Question #98083
A study was conducted to determine student, faculty, and administration
attitudes on a new University parking policy. The distribution of those favouring or opposing the policy is shown in the following table:

Student Faculty Administration
Favour 252 107 43
Oppose 139 81 40

Do the data provide sufficient evidence to indicate that attitudes regarding the parking policy do not depend on the category of the respondent?
1
Expert's answer
2019-11-12T11:43:52-0500
StudentFacultyAdministrationTotalFavor25210743402Oppose1398140260Total39118883662\def\arraystretch{1.5} \begin{array}{c:c:c:c:c: } & Student & Faculty& Administration & Total \\ \hline Favor & 252 & 107 & 43 & 402 \\ \hdashline Oppose & 139 & 81& 40 & 260 \\ \hdashline Total & 391 & 188 & 83 & 662 \end{array}

Given

r=Number of rows in table=2,r=\text{Number of rows in table}=2,

c=Number of columns in table=3c=\text{Number of columns in table}=3

α=Significance level=0.05\alpha=\text{Significance level}=0.05

H0:H_0: attitudes regarding the parking policy are independent on the category of the respondent.

H1:H_1: attitudes regarding the parking policy are dependent on the category of the respondent.


The expected frequencies EE are the product of the column and row total, divided by the table total


E11=r1×c1n=402×391662237.435E_{11}={r_1\times c_1 \over n}={402\times 391 \over 662}\approx237.435

E12=r1×c2n=402×188662114.163E_{12}={r_1\times c_2 \over n}={402\times 188 \over 662}\approx114.163

E13=r1×c3n=402×8366250.402E_{13}={r_1\times c_3 \over n}={402\times 83 \over 662}\approx50.402

E21=r2×c1n=260×391662153.565E_{21}={r_2\times c_1 \over n}={260\times 391 \over 662}\approx153.565

E22=r2×c2n=260×18866273.837E_{22}={r_2\times c_2 \over n}={260\times 188 \over 662}\approx73.837


E23=r2×c3n=260×8366232.598E_{23}={r_2\times c_3 \over n}={260\times 83 \over 662}\approx32.598

The chi-square subtotals are the squared differences between the observed and expected frequencies, divided by the expected frequency.


X=(OE)2E\Chi=\sum{(O-E)^2 \over E}

=(252237.435)2237.435+(107114.163)2114.163+(4350.402)250.402+={(252-237.435)^2 \over 237.435}+{(107-114.163)^2 \over 114.163}+{(43-50.402)^2 \over 50.402}+

+(139153.565)2153.565+(8173.837)273.837+(4032.598)232.598+{(139-153.565)^2 \over 153.565}+{(81-73.837)^2 \over 73.837}+{(40-32.598)^2 \over 32.598}\approx


6.187\approx6.187

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme.

The P-value is the number (or interval) in the column title of the chi-square distribution table in the row

df=(r1)(c1)=(21)(31)=2:df=(r-1)(c-1)=(2-1)(3-1)=2:


0.025<P<0.050.025<P<0.05

If the P-value is less or equal to the significance level, then the null hypothesis is rejected:


P<0.05=α=>Reject H0P<0.05=\alpha=>\text{Reject }H_0

There is sufficient evidence to support the claim that the attitudes regarding the parking policy are dependent on the category of the respondent.

Answer: Attitudes regarding the parking policy depend on the category of the respondent.


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