduates in the sample to have student loans if \(64 \%\) is the correct rate? Do not round off.

Here are the conviction rates with the "stand your ground" data mentioned in the previous exercise. "White shooter on nonwhite" means that a white assailant shot a minority victim. $$ \begin{array}{lr} & \text { Conviction Rate } \\ \hline \text { White shooter on white } & 35 / 97=36.1 \% \\ \text { White shooter on nonwhite } & 3 / 21=14.3 \% \\ \text { Nonwhite shooter on white } & 8 / 22=36.4 \% \\ \text { Nonwhite shooter on nonwhite } & 11 / 45=24.4 \% \end{array} $$ a. Which has the higher conviction rate: white shooter on nonwhite or nonwhite shooter on white? b. Create a two-way table using White Shooter on Nonwhite and NonWhite Shooter on White across the top and Convicted and Not Con- victed on the side. c. Test the hypothesis that race and conviction rate (for these two groups) are independent at the \(0.05\) level. d. Because some of the expected counts are pretty low, try a Fisher's Exact Test with the data, reporting the p-value (two-tailed) and the conclusion.

Fill in the blank by choosing one of the options given: Chi-square goodness- of-fit tests are applicable if the data consist of (one categorical variable, two categorical variables, one numerical variable, or two numerical variables).

The following table shows the average number of vehicles sold in the United States monthly (in millions) for the years 2001 through 2018 . Data on all monthly vehicle sales for these years were obtained and the average number per month was calculated. Would it be appropriate to do a chi-square analysis of this data set to see if vehicle sales are distributed equally among the months of the year? If so, do the analysis. If not, explain why it would be inappropriate to do so. (Source: www.fred.stlouisfed.org) $$ \begin{array}{|l|l|} \hline \text { Month } & \text { Avg Sales per Month (in millions) } \\ \hline \text { Jan } & 15.7 \\ \hline \text { Feb } & 15.7 \\ \hline \text { Mar } & 15.8 \\ \hline \text { Apr } & 15.8 \\ \hline \text { May } & 15.8 \\ \hline \text { June } & 15.7 \\ \hline \text { July } & 16.1 \\ \hline \text { Aug } & 16.1 \\ \hline \text { Sept } & 15.8 \\ \hline \text { Oct } & 15.9 \\ \hline \text { Nov } & 15.9 \\ \hline \text { Dec } & 15.9 \\ \hline \end{array} $$

a. In Chapter 8 , you learned some tests of proportions. Are tests of proportions used for categorical or numerical data? b. In this chapter, you are learning to use chi-square tests. Do these tests apply to categorical or numerical data?

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