# Find the expected count and the contribution to the chi-square statistic for the (Control, Disagree) cell in the two-way table below. begin{array}{|c|c|c|}hline&text{Strongly Agree}&text{Agree}&text{Neutral}&text{Disagree}&text{Strongly Disagree}hlinetext{Control} &38&47&2&12&11 hline text{Treatment}&60&45&9&4&2 hline end{array} Round your answer for the excepted count to one decimal place, and your answer for the contribution to the chi-square statistic to three decimal places. Expected count ? Contribution to the chi-square statistic ?

Question
Two-way tables
Find the expected count and the contribution to the chi-square statistic for the (Control, Disagree) cell in the two-way table below. $$\begin{array}{|c|c|c|}\hline&\text{Strongly Agree}&\text{Agree}&\text{Neutral}&\text{Disagree}&\text{Strongly Disagree}\\\hline\text{Control} &38&47&2&12&11\\ \hline \text{Treatment}&60&45&9&4&2 \\ \hline \end{array}\\$$
Round your answer for the excepted count to one decimal place, and your answer for the contribution to the chi-square statistic to three decimal places.
Expected count ?
Contribution to the chi-square statistic ?

2021-01-03
Step 1
Introduction:
The row and column totals are obtained as follows, for the given observed frequencies: $$\begin{array}{|c|c|c|}\hline&\text{Strongly Agree}&\text{Agree}&\text{Neutral}&\text{Disagree}&\text{Strongly Disagree}&\text{Total}\\\hline\text{Control} &38&47&2&12&11&110\\ \hline \text{Treatment}&60&45&9&4&2&120 \\ \hline \text{Total}&98&92&11&16&13&230 \\\hline \end{array}\\$$
Step 2
Calculation:
Consider the observation in the ith row and jth column of the table, that is, in the cell (i, j). Then, the expected frequency of the cell (i, j), if it is assumed that the two categories are independent, is $$\frac{[\text{(total of row i)} \cdot \text{(total of column j)}]}{\text{(grand total)}}$$
$$=\frac{(110 \cdot 16)}{230}$$
$$\approx 7.7$$
Thus, the expected count for the cell (Control, Disagree) is 7.7.
Now, the formula for the chi square test statistic is:
$$x^2=\sum_i\sum_j \frac{(O_{ij}-E_{ij})^2}{E_{ij}}$$
where
$$O_{ij} \text{is the observed frequency in cell} (i,j)$$
$$E_{ij} \text{is the expected frequency in cell} (i,j)$$
$$\frac{(O_{ij}-E_{ij})^2}{E_{ij}}$$
For the cell (Control, Disagree), that is, for the observation in row 1 and column 4, the component of the test statistic will be:
$$\frac{(O_{ij}-E_{ij})^2}{E_{ij}}$$
$$=\frac{(12-7.7)^2}{7.7}$$
$$\approx 2.416$$
Thus, the contribution to the chi-square statistic for the cell (Control, Disagree) is 2.416.

