Market researchers wonder if people in different countries have different preferred sports to watch during the winter Olympics. They take a random sample of people from each country and survey them about their favorite sport to watch. Here are the responses and partial results of a chi-square test (expected counts appear below observed counts):

Chi-square test: Country vs. favorite winter Olympics sport
\begin{tabular}{lrrr}
& China & India & Total \
\hline Bobsleigh & 8 & 10 & 18 \
& 9 & 9 & \
Curling & 28 & 61 & 89 \
& 44.5 & 44.5 & \
Hockey & 39 & 29 & 68 \
& 34 & 34 & \
Speed skating & 75 & 50 & 125 \
& 62.5 & 62.5 & \
Total & 150 & 150 & 300
\end{tabular}

They want to use these results to carry out a $\chi^{2}$ test for homogeneity. Assume that all conditions for inference were met.

What are the values of the test statistic and P-value for their test?

Choose 1 answer:
(A)
$$
\begin{array}{l}
\chi^{2}=6.309 \
0.05

Q: Market researchers wonder if people in different countries have different preferred sports to watch during the winter Olympics. They take a random sample of people from each country and survey them about their favorite sport to watch. Here are the responses and partial results of a chi-square test (expected counts appear below observed counts): Chi-square test: Country vs. favorite winter Olympics sport \begin{tabular}{lrrr} & China & India & Total \\ \hline Bobsleigh & 8 & 10 & 18 \\ & 9 & 9 & \\ Curling & 28 & 61 & 89 \\ & 44.5 & 44.5 & \\ Hockey & 39 & 29 & 68 \\ & 34 & 34 & \\ Speed skating & 75 & 50 & 125 \\ & 62.5 & 62.5 & \\ Total & 150 & 150 & 300 \end{tabular} They want to use these results to carry out a $\chi^{2}$ test for homogeneity. Assume that all conditions for inference were met. What are the values of the test statistic and P-value for their test? Choose 1 answer: (A) \[ \begin{array}{l} \chi^{2}=6.309 \\ 0.05<\text { P-value }<0.10 \end{array} \] (B) \[ \begin{array}{l} \chi^{2}=6.309 \\ 0.15<\text { P-value }<0.20 \end{array} \] (C) \[ \begin{array}{l} \chi^{2}=18.928 \\ \text { P-value }<0.0005 \end{array} \] () \[ \begin{array}{l} \chi^{2}=18.928 \\ 0.0005<\text { P-value }<0.001 \end{array} \]

∻Solution by Steps∻ ‖ step 1 ⋮ To calculate the chi-square test statistic, we use the formula $X^2 = \sum \frac{(O - E)^2}{E}$, where $O$ represents the observed frequency and $E$ represents the expected frequency. ‖ ‖ step 2 ⋮ For Bobsleigh, the calculation is $\frac{(8-9)^2}{9} + \frac{(10-9)^2}{9}$. ‖ ‖ step 3 ⋮ For Curling, the calculation is $\frac{(28-44.5)^2}{44.5} + \frac{(61-44.5)^2}{44.5}$. ‖ ‖ step 4 ⋮ For Hockey, the calculation is $\frac{(39-34)^2}{34} + \frac{(29-34)^2}{34}$. ‖ ‖ step 5 ⋮ For Speed skating, the calculation is $\frac{(75-62.5)^2}{62.5} + \frac{(50-62.5)^2}{62.5}$. ‖ ‖ step 6 ⋮ Summing the results from steps 2 to 5 gives us the total chi-square test statistic. ‖ ‖ step 7 ⋮ To find the P-value, we compare the calculated test statistic to the chi-square distribution with the appropriate degrees of freedom. The degrees of freedom are calculated as $(r - 1)(c - 1)$, where $r$ is the number of rows minus 1 and $c$ is the number of columns minus 1. ‖ ‖ step 8 ⋮ In this case, the degrees of freedom are $(4 - 1)(2 - 1) = 3$. ‖ ‖ step 9 ⋮ Using a chi-square distribution table or calculator with 3 degrees of freedom, we find the range in which our test statistic falls to determine the P-value. ‖ ***C*** ∻Key Concept∻ ⚹Chi-square test for homogeneity⚹ ∻Explanation∻ ⚹The chi-square test for homogeneity is used to determine if there are significant differences in the distribution of categorical variables across different populations. The test statistic is calculated by summing the squared differences between observed and expected frequencies, divided by the expected frequencies. The P-value is then found by comparing the test statistic to the chi-square distribution with the appropriate degrees of freedom.◊What is the formula to calculate the test statistic for a chi-square test for homogeneity?⍭ Generate me a similar question◊⚹

