Respuesta :
Answer:
a)
The proportions of the females and males who took the survey who are annoyed by the behavior in question are 0.3301 and_0.3791 respectively
b)
D) The data come from a Population that is Normally distributed
F) the samples are independent
c)
Null hypothesis: H₀: [tex]p_{m} ^{-} - p^{-} _{fm} \leq 0[/tex]
Alternative Hypothesis H₁: [tex]p_{m} ^{-} - p^{-} _{fm} \geq 0[/tex]
Step-by-step explanation:
Given data Among the 530 males surveyed, 175 responded "Yes
The sample proportion of males
[tex]p_{m} ^{-} = \frac{x}{n} = \frac{175}{530} = 0.3301[/tex]
The sample proportion of 575 Females surveyed, 218 responded "Yes
[tex]p_{fem} ^{-} = \frac{x}{n} = \frac{218}{575} = 0.3791[/tex]
a)
The proportions of the females and males who took the survey who are annoyed by the behavior in question are 0.3301 and_0.3791 respectively
b)
The data come from a Population that is normally distributed.
The two samples are independent
c)
Null hypothesis: H₀: [tex]p_{m} ^{-} - p^{-} _{fm} \geq 0[/tex]
Alternative Hypothesis H₁: [tex]p_{m} ^{-} - p^{-} _{fm} \leq 0[/tex]
We will use two samples Z - hypothesis test
[tex]Z = \frac{p^{-} _{m} - p^{-} _{fem} }{se(p^{-} _{m}-p^{-} _{fem} ) }[/tex]
[tex]Se(p_{m} -p_{fem}) = \sqrt{\frac{p_{m}(1-p_{m}) }{n_{m} }+\frac{p_{fem} (1-p_{fem} )}{n_{fem} } }[/tex]
[tex]Se(p_{m} -p_{fem}) = \sqrt{\frac{0.3301(1-0.3301) }{530 }+\frac{0.3791(1-0.3791)}{575} }[/tex]
[tex]Se(p_{m} ^{-} - p^{-} _{fem} ) = \sqrt{0.000826} = 0.0287[/tex]
The test statistic
[tex]Z = \frac{0.3301-0.3791}{0.02875}[/tex]
Z = -1.7043
|Z| = |-1.7043|
The tabulated value Z₀.₉₅ = 1.96
The calculated Z= 1.7043 < 1.96 at 0.05 level of significance
The null hypothesis is accepted
Conclusion:-
The evidence suggest not a higher proportion of females are annoyed by this behavior
