Comparing texting while driving behaviors between differrent bike helmet usage groups
In J Lab 4 you investigated the relationship between texting while driving and wearing a helmet while biking. You explored the difference in the proportions of students who texted while driving every day in the last 30 days between students who never wore a helmet and students who reported wearing a helmet at least some of the time in the last 12 months. You estimated the difference in proportions using a 95% confidence interval, and you tested the hypothesis that the difference in proportions was 0.
In this lab you will continue to investigate between texting while driving and wearing a helmet while biking. However, you will take a different approach to the analysis this time. Instead of a binary response (texted every day vs. did not), you will consider a related response variable with three levels (never texted, texted 1-19 days, and texted 20 or more days). Also, instead of comparing the values of the response across two groups (wore a helmet at least some of the time vs. never wore a helmet), you will compare the values of the response across three groups (always wore a helmet, sometimes wore a helmet but not always, and never wore a helmet).
Filtering data and recoding variables
Start by filtering the data down to the relevant observations by excluding the following cases:
- Students who did not drive in the last 30 days
- Students who did not respond to the texting question
- Students who did not ride a bike in the last 12 months
- Students who did not respond to the helmet question
You may refer to the instructions in J Lab 2 or J Lab 4 if you need a refresher on how to filter data in Jamovi.
Next, you will recode the helmet_12m and text_while_driving_30d variables. Create two new variables, helmet_ind and text_ind that each have three levels. Start with the texting variable. As in previous labs, you can recode the variable by clicking on it’s name in the spreadsheet in Jamovi, then clicking on the Transform button in the Data tab. Set the name of the transformed variable to text_ind. Then, select “Create New Transform”.
In the Transform interface that pops up, click + recode condition three times and add the following recode conditions:
if $source == "0" use "0"if $source == "30" use "20-30"if $source == "20-29" use "20-30"else use "1-19"
Figure 1 shows the last three of these conditions set up correctly (the first one is hidden in this view).
Once you have recoded the text_while_driving_30d variable, compare the values of the new text_ind variable to the original text_while_driving_30d variable to make sure it was done correctly.
Next recode the helmet_12m variable. The new helmet_ind variable should have the value “always” for students that reported always wearing a helmet, “never” for students that reported never wearing a helmet, and “sometimes” for everyone else. Compare the values of the new variable to the original helmet_12m variable to make sure the recoding was done correctly.
Comparing texting distributions between helmet groups
Create a contingency table that shows the counts of students in each of the three texting while driving groups for each of the three bike helmet groups. Add conditional proportions to the table to show the proportions within each helmet group. Also create a stacked bar plot that shows the distribution of the text_ind variable for each helmet_ind group. You learned how to create these numerical and visual summaries in J Lab 2, and you created similar tables and plots in J Lab 4. You should refer to those labs if you need a refresher. Your results should be similar to the following:
Contingency Tables
──────────────────────────────────────────────────────────────────────────────────
helmet_ind 0 1-19 20-30 Total
──────────────────────────────────────────────────────────────────────────────────
always Observed 172 38 19 229
% within row 75.10917 16.59389 8.29694 100.00000
never Observed 2566 1178 643 4387
% within row 58.49100 26.85206 14.65694 100.00000
sometimes Observed 521 191 67 779
% within row 66.88062 24.51861 8.60077 100.00000
Total Observed 3259 1407 729 5395
% within row 60.40778 26.07970 13.51251 100.00000
──────────────────────────────────────────────────────────────────────────────────
In the contingency table and the plot in Figure 1, it would be better to order the groups using increasing helmet usage and increasing texting frequency. Unfortunately, Jamovi does not currently allow you to reorder the levels of a transformed variable in an efficient way. It is possible to copy the data from the new helmet_ind or text_ind variable into a new variable and then to reorder the levels in the desired order. However, we will not do that here.
Based on the plot and the contingency table, does it appear that there is an association between texting while driving and helmet usage? How does the information in the table and the plot compare to the ones you created in J Lab 4?
Hypothesis tests
Next you will conduct tests of the following hypotheses:
- \(H_0:\) There is no difference in rates of texting while driving between students with different bike helmet usage
- \(H_A:\) There is a difference in rates of texting while driving between students with different bike helmet usage
You will conduct both a hypothesis test using a mathematical model (a Chi-squared test) and a hypothesis test using random permutation.