Question

Market researchers wonder if people in different countries have different preferred sports to watch during the winter Olympics. They take a random sample of people from each country and survey them about their favorite sport to watch. Here are the responses and partial results of a chi-square test (expected counts appear below observed counts):

Chi-square test: Country vs. favorite winter Olympics sport
\begin{tabular}{lrrr} 
& China & India & Total \
\hline Bobsleigh & 8 & 10 & 18 \
& 9 & 9 & \
Curling & 28 & 61 & 89 \
& 44.5 & 44.5 & \
Hockey & 39 & 29 & 68 \
& 34 & 34 & \
Speed skating & 75 & 50 & 125 \
& 62.5 & 62.5 & \
Total & 150 & 150 & 300
\end{tabular}

They want to use these results to carry out a $\chi^{2}$ test for homogeneity. Assume that all conditions for inference were met.

What are the values of the test statistic and P-value for their test?

Choose 1 answer:
(A)
$$
\begin{array}{l}
\chi^{2}=6.309 \
0.05<\text { P-value }<0.10
\end{array}
$$
(B)
$$
\begin{array}{l}
\chi^{2}=6.309 \
0.15<\text { P-value }<0.20
\end{array}
$$
(C)
$$
\begin{array}{l}
\chi^{2}=18.928 \
\text { P-value }<0.0005
\end{array}
$$
()
$$
\begin{array}{l}
\chi^{2}=18.928 \
0.0005<\text { P-value }<0.001
\end{array}
$$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To calculate the chi-square test statistic, we use the formula $X^2 = \sum \frac{(O - E)^2}{E}$, where $O$ represents the observed frequency and $E$ represents the expected frequency. ‖ ‖ step 2 ⋮ For Bobsleigh, the calculation is $\frac{(8-9)^2}{9} + \frac{(10-9)^2}{9}$. ‖ ‖ step 3 ⋮ For Curling, the calculation is $\frac{(28-44.5)^2}{44.5} + \frac{(61-44.5)^2}{44.5}$. ‖ ‖ step 4 ⋮ For Hockey, the calculation is $\frac{(39-34)^2}{34} + \frac{(29-34)^2}{34}$. ‖ ‖ step 5 ⋮ For Speed skating, the calculation is $\frac{(75-62.5)^2}{62.5} + \frac{(50-62.5)^2}{62.5}$. ‖ ‖ step 6 ⋮ Summing the results from steps 2 to 5 gives us the total chi-square test statistic. ‖ ‖ step 7 ⋮ To find the P-value, we compare the calculated test statistic to the chi-square distribution with the appropriate degrees of freedom. The degrees of freedom are calculated as $(r - 1)(c - 1)$, where $r$ is the number of rows minus 1 and $c$ is the number of columns minus 1. ‖ ‖ step 8 ⋮ In this case, the degrees of freedom are $(4 - 1)(2 - 1) = 3$. ‖ ‖ step 9 ⋮ Using a chi-square distribution table or calculator with 3 degrees of freedom, we find the range in which our test statistic falls to determine the P-value. ‖ ***C*** ∻Key Concept∻ ⚹Chi-square test for homogeneity⚹ ∻Explanation∻ ⚹The chi-square test for homogeneity is used to determine if there are significant differences in the distribution of categorical variables across different populations. The test statistic is calculated by summing the squared differences between observed and expected frequencies, divided by the expected frequencies. The P-value is then found by comparing the test statistic to the chi-square distribution with the appropriate degrees of freedom.◊What is the formula to calculate the test statistic for a chi-square test for homogeneity?⍭ Generate me a similar question◊⚹

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