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A child psychologist believes that children perform better on tests when they are given perceived freedom of choice. To test this belief, the psychologist carried out an experiment in which third graders were randomly assigned to two groups, A and B. Each child was given the same simple logic test. However, in group B, each child was given the freedom to choose a text booklet from many with various drawings on the covers. The performance of each child was rated as Very Good, Good, and Fair. The results are summarized in the table provided. Test, at the $5$ \% level of significance, whether there is sufficient evidence in the data to support the psychologist's belief. $\newline$ \begin{tabular}{|l|l|l|l|} $\newline$ \hline \multicolumn{ $2$ }{c}{} & \multicolumn{ $2$ }{c|}{ Group } \\ $\newline$ \hline \multicolumn{ $2$ }{|c|}{ A } & B \\ $\newline$ \hline \multirow{ $2$ }{*}{\begin{tabular}{l} $\newline$ Performanc $\newline$ \end{tabular}} & Very Good & $32$ & $29$ \\ $\newline$ \cline { $2$ - $4$ } & Good & $55$ & $51$ \\ $\newline$ \cline { $2$ - $4$ } & Fair & $10$ & $13$ \\ $\newline$ \hline $\newline$ \end{tabular}

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Math Problems

Algebra 2

Interpret confidence intervals for population means

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Q. A child psychologist believes that children perform better on tests when they are given perceived freedom of choice. To test this belief, the psychologist carried out an experiment in which third graders were randomly assigned to two groups, A and B. Each child was given the same simple logic test. However, in group B, each child was given the freedom to choose a text booklet from many with various drawings on the covers. The performance of each child was rated as Very Good, Good, and Fair. The results are summarized in the table provided. Test, at the

5

\% level of significance, whether there is sufficient evidence in the data to support the psychologist's belief.

\newline

\begin{tabular}{|l|l|l|l|}

\newline

\hline \multicolumn{

2

}{c}{} & \multicolumn{

2

}{c|}{ Group } \\

\newline

\hline \multicolumn{

2

}{|c|}{ A } & B \\

\newline

\hline \multirow{

2

}{*}{\begin{tabular}{l}

\newline

Performanc

\newline

\end{tabular}} & Very Good &

32

29

\newline

\cline {

2

4

} & Good &

55

51

\newline

\cline {

2

4

} & Fair &

10

13

\newline

\hline

\newline

\end{tabular}

State Hypotheses: State the null and alternative hypotheses. $\newline$ The null hypothesis ( $H_0$ ) is that there is no difference in performance between the two groups. The alternative hypothesis ( $H_1$ ) is that there is a difference in performance, specifically that group B performs better.
Set up Table: Set up the contingency table based on the data provided. $\newline$ The contingency table is as follows: $\newline$ | Performance | Group A | Group B | $\newline$ |-------------|---------|---------| $\newline$ | Very Good | $32$ | $29$ | $\newline$ | Good | $55$ | $51$ | $\newline$ | Fair | $10$ | $13$ |
Calculate Expected Frequencies: Calculate the expected frequencies for each cell in the contingency table assuming the null hypothesis is true. $\newline$ To do this, we use the formula: Expected frequency $= (\text{Row total} \times \text{Column total}) / \text{Grand total}$ $\newline$ First, we calculate the row and column totals and the grand total. $\newline$ Row totals: $\newline$ Very Good: $32 + 29 = 61$ $\newline$ Good: $55 + 51 = 106$ $\newline$ Fair: $10 + 13 = 23$ $\newline$ Column totals: $\newline$ Group A: $32 + 55 + 10 = 97$ $\newline$ Group B: $29 + 51 + 13 = 93$ $\newline$ Grand total: $97 + 93 = 190$ $\newline$ Now we calculate the expected frequencies for each cell. $\newline$ Expected frequency for Very Good in Group A: $(61 \times 97) / 190$ $\newline$ Expected frequency for Very Good in Group B: $(61 \times 93) / 190$ $\newline$ Expected frequency for Good in Group A: $(106 \times 97) / 190$ $\newline$ Expected frequency for Good in Group B: $32 + 29 = 61$ $0$ $\newline$ Expected frequency for Fair in Group A: $32 + 29 = 61$ $1$ $\newline$ Expected frequency for Fair in Group B: $32 + 29 = 61$ $2$
Calculate Chi-Square Statistic: Calculate the chi-square test statistic. $\newline$ The chi-square test statistic is calculated using the formula: $\chi^2 = \Sigma[(O - E)^2 / E]$ , where $O$ is the observed frequency and $E$ is the expected frequency. $\newline$ We will calculate this for each cell and sum them up. $\newline$ Let's calculate the expected frequencies and the chi-square statistic for each cell: $\newline$ For Very Good in Group A: $E = (61 \times 97) / 190 = 31.3$ $\newline$ For Very Good in Group B: $E = (61 \times 93) / 190 = 29.7$ $\newline$ For Good in Group A: $E = (106 \times 97) / 190 = 54.4$ $\newline$ For Good in Group B: $E = (106 \times 93) / 190 = 51.6$ $\newline$ For Fair in Group A: $E = (23 \times 97) / 190 = 11.8$ $\newline$ For Fair in Group B: $E = (23 \times 93) / 190 = 11.2$ $\newline$ Now we calculate the chi-square statistic for each cell: $\newline$ For Very Good in Group A: $\chi^2 = (32 - 31.3)^2 / 31.3$ $\newline$ For Very Good in Group B: $O$ $0$ $\newline$ For Good in Group A: $O$ $1$ $\newline$ For Good in Group B: $O$ $2$ $\newline$ For Fair in Group A: $O$ $3$ $\newline$ For Fair in Group B: $O$ $4$ $\newline$ Summing these up gives us the total chi-square statistic.
Calculate Degrees of Freedom: Calculate the degrees of freedom for the chi-square test. The degrees of freedom ( extit{df}) for a contingency table is calculated as $\textit{df} = (\textit{number of rows} - 1) \times (\textit{number of columns} - 1)$ . In this case, $\textit{df} = (3 - 1) \times (2 - 1) = 2 \times 1 = 2$ .
Determine Critical Value: Determine the critical value from the chi-square distribution table for $2$ degrees of freedom at the $5\%$ level of significance. $\newline$ The critical value for $\chi^2$ with $2$ degrees of freedom at the $5\%$ level is approximately $5.991$ .
Compare Statistic to Critical Value: Compare the calculated chi-square test statistic to the critical value. If the calculated chi-square statistic is greater than the critical value, we reject the null hypothesis. If it is less, we fail to reject the null hypothesis. $\newline$ However, we have not actually calculated the chi-square statistic in the previous steps. We need to go back and perform the calculations for each cell and sum them up to get the total chi-square statistic.