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 ?

2020-11-27
Step 1
a) The expected frequencies are calculated by using the following formula:
$$\text{Expected frequency}=\frac{\text{Row total} \cdot \text{Column total}}{\text{Overall total}}$$
Step 2
Then, expected frequencies are calculated as follows: $$\begin{array}{|c|c|c|}\hline&\text{Strongly Agree}&\text{Agree}&\text{Neutral}&\text{Disagree}&\text{Strongly Disagree}\\\hline\text{Control} &46.9&44.0&5.3&7.7&6.2\\ \hline \text{Treatment}&51.1&48.0&5.7&8.3&6.8 \\ \hline \end{array}\\$$ The formula for finding the test statistic is obtained as follows: $$x^2=\sum \frac{(O_i-E_i)^2}{E_i}$$ The chi-square statistic is calculated as follows:
$$\begin{array}{|c|c|c|}\hline\text{Observed count (O)}&\text{Expected count (E)}&\frac{(O-E)^2}{E}\\\hline 38 &46.9&1.678\\ \hline 47&44.0&0.205 \\ \hline 2&5.3&2.021 \\ \hline 12&7.7&2.470 \\ \hline 11&6.2&3.679 \\ \hline 60&51.1&1.539 \\ \hline 45&48.0&0.188 \\ \hline 9&5.7&1.853 \\ \hline 4&8.3&2.264 \\ \hline 2&6.8&3.372 \\ \hline \text{Total}: 230&&x^2=19.269 \\ \hline \end{array}\\$$ Thus , the chi-square statistic is 19.269

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.
(d) What is your conclusion about significance?
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.
Round your answer to three decimal places.
$$P(\text{more than high school or worse off})=?$$
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.
Round your answer to four decimal places.
$$P(\text{graduated or male})=$$ 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)
$$\begin{array}{|c|c|c|}\hline &\text{Baseball}&\text{Basketball}&\text{Football}&\text{Hockey}&\text{Other}&\text{Total}\\\hline \text{Boys} &18&14&20&6&2&60\\ \hline \text{Girls}&14&16&13&5&12&60\\ \hline \text{Total}&32&30&33&11&14&120\\ \hline \end{array}\\$$
$$\begin{array}{|c|c|c|}\hline&\text{morning}&\text{afternoon}&\text{total}\\\hline\text{grade 6} &15 & 8 & 23\\ \hline\text{grade 8}& 18 & 21&39\\\hline\text{grade 10}& 12 & 26&38\\ \hline \text{total}&45&55&100 \\ \hline \end{array}$$