Math  /  Data & Statistics

Question\begin{tabular}{l|ccc|c} & HS & Some & College & Total \\ \hline Agree & 364 & 164 & 197 & 725 \\ Disagree & 558 & 471 & 787 & 1816 \\ \begin{tabular}{l} Don't \\ know \end{tabular} & 15 & 28 & 30 & 73 \\ \hline Total & 937 & 663 & 1014 & 2614 \\ \hline \end{tabular}
Table 1 Educational level and belief in One True Love
Round your answer for the chi-square statistic to one decimal place, and your answer for the pp-value to three decimal places. chi-square statistic == \square pp-value == \square
Conclusion: \square H0H_{0}

Studdy Solution

STEP 1

1. The data is presented in a contingency table format.
2. We are testing the independence of educational level and belief in One True Love.
3. The null hypothesis H0 H_0 is that educational level and belief in One True Love are independent.
4. We will use the chi-square test for independence.

STEP 2

1. Calculate the expected frequencies.
2. Compute the chi-square statistic.
3. Determine the degrees of freedom.
4. Find the p-value.
5. Draw a conclusion regarding H0 H_0 .

STEP 3

Calculate the expected frequencies for each cell using the formula:
Eij=(Row Totali)×(Column Totalj)Grand TotalE_{ij} = \frac{( \text{Row Total}_i ) \times ( \text{Column Total}_j )}{\text{Grand Total}}
For example, for the cell (Agree, HS):
E11=(725)×(937)2614=259.8E_{11} = \frac{(725) \times (937)}{2614} = 259.8
Calculate for all cells.

STEP 4

Continue calculating expected frequencies for all cells:
E12=(725)×(663)2614=183.8E_{12} = \frac{(725) \times (663)}{2614} = 183.8
E13=(725)×(1014)2614=281.4E_{13} = \frac{(725) \times (1014)}{2614} = 281.4
E21=(1816)×(937)2614=651.2E_{21} = \frac{(1816) \times (937)}{2614} = 651.2
E22=(1816)×(663)2614=460.2E_{22} = \frac{(1816) \times (663)}{2614} = 460.2
E23=(1816)×(1014)2614=704.6E_{23} = \frac{(1816) \times (1014)}{2614} = 704.6
E31=(73)×(937)2614=26.0E_{31} = \frac{(73) \times (937)}{2614} = 26.0
E32=(73)×(663)2614=18.5E_{32} = \frac{(73) \times (663)}{2614} = 18.5
E33=(73)×(1014)2614=28.4E_{33} = \frac{(73) \times (1014)}{2614} = 28.4

STEP 5

Compute the chi-square statistic using:
χ2=(OijEij)2Eij\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}
Calculate for each cell and sum them up.

STEP 6

Calculate each component:
χ112=(364259.8)2259.8=42.5\chi^2_{11} = \frac{(364 - 259.8)^2}{259.8} = 42.5
χ122=(164183.8)2183.8=2.3\chi^2_{12} = \frac{(164 - 183.8)^2}{183.8} = 2.3
χ132=(197281.4)2281.4=24.5\chi^2_{13} = \frac{(197 - 281.4)^2}{281.4} = 24.5
χ212=(558651.2)2651.2=13.8\chi^2_{21} = \frac{(558 - 651.2)^2}{651.2} = 13.8
χ222=(471460.2)2460.2=0.3\chi^2_{22} = \frac{(471 - 460.2)^2}{460.2} = 0.3
χ232=(787704.6)2704.6=9.9\chi^2_{23} = \frac{(787 - 704.6)^2}{704.6} = 9.9
χ312=(1526.0)226.0=4.7\chi^2_{31} = \frac{(15 - 26.0)^2}{26.0} = 4.7
χ322=(2818.5)218.5=4.9\chi^2_{32} = \frac{(28 - 18.5)^2}{18.5} = 4.9
χ332=(3028.4)228.4=0.1\chi^2_{33} = \frac{(30 - 28.4)^2}{28.4} = 0.1
Sum these values to get the total chi-square statistic.

STEP 7

Sum the chi-square components:
χ2=42.5+2.3+24.5+13.8+0.3+9.9+4.7+4.9+0.1=102.9\chi^2 = 42.5 + 2.3 + 24.5 + 13.8 + 0.3 + 9.9 + 4.7 + 4.9 + 0.1 = 102.9

STEP 8

Determine the degrees of freedom:
df=(number of rows1)×(number of columns1)=(31)×(31)=4df = ( \text{number of rows} - 1 ) \times ( \text{number of columns} - 1 ) = (3 - 1) \times (3 - 1) = 4

STEP 9

Using a chi-square distribution table or calculator, find the p-value for χ2=102.9\chi^2 = 102.9 with df=4df = 4.
The p-value is approximately p<0.001p < 0.001.

STEP 10

Conclusion: Since the p-value is less than 0.05, we reject H0 H_0 .
Conclusion: Reject H0 H_{0} .

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