<p><div><div><div><div data-width="100" data-show-faces="false" data-send="true" data-layout="button_count" data-height="21" data-href="https://www.r-bloggers.com/frequency-and-chi-square-test-for-independence-exercises/"></div></div><div><a data-url="https://www.r-bloggers.com/frequency-and-chi-square-test-for-independence-exercises/" data-counturl="https://www.r-bloggers.com/frequency-and-chi-square-test-for-independence-exercises/" data-count="horizontal" data-text="Frequency and chi-square test for independence Exercises" class="twitter-share-button" data-via="rbloggers" href="https://twitter.com/share"></a></div><div></div></div><div></div></div>
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(This article was first published on <strong><a href="http://r-exercises.com/2016/11/30/frequency-and-chi-square-test-for-independance/"> R-exercises</a></strong>, and kindly contributed to <a rel="nofollow" href="https://www.r-bloggers.com/">R-bloggers)</a> </div>
<p><a target="_blank" rel="nofollow" href="http://r-exercises.com/wp-content/uploads/2016/11/Occasional-Table-Group.jpg?x61433"><img sizes="(max-width: 225px) 100vw, 225px" alt="occasional-table-group" class="wp-image-3037" src="https://i0.wp.com/r-exercises.com/wp-content/uploads/2016/11/Occasional-Table-Group-300x168.jpg?resize=225%2C126" data-recalc-dims="1"></a>In this exercise, we cover some basics on frequency tables. We also briefly look at chi-square test for independence to find relationships of two variables. Before proceeding, it might be helpful to look over the help pages for the <code>table</code>, <code>summary</code>, and <code>margin.table</code> functions.</p>
<p>Answers to the exercises are available <a target="_blank" rel="nofollow" href="http://r-exercises.com/2016/11/30/frequency-and-chi-square-test-for-independance-2/">here</a>.</p>
<p>If you obtained a different (correct) answer than those listed on the solutions page, please feel free to post your answer as a comment on that page.</p>
<p><strong>Exercise 1</strong><br>
use the <code>attach()</code> command to load the trees dataset in R</p>
<p><strong>Exercise 2</strong><br>
Use the <code>table()</code> command with the arguments: trees$Height and trees$Volume. This will generate a two-way frequency table. Store this in variable mytable.</p>
<p><strong>Exercise 3</strong><br>
If you are familiar with excel pivot tables, then you will know this function. Use the margin.table( ) function to get the Height frequencies summed over Volume</p>
<p><strong>Exercise 4</strong><br>
Use the <code>margin.table( )</code> function to get the Volume frequencies summed over Height.</p>
<p><strong>Exercise 5</strong><br>
Now use the <code>table()</code> function again but using all the features of the trees dataset, that includes girth, height and volume. This will print out a multidimensional 3 way frequency table.</p>
<p><strong>Exercise 6</strong><br>
Suppose you have a variable ‘a’ that stores a second sample of heights of trees.<br>
<code><br>
a=c(70, 65, 63, 72, 80, 83, 66, 75, 80, 75, 79, 76, 76, 69, 75, 74, 85, 8, 71, 63, 78, 80, 74, 72, 77, 81, 82, 80, 86, 80, 87)<br>
</code><br>
Use the cbind() to add the a column to your trees dataset. Store the results back into trees.</p>
<p><strong>Exercise 7</strong><br>
Now create a 2 way frequency table between Height and a as the arguments. Store this table in mytable_2.</p>
<p><strong>Exercise 8</strong></p>
<p>Use the margin.table() function again from Q3 and get Height frequencies summer over a. What differences do you observe from the results of Q3.</p>
<p><strong>Exercise 9</strong><br>
Chi Square test for independance:<br>
a)Print the results of the summary() function on mytable. Note the Chi Square test for independance results and P value<br>
b)Print the results of the summary() function on mytable_2. Note the Chi Square test for independance results and P value.</p>
<p><strong>Exercise 10</strong><br>
What did the chi square test for independance help you to see?</p>
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