Samples of 185 male and 170 female residents are chosen from a large metropolita
ID: 3219489 • Letter: S
Question
Samples of 185 male and 170 female residents are chosen from a large metropolitan area. Each person’s opinion (F=For, A=Against) on whether tax dollars should be used to support a local hospital was determined. The following table gives the frequencies in the various groups. Can we conclude with alpha=.05 that there is a difference between the proportions of males and females that support the local hospital?
0.
1.
2.
3.
4.
Opinion on tax support for hospital
Gender
F
A
Total
Male
110
75
185
Female
78
92
170
Opinion on tax support for hospital
Gender
F
A
Total
Male
110
75
185
Female
78
92
170
Explanation / Answer
here we use chi-square test for tesing difference between the proportions of males and females that support the local hospital.
and chi-square=sum((O-E)2/E) with (r-1)(c-1) df
here number of row=r=2
number of column=c=2
so df=(r-1)(c-1)=(2-1)*(2-1)=1
critical chi-square(0.05,1)=3.84
Null hypothesis
H0: there is no difference between the proportions of males and females that support the local hospital
alternate hypothesis H1:
there is difference between the proportions of males and females that support the local hospital
the calculated chi-squre=6.56 with 1 df and it is more than critical chi-square=3.84, so we fail to accept null hypothesis and conclude in favour of alternate hypothesis and conclude that there is difference between the proportions of males and females that support the local hospital.
the following way calculation has been done.
Opinion on tax support for hospital F A Total Gender Male 110 75 185 Female 78 92 170 Toal 188 167 355Related Questions
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