Before you test these hypotheses using a mathematical model, you must check the technical conditions for the test. In particular, you need to check that there are at least 5 expected counts in each cell of the contingency table (assuming that the null hypothesis is true). Since the smallest helmet group is those who “always” wear a helmet, and the smallest texting group is those who texted “20-30 days”, you can focus on the cell of the table that corresponds to the intersection of these two groups. Overall, about 13.5% of students texted while driving 20-30 days (Figure 1). If \(H_0\) is true, we expect the percentage of students who texted while driving 20-30 days to be the same (13.5%) for all three helmet groups. Thus, since there are 229 students that always wore a helmet (Figure 1), we expect \(0.135\times 229= 30.9\) of these students to text while driving 20-30 days. This is greater than 5, so the technical condition is met.
You can also have Jamovi calculate expected counts for the entire contingency table (assuming that the null hypothesis is true). These can be added in the same interface that allowed you to add row percentages to the table. Simply click on the checkbox next to “Expected counts” to add them to the table. Does the expected count for students who always wore a helmet and texted while driving for 20-30 days match the expected count of 30.9 you calculated above?
Once you have constructed a contingency table in Jamovi and checked the technical conditions for the test, it is easy to conduct the model-based Chi-squared test. In fact, you may have noticed the results of the test have already appeared in Jamovi, since they are included in the Contingency Tables analysis results by default. If you do not see the results of the Chi-squared test, make sure the box next to \(X^2\) is checked in the Statistics section of the Contingency Tables interface. Your results should be similar to the following:
χ² Tests
──────────────────────────────────────
Value df p
──────────────────────────────────────
χ² 48.66734 4 < .0000001
N 5395
──────────────────────────────────────
The table shows the value of the Chi-squared statistic, \(X^2=48.7\). It also shows the degrees of freedom for the Chi-squared distribution that approximates the null-distribution. Recall that the degrees of freedom are calculated as \((r-1)\times(c-1)\), where \(r\) is the number of rows in the contingency table and \(c\) is the number of columns (not including marginal totals). Since the table has 3 rows and 3 tables, the degrees of freedom are \(2\times2=4\).
The p-value is very small (<0.001),so we reject the null hypothesis at the \(\alpha=0.05\) significance level. There is convincing evidence that there is a difference in the rates of texting while driving between students with different levels of bike helmet usage.
You can also conduct a simulation-based hypothesis test using the Randomize module in Jamovi. The module creates simulated samples by randomly permuting (shuffling) the values of the response variable, text_ind. Since the values of the response are randomly assigned to the cases, the distribution of the text_ind variable will not tend to differ between helmet usage groups. Thus, the simulated samples are examples of samples from a population in which the null hypothesis is true. We can create a null distribution for the \(X^2\) statistic by calculating the values of the statistic for the simulated samples. This is similar to the approach you took in J Lab 4 to perform a hypothesis test for the difference in proportions.
Select Contingency Table - Hypothesis Test from the Randomize module. Set up the analysis similar to how you built the contingency table above by dragging helmet_ind into the “Rows” box and text_ind into the “Columns” box. Under Simulations, specify 1000 simulations and select “Histogram” for the plot. You may also seed the random number generator with 8675309 if you want your results to match the ones in this lab.
Figure 2 shows the interface for the Contingency Table - Hypothesis Test analysis with the correct options selected, along with the results of the analysis.
From the histogram in Figure 2, you can see that there were no simulated samples with \(X^2\) values as large as the observed value of \(X^2=48.7\). Thus the p-value is approximately \(0/1000 = 0\). The results of the simulation-based test match the results of the model-based Chi-squared test.
Student height
Next you will learn how to use Jamovi to analyze a single mean. We will use the height variable in the yrbss data, which gives the height of the student in meters.
Another study claims that the average height of a 15 year old in the United States is 67 inches (1.70 meters). You will evaluate this claim using the yrbss data by conducting a hypothesis test and calculating a 95% confidence interval for the mean height of a 15 year old in the United States.
Hypothesis test for a single mean
You will test the following hypotheses:
- \(H_0: \mu = 1.70\)
- \(H_A: \mu \neq 1.70\)
where \(\mu\) is the average height of a 15 year old in the United States in meters.
Start by filtering the data to retain only 15 year olds. Deactivate the filters that you created for the first part of the lab. Then create a new filter that only retains the 15 year olds. Finally, create another new filter that removes any cases for which the height variable is missing. Activate both new filters and check the data to make sure they are working the way you would expect.
