Math  /  Data & Statistics

QuestionIn Bloomington, they randomly sampled 210 female voters, and 440 male voters. They collected data on the respondent's opinion on an environmental bond issue. We want to know whether there is good evidence that one's gender influences whether a person is for or against the bond issue. Use α=0.1\alpha=0.1. \begin{tabular}{|c|c|c|c|} \hline & For Bond Issue & Against Bond Issue & Total \\ \hline Men & 51 & 159 & 210 \\ \hline Women & 39 & 401 & 440 \\ \hline Total & 90 & 560 & 650 \\ \hline \end{tabular} a) What is the correct null hypothesis? Ho:p1p2p3p4H_{o}: p_{1} \neq p_{2} \neq p_{3} \neq p_{4} Ho:p1=p2=p3=p4H_{o}: p_{1}=p_{2}=p_{3}=p_{4} HoH_{o} : Gender and Bond Issue Attitudes are dependent. HoH_{o} : Gender and Bond Issue Attitudes are Independent. b) Fill in the expected values, round answers to 1 decimal place. \begin{tabular}{|c|c|c|} \hline \begin{tabular}{c} Expected \\ Values \end{tabular} & For Bond Issue & \begin{tabular}{c} Against \\ Bond Issue \end{tabular} \\ \hline Men & & \\ \hline Women & & \\ \hline \end{tabular} c) Chi Square Test Statistic == \square Round answer to 1 decimal place.
Degree's of Freedom == \square d) The pp-value == \square Round answer to 3 decimal places.

Studdy Solution

STEP 1

1. We are conducting a chi-square test for independence.
2. The significance level α\alpha is 0.1.
3. The null hypothesis H0H_0 is that gender and bond issue attitudes are independent.

STEP 2

1. Define the null hypothesis.
2. Calculate the expected values.
3. Compute the chi-square test statistic.
4. Determine the degrees of freedom.
5. Calculate the p-value.

STEP 3

Define the null hypothesis:
H0:Gender and Bond Issue Attitudes are Independent. H_0: \text{Gender and Bond Issue Attitudes are Independent.}

STEP 4

Calculate the expected values using the formula:
E=(Row Total)×(Column Total)Grand Total E = \frac{(\text{Row Total}) \times (\text{Column Total})}{\text{Grand Total}}
For Men, For Bond Issue:
E=210×90650=29.1 E = \frac{210 \times 90}{650} = 29.1
For Men, Against Bond Issue:
E=210×560650=180.9 E = \frac{210 \times 560}{650} = 180.9
For Women, For Bond Issue:
E=440×90650=60.9 E = \frac{440 \times 90}{650} = 60.9
For Women, Against Bond Issue:
E=440×560650=379.1 E = \frac{440 \times 560}{650} = 379.1

STEP 5

Compute the chi-square test statistic using the formula:
χ2=(OE)2E \chi^2 = \sum \frac{(O - E)^2}{E}
Where OO is the observed frequency and EE is the expected frequency.
For Men, For Bond Issue:
(5129.1)229.1=17.0 \frac{(51 - 29.1)^2}{29.1} = 17.0
For Men, Against Bond Issue:
(159180.9)2180.9=2.7 \frac{(159 - 180.9)^2}{180.9} = 2.7
For Women, For Bond Issue:
(3960.9)260.9=8.4 \frac{(39 - 60.9)^2}{60.9} = 8.4
For Women, Against Bond Issue:
(401379.1)2379.1=1.2 \frac{(401 - 379.1)^2}{379.1} = 1.2
Summing these values gives:
χ2=17.0+2.7+8.4+1.2=29.3 \chi^2 = 17.0 + 2.7 + 8.4 + 1.2 = 29.3

STEP 6

Determine the degrees of freedom using the formula:
Degrees of Freedom=(Number of Rows1)×(Number of Columns1) \text{Degrees of Freedom} = (\text{Number of Rows} - 1) \times (\text{Number of Columns} - 1)
Degrees of Freedom=(21)×(21)=1 \text{Degrees of Freedom} = (2 - 1) \times (2 - 1) = 1

STEP 7

Calculate the p-value using the chi-square distribution table or calculator for χ2=29.3\chi^2 = 29.3 and 1 degree of freedom.
The p-value is approximately:
p=0.000 p = 0.000
The solution is:
a) Null Hypothesis: H0:Gender and Bond Issue Attitudes are Independent. H_0: \text{Gender and Bond Issue Attitudes are Independent.}
b) Expected Values:
Expected ValuesFor Bond IssueAgainst Bond IssueMen29.1180.9Women60.9379.1\begin{array}{|c|c|c|} \hline \text{Expected Values} & \text{For Bond Issue} & \text{Against Bond Issue} \\ \hline \text{Men} & 29.1 & 180.9 \\ \hline \text{Women} & 60.9 & 379.1 \\ \hline \end{array}
c) Chi Square Test Statistic =29.3 = 29.3
Degrees of Freedom =1 = 1
d) The p p -value =0.000 = 0.000

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