Respuesta :
Answer:
Null hypothesis:[tex]p_{Y} - p_{O}=0[/tex]
Alternative hypothesis:[tex]p_{Y} - p_{O} \neq 0[/tex]
[tex]z=\frac{0.248-0.0543}{\sqrt{0.179(1-0.179)(\frac{1}{165}+\frac{1}{92})}}=3.883[/tex]
[tex]p_v =2*P(Z>3.883)=5.15x10^{-5}[/tex]
Reject the null hypothesis because there is sufficient evidence to support the claim that there is a difference in the effectiveness of the methods
The 99% confidence interval would be given (0.0877;0.2997).
Step-by-step explanation:
1) Data given and notation
[tex]X_{Y}=41[/tex] represent the number of live births in women under the age of 38
[tex]X_{O}=5[/tex] represent the number of residents of live births in women aged 38 and older
[tex]n_{Y}=165[/tex] sample of women under the age of 38
[tex]n_{O}=92[/tex] sample of women aged 38 and older
[tex]p_{Y}=\frac{41}{165}=0.248[/tex] represent the proportion of live births in women under the age of 38
[tex]p_{O}=\frac{5}{92}=0.0543[/tex] represent the proportion of live births in women aged 38 and older
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the value for the test (variable of interest)
[tex]\alpha=0.01[/tex] significance level given
2) Concepts and formulas to use
We need to conduct a hypothesis in order to check if there evidence of a difference in the effectiveness of the clinic's methods for older women, the system of hypothesis would be:
Null hypothesis:[tex]p_{Y} - p_{O}=0[/tex]
Alternative hypothesis:[tex]p_{Y} - p_{O} \neq 0[/tex]
We need to apply a z test to compare proportions, and the statistic is given by:
[tex]z=\frac{p_{Y}-p_{O}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{Y}}+\frac{1}{n_{O}})}}[/tex] (1)
Where [tex]\hat p=\frac{X_{Y}+X_{O}}{n_{Y}+n_{O}}=\frac{41+5}{165+92}=0.179[/tex]
3) Calculate the statistic
Replacing in formula (1) the values obtained we got this:
[tex]z=\frac{0.248-0.0543}{\sqrt{0.179(1-0.179)(\frac{1}{165}+\frac{1}{92})}}=3.883[/tex]
4) Statistical decision
Since is a two sided test the p value would be:
[tex]p_v =2*P(Z>3.883)=5.15x10^{-5}[/tex]
Comparing the p value with the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that there is a significant difference between the two propotions analyzed.
Reject the null hypothesis because there is sufficient evidence to support the claim that there is a difference in the effectiveness of the methods
5) Confidence interval for the difference of proportions
The confidence interval for the difference of two proportions would be given by this formula
[tex](\hat p_Y -\hat p_O) \pm z_{\alpha/2} \sqrt{\frac{\hat p_Y(1-\hat p_Y)}{n_Y} +\frac{\hat p_O (1-\hat p_O)}{n_O}}[/tex]
For the 99% confidence interval the value of [tex]\alpha=1-0.99=0.01[/tex] and [tex]\alpha/2=0.005[/tex], with that value we can find the quantile required for the interval in the normal standard distribution.
[tex]z_{\alpha/2}=2.58[/tex]
And replacing into the confidence interval formula we got:
[tex](0.248-0.0543) - 2.58 \sqrt{\frac{0.248(1-0.248)}{165} +\frac{0.0543(1-0.0543)}{92}}=0.0877/tex]
[tex](0.248-0.0543) + 2.58 \sqrt{\frac{0.248(1-0.248)}{165} +\frac{0.0543(1-0.0543)}{92}}=0.2997[/tex]
And the 99% confidence interval would be given (0.0877;0.2997).
There is 99% confidence that the proportion of successful live births at th clinic is between 8.77% and 29.97% for mothrs under 38 for those 38 and older.