Next you should explore the distributions of heights for the 15 year olds. Create a histogram of the heights, and a table of descriptive statistics. Include the following in the table: the number of cases, the mean, the median, the standard deviation, the minimum, the maximum, Q1, and Q2. You learned how to create histograms and tables of descriptive statistics for a numeric variable in J Lab 2. Your results should be similar to the following:
DESCRIPTIVES
Descriptives
───────────────────────────────────
height
───────────────────────────────────
N 2870
Mean 1.678927
Median 1.680000
Standard deviation 0.1004759
Minimum 1.270000
Maximum 2.110000
25th percentile 1.600000
75th percentile 1.750000
───────────────────────────────────
Your hypothesis test will be based on a mathematical model–a T-distribution. However, before proceeding with the hypothesis test, you must check to see that the conditions for the test are met. You can use the table and the histogram you just created to check the conditions. Since there are at least 30 observations (see the table) and no extreme outliers (see the histogram), the conditions for the test are met.
Before you can conduct the hypothesis test you need to compute the value of the T-statistic, which is calculated as follows:
\[T = \frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}\]
where \(\mu_0\) is the value of the mean under the null hypothesis, \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. Most of this information is included in the table you created. Thus,
\[T = \frac{1.679-1.70}{\frac{0.1005}{\sqrt{2870}}}=-11.19\]
Next you have to calculate the degrees of freedom for the T-distribution that you will use as a model of the null distribution. The degrees of freedom for the T-distribution is given by \(df=n-1\). Thus, the degrees of freedom for our test are \(df=2870-1=2869\).
You can conduct the hypothesis test using the Model-Based Inference analysis in the Randomize module in Jamovi. In the interface, select Use T distribution and enter 2869 for the degrees of freedom. Then check the box to calculate the p-value, enter the observed value of \(T\), and select “Both” tails. Figure 3 shows the interface after it has been set up correctly for the hypothesis test along with the results.
The p-value is very small (<0.001), so we reject the null hypothesis at the \(\alpha=0.05\) significance level. There is convincing evidence that the average height of a 15 year old in the United States is different from 1.70 meters.
Confidence interval for a mean
Next you will calculate a 95% confidence interval for the average height of a 15 year old in the United States. You will calculate a 95% confidence interval in two ways: using a mathematical model (a T-distribution) and using randomization (bootstrapping).
You have already checked the technical conditions for using a T-distribution. You can use the Model-Based Inference analysis in the Randomize module in Jamovi to calculate the appropriate multiplier for the 95% confidence interval. This time, instead of checking the box for calculating a p-value, check the box for calculating a CI multiplier. Also, specify that the confidence level is 95%. Figure 4 shows the interface after it has been set up correctly for the confidence interval along with the results.
Since the multiplier is \(T^{\ast}_{2869}=1.961\), the 95% confidence interval is calculated as
\[\bar{x} \pm T^{\ast}_{df}\times \frac{s}{\sqrt{n}} = 1.679 \pm 1.961\times \frac{0.1005}{\sqrt{2870}}\]
This results in a 95% confidence interval of \((1.675, 1.683)\) for the mean height of 15 year olds in the United States. Note that this confidence interval does not include the value of 1.70 suggested by the other study, which is consistent with the results of the hypothesis test you just conducted.
Next, you will calculate a 95% confidence interval using a bootstrap distribution of means. You will use the Single Mean - Confidence Interval analysis in the Randomize module to calculate a 95% bootstrap percentile confidence interval. Set the confidence level to 95%, select “bootstrap percentile” for the type of interval, specify that you want to do 1000 bootstraps, with a histogram as the plot type. You may also seed the random number generator with 8675309 if you want your results to match the ones in this lab. Figure 5 shows the interface after it has been set up correctly for the confidence interval along with the results.
The histogram shows the mean heights from 1,000 bootstrap samples. A bootstrap sample is randomly selected from the original sample with replacement. The variability of the bootstrap distribution approximates the variability in the sampling distribution for the mean. The 95% confidence interval is calculated as the 2.5th and 97.5th percentiles of the bootstrap distribution. In Figure 5 you can see that the 95% confidence interval is approximately \((1.675, 1.682)\), which is very close to the 95% confidence interval you calculated using a mathematical model. If your results do not show enough significant figures, you can increase the number of significant figures displayed in the results by clicking on the icon that looks like three vertically stacked dots at the top right of the Jamovi window. Under Results, set the Number format to “5 sf”.
