Answer:
Step-by-step explanation:
From this study:
The null hypothesis:
[tex]H_o : p_1 =p_2[/tex]
The altenative is:
[tex]H_a : p_1 \ne p_2[/tex]
This test is a two-tailed test.
However; we are told that the wives have 44 success out of 66, then the number of failures will be 22.
Then;
[tex]\hat p_1 = \dfrac{44}{66}[/tex]
[tex]\hat p_1 = 0.6667[/tex]
Similarly, the husbands have 18 success out of 46, then the number of failures will be 28
Then:
[tex]\hat p_2 = \dfrac{18}{46}[/tex]
[tex]\hat p_2 = 0.3913[/tex]
The pooled proportion [tex]p = \dfrac{18+44}{66+46}[/tex]
[tex]p = \dfrac{62}{112}[/tex]
p = 0.55357
The estimated standard error S.E is:
[tex]= \sqrt{\dfrac{ \bar p(1- \bar p)}{n_1} +\dfrac{ \bar p(1-\bar p)}{n_2}}[/tex]
= [tex]\sqrt{ 0.55357(1-0.55357) \Big( \dfrac{1}{66} + \dfrac{1}{46} \Big)}[/tex]
[tex]=\sqrt{ 0.55357(0.44643) \Big(0.01515 + 0.021739 \Big)}[/tex]
[tex]=\sqrt{ 0.00911638798}[/tex]
= 0.0955
The Z test statitics can now be computed as:
[tex]Z = \dfrac{ \hat p_1 - \hat p_2}{\sqrt{\dfrac{ \bar p(1- \bar p)}{n_1} +\dfrac{ \bar p(1-\bar p)}{n_2}}}[/tex]
[tex]Z = \dfrac{0.6667 -0.3913}{0.0955}[/tex]
Z = 2.88
Th p -value from the test statistics is:
p-value = 2P(Z > 2.88)
p- value = 2 P (1 - Z < 2.88)
p-value = 2 ( 1 - 0.998)
p-value = 2 ( 0.002)
p -alue = 0.004
Decision Rule:
Thus, at 0.01 significance level, we reject the null hypothesis because, p-value is less than that (i.e. significance level)
Conclusion:
We conclude that there is a significant difference between the proportions.
