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. = ~ N(0,1)
where, = sample proportion of girls born = = 0.85
n = sample of babies = 400
p = population percentage of girls born
<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>
<u>So, 99% confidence interval for the population proportion, p is ;</u>
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 < < 2.58) = 0.99
P( < < ) = 0.99
P( < p < ) = 0.99
<u>99% confidence interval for p</u> = [ , ]
= [ , ]
= [0.804 , 0.896]
Therefore, 99% confidence interval for the percentage of girls born is [0.804 , 0.896].
(b) <em>Let p = population proportion of girls born.</em>
So, Null Hypothesis, : p = 0.50 {means that the proportion of girls is equal to 0.50}
Alternate Hypothesis, : p 0.50 {means that the proportion of girls is significantly different from 0.50}
The test statistics that will be used here is <u>One-sample z proportion test</u> <u>statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of girls born = = 0.85
n = sample of babies = 400
So, <u><em>the test statistics</em></u> =
= 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 <u>we reject our null hypothesis</u>.
Therefore, we conclude that the proportion of girls is significantly different from 0.50.