Respuesta :
Answer:
(a) 99% confidence interval for the percentage of girls born is [0.804 , 0.896].
(b) Yes, the proportion of girls is significantly different from 0.50.
Step-by-step explanation:
We are given that a clinical trial tests a method designed to increase the probability of conceiving a girl.
In the study 400 babies were born, and 340 of them were girls.
(a) Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;
P.Q. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of girls born = [tex]\frac{340}{400}[/tex] = 0.85
n = sample of babies = 400
p = population percentage of girls born
Here for constructing 99% confidence interval we have used One-sample z proportion statistics.
So, 99% confidence interval for the population proportion, p is ;
P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level
of significance are -2.58 & 2.58}
P(-2.58 < [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < 2.58) = 0.99
P( [tex]-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < [tex]{\hat p-p}[/tex] < [tex]2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99
P( [tex]\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] < p < [tex]\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ) = 0.99
99% confidence interval for p = [[tex]\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] , [tex]\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex]]
= [ [tex]0.85-2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }[/tex] , [tex]0.85+2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } }[/tex] ]
= [0.804 , 0.896]
Therefore, 99% confidence interval for the percentage of girls born is [0.804 , 0.896].
(b) Let p = population proportion of girls born.
So, Null Hypothesis, [tex]H_0[/tex] : p = 0.50 {means that the proportion of girls is equal to 0.50}
Alternate Hypothesis, [tex]H_A[/tex] : p [tex]\neq[/tex] 0.50 {means that the proportion of girls is significantly different from 0.50}
The test statistics that will be used here is One-sample z proportion test statistics;
T.S. = [tex]\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }[/tex] ~ N(0,1)
where, [tex]\hat p[/tex] = sample proportion of girls born = [tex]\frac{340}{400}[/tex] = 0.85
n = sample of babies = 400
So, the test statistics = [tex]\frac{0.85-0.50}{\sqrt{\frac{0.85(1-0.85)}{400} } }[/tex]
= 19.604
Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is way more than the critical value of z as 19.604 > 2.3263, so we have sufficient evidence to reject our null hypothesis due to which we reject our null hypothesis.
Therefore, we conclude that the proportion of girls is significantly different from 0.50.
