Math  /  Data & Statistics

Question\begin{tabular}{|c|c|c|c|c|} \hline & & Closing price & novement from Tu Wednesday & day to \\ \hline & & Closing price up on Wednesday & Closing price not up on Wednesday & Total \\ \hline \multirow{7}{*}{Closing price movement from Monday to Tuesday} & \multirow{3}{*}{Closing price up on Tuesday} & 134 & 133 & \\ \hline & & 124.42 & 142.58 & 267 \\ \hline & & 0.738 & 0.644 & \\ \hline & \multirow{3}{*}{Closing price not up on Tuesday} & 99 & 134 & \\ \hline & & 108.58108.58 & 124.58 & 233 \\ \hline & & 0.845 & 0.712 & \\ \hline & Total & 233 & 267 & 500 \\ \hline \end{tabular}
Part 2 Answer the following to summarize the test of the hypothesis that there is no association between the two variables closing price movement from Monday to Tuesday and closing price movement from Tuesday to Wednesday. For your test, use the 0.05 level of significance. (a) Determine the type of test statistic to use.
Type of test statistic: \square Degrees of freedom: \square (b) Find the value of the test statistic. (Round to two or more decimal places.) \square

Studdy Solution
Calculate the chi-square test statistic:
χ2=(OE)2E\chi^2 = \sum \frac{(O - E)^2}{E}
Where O is the observed frequency and E is the expected frequency.
χ2=(134124.42)2124.42+(133142.58)2142.58+(99108.58)2108.58+(134124.42)2124.42\chi^2 = \frac{(134 - 124.42)^2}{124.42} + \frac{(133 - 142.58)^2}{142.58} + \frac{(99 - 108.58)^2}{108.58} + \frac{(134 - 124.42)^2}{124.42}
χ2=0.7373+0.6444+0.8452+0.7373=2.9642\chi^2 = 0.7373 + 0.6444 + 0.8452 + 0.7373 = 2.9642
The value of the test statistic is 2.96 (rounded to two decimal places).

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