Respuesta :
Answer:
[tex]z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154[/tex]
[tex]p_v =2*P(z<-1.154)=0.249[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of girls born is not significantly different from 0.467
Step-by-step explanation:
Data given and notation
n=150 represent the random sample taken
X=63 represent the number of girls born
[tex]\hat p=\frac{63}{150}=0.42[/tex] estimated proportion of girls born
[tex]p_o=0.467[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the true proportion if girls is 0.467.:
Null hypothesis:[tex]p=0.467[/tex]
Alternative hypothesis:[tex]p \neq 0.467[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z<-1.154)=0.249[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of girls born is not significantly different from 0.467
So, we can conclude that the percentage of girls born in china is [tex]46.7\%[/tex]
Test value Statistics:
The formula for the test value statistics is,
[tex]Z=\frac{\hat {p}-p}{\sqrt{\frac{p(1-P}{n} } }[/tex]
Given that,
[tex]n=150\\\hat{p}=\frac{x}{n} \\\hat{p} =\frac{60}{150} \\=0.40[/tex]
[tex]H_0:p=0.467\\H_1:p\neq 0.47[/tex]
Now, substituting the given values into the above formula we get,
[tex]Z=\frac{0.467-0.40}{\frac{0.40(1-0.40)}{150} } \\Z=1.675[/tex]
Now, calculating the P-value as,
[tex]P-value=2(1-p(Z\leq z))\\=2(1-0.953)\\=0.094[/tex]
Since P-value is greater than [tex]0.05[/tex] the level of significance we do not reject [tex]H_0[/tex]
Learn more about the topic test value statistics
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