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 & & $108.58$ & 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

STEP 1

1. We are dealing with categorical data in a contingency table.
2. We need to test for association between two variables.
3. The significance level is 0.05.
4. The sample size is large enough for a chi-square test.

STEP 2

1. Identify the appropriate test for association.
2. Calculate the degrees of freedom.
3. Calculate the expected frequencies.
4. Calculate the chi-square test statistic.

STEP 3

The appropriate test for association between two categorical variables is the chi-square test of independence.
Type of test statistic: Chi-square

STEP 4

Calculate the degrees of freedom: df = (number of rows - 1) × (number of columns - 1) df = (2 - 1) × (2 - 1) = 1
Degrees of freedom: 1

STEP 5

Calculate the expected frequencies for each cell:
$E_{ij} = \frac{(row total)(column total)}{grand total}$
For "Up on Tuesday" and "Up on Wednesday": $E_{11} = \frac{267 \times 233}{500} = 124.42$
For "Up on Tuesday" and "Not up on Wednesday": $E_{12} = \frac{267 \times 267}{500} = 142.58$
For "Not up on Tuesday" and "Up on Wednesday": $E_{21} = \frac{233 \times 233}{500} = 108.58$
For "Not up on Tuesday" and "Not up on Wednesday": $E_{22} = \frac{233 \times 267}{500} = 124.42$

STEP 6

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

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Studdy Solution

STEP 1

1. We are dealing with categorical data in a contingency table.
2. We need to test for association between two variables.
3. The significance level is 0.05.
4. The sample size is large enough for a chi-square test.

STEP 2

1. Identify the appropriate test for association.
2. Calculate the degrees of freedom.
3. Calculate the expected frequencies.
4. Calculate the chi-square test statistic.

STEP 3

The appropriate test for association between two categorical variables is the chi-square test of independence.
Type of test statistic: Chi-square

STEP 4

Calculate the degrees of freedom: df = (number of rows - 1) × (number of columns - 1) df = (2 - 1) × (2 - 1) = 1
Degrees of freedom: 1

STEP 5

STEP 6

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Studdy Solution

STEP 1

1. We are dealing with categorical data in a contingency table.2. We need to test for association between two variables.3. The significance level is 0.05.4. The sample size is large enough for a chi-square test.

STEP 2

1. Identify the appropriate test for association.2. Calculate the degrees of freedom.3. Calculate the expected frequencies.4. Calculate the chi-square test statistic.

STEP 3

The appropriate test for association between two categorical variables is the chi-square test of independence.Type of test statistic: Chi-square

STEP 4

Calculate the degrees of freedom: df = (number of rows - 1) × (number of columns - 1) df = (2 - 1) × (2 - 1) = 1Degrees of freedom: 1

STEP 5

STEP 6

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. We are dealing with categorical data in a contingency table.
2. We need to test for association between two variables.
3. The significance level is 0.05.
4. The sample size is large enough for a chi-square test.

1. Identify the appropriate test for association.
2. Calculate the degrees of freedom.
3. Calculate the expected frequencies.
4. Calculate the chi-square test statistic.

The appropriate test for association between two categorical variables is the chi-square test of independence.
Type of test statistic: Chi-square

Calculate the degrees of freedom: df = (number of rows - 1) × (number of columns - 1) df = (2 - 1) × (2 - 1) = 1
Degrees of freedom: 1