### Relevant Questions

Find the expected count and the contribution to the chi-square statistic for the (Control, Disagree) cell in the two-way table below.
$$\begin{array}{|c|c|c|}\hline&\text{Strongly Agree}&\text{Agree}&\text{Neutral}&\text{Disagree}&\text{Strongly Disagree}\\\hline\text{Control} &38&47&2&12&11\\ \hline \text{Treatment}&60&45&9&4&2 \\ \hline \end{array}\\$$
Round your answer for the excepted count to one decimal place, and your answer for the contribution to the chi-square statistic to three decimal places.
Expected count ?
Contribution to the chi-square statistic ?
Find the expected count and the contribution to the chi-square statistic for the (Group 1, No) cell in the two-way table below. $$\begin{array}{|c|c|c|}\hline&\text{Yes}&\text{No}&\text{Total}\\\hline\text{Group 1} &56 & 42 & 98\\ \hline \ \text{Group 2}&135&67&202 \\ \hline \text{Group 3}&66&23&89 \\ \hline \text{Total}&257&132&389 \\ \hline \end{array}$$
Round your answer for the excepted count to one decimal place, and your answer for the contribution to the chi-square statistic to three decimal places.
Expected count =?
contribution to the chi-square statistic = ?
Find the expected count and the contribution to the chi-square statistic for the (Group 1, Yes) cell in the two-way table below.
$$\begin{array}{|c|c|c|}\hline&\text{Yes}&\text{No}&\text{Total}\\\hline\text{Group 1} &710 & 277 & 987\\ \hline\text{Group 2}& 1175 & 323&1498\\\hline \ \text{Total}&1885&600&2485 \\ \hline \end{array}$$
Round your answer for the excepted count to one decimal place, and your answer for the contribution to the chi-square statistic to three decimal places.
Expected count=?
contribution to the chi-square statistic=?
How are the smoking habits of students related to their parents' smoking? Here is a two-way table from a survey of student s in eight Arizona high schools:
$$\begin{array}{c|c}&\text{Student smokes}&\text{Student does not smoke}&\text{Total}\\\hline\text{Both parents smoke}&400&1380&400+1380=1780\\\hline\text{One parent smokes}&416&1823&416+1823=2239\\\hline\text{Neither parent smokes}&188&1168&188+1168=1356\\\hline\text{Total}&400+416+188=1004&1380+1823+1168=4371&1004+4371=5375\end{array}$$
(a) Write the null and alternative hypotheses for the question of interest.
(b) Find the expected cell counts. Write a sentence that explains in simple language what "expected counts" are.
(c) Find the chi-square statistic, its degrees of freedom, and the P-value.
1950 randomly selected adults were asked if they think they are financially better off than their parents. The following table gives the two-way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as, or worse off than their parents
$$\begin{array}{|c|c|c|}\hline &\text{Less Than High School}&\text{High School}&\text{More Than High School}\\\hline \text{Better off} &140&440&430\\ \hline \text{Same as}&60&230&110\\ \hline \text{Worse off}&180&280&80\\ \hline\end{array}\\$$
Suppose one adult is selected at random from these 1950 adults. Find the following probablity.
$$P(\text{more than high school or worse off})=?$$
The following table gives a two-way classification of all basketball players at a state university who began their college careers between 2004 and 2008, based on gender and whether or not they graduated.
$$\begin{array}{|c|c|c|}\hline &\text{Graduated}&\text{Did not Graduate}\\\hline \text{Male} &129&51\\ \hline \text{Female}&134&36 \\ \hline \end{array}\\$$
If one of these players is selected at random, find the following probability.
$$P(\text{graduated or male})=$$ Enter your answer in accordance to the question statement
The accompanying two-way table was constructed using data in the article “Television Viewing and Physical Fitness in Adults” (Research Quarterly for Exercise and Sport, 1990: 315–320). The author hoped to determine whether time spent watching television is associated with cardiovascular fitness. Subjects were asked about their television-viewing habits and were classified as physically fit if they scored in the excellent or very good category on a step test. We include MINITAB output from a chi-squared analysis. The four TV groups corresponded to different amounts of time per day spent watching TV (0, 1–2, 3–4, or 5 or more hours). The 168 individuals represented in the first column were those judged physically fit. Expected counts appear below observed counts, and MINITAB displays the contribution to $$\displaystyle{x}^{{{2}}}$$ from each cell.
State and test the appropriate hypotheses using $$\displaystyle\alpha={0.05}$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{r}{r}{a}{y}\right\rbrace}{\left\lbrace{\left|{c}\right|}{c}{\mid}\right\rbrace}{h}{l}\in{e}&{a}\mp,\ {1}&{a}\mp,\ {2}&{a}\mp,\ {T}{o}{t}{a}{l}\backslash{h}{l}\in{e}{1}&{a}\mp,\ {35}&{a}\mp,\ {147}&{a}\mp,\ {182}\backslash{h}{l}\in{e}&{a}\mp,\ {25.48}&{a}\mp,\ {156.52}&{a}\mp,\backslash{h}{l}\in{e}{2}&{a}\mp,\ {101}&{a}\mp,\ {629}&{a}\mp,\ {730}\backslash{h}{l}\in{e}&{a}\mp,\ {102.20}&{a}\mp,\ {627.80}&{a}\mp,\backslash{h}{l}\in{e}{3}&{a}\mp,\ {28}&{a}\mp,\ {222}&{a}\mp,\ {250}\backslash{h}{l}\in{e}&{a}\mp,\ {35.00}&{a}\mp,\ {215.00}&{a}\mp,\backslash{h}{l}\in{e}{4}&{a}\mp,\ {4}&{a}\mp,\ {34}&{a}\mp,\ {38}\backslash{h}{l}\in{e}&{a}\mp,\ {5.32}&{a}\mp,\ {32.68}&{a}\mp,\backslash{h}{l}\in{e}{T}{o}{t}{a}{l}&{a}\mp,\ {168}&{a}\mp,\ {1032}&{a}\mp,\ {1200}\backslash{h}{l}\in{e}$$
$$\displaystyle{C}{h}{i}{s}{q}={a}\mp,\ {3.557}\ +\ {0.579}\ +\ {a}\mp,\ {0.014}\ +\ {0.002}\ +\ {a}\mp,\ {1.400}\ +\ {0.228}\ +\ {a}\mp,\ {0.328}\ +\ {0.053}={6.161}$$
$$\displaystyle{d}{f}={3}$$
The following is a two-way table showing preferences for an award (A, B, C) by gender for the students sampled in survey. Test whether the data indicate there is some association between gender and preferred award.
$$\begin{array}{|c|c|c|}\hline &\text{A}&\text{B}&\text{C}&\text{Total}\\\hline \text{Female} &20&76&73&169\\ \hline \text{Male}&11&73&109&193 \\ \hline \text{Total}&31&149&182&360 \\ \hline \end{array}\\$$
Chi-square statistic=?
p-value=?
Conclusion: (reject or do not reject $$H_0$$)
Does the test indicate an association between gender and preferred award? (yes/no)
$$\begin{array}{|c|c|c|}\hline &\text{Baseball}&\text{Basketball}&\text{Football}&\text{Hockey}&\text{Other}&\text{Total}\\\hline \text{Boys} &18&14&20&6&2&60\\ \hline \text{Girls}&14&16&13&5&12&60\\ \hline \text{Total}&32&30&33&11&14&120\\ \hline \end{array}\\$$
$$\begin{array}{|c|c|c|}\hline&\text{morning}&\text{afternoon}&\text{total}\\\hline\text{grade 6} &15 & 8 & 23\\ \hline\text{grade 8}& 18 & 21&39\\\hline\text{grade 10}& 12 & 26&38\\ \hline \text{total}&45&55&100 \\ \hline \end{array}$$