Question

Urgent

all answers to 5 significant figures please

4. Suppose the researchers in question 2 are convinced that women really can identify the sex of an unborn baby at a rate better than 50%, (a) find the sample size they would require to reduce the 95% CI to the point where it excludes the value of 50%, ie, to ensure that the lower confidence limit >50%. (You can use a trial and error approach or notice that the sample proportion is 57/104 -0.548077, which implies that the margin of error, E for the new confidence interval must be less than 0.548077-0.5- 0.048077. Then use this value of E in the usual equation for estimating sample size, n.) (b) Remembering to round the estimate of n up to the next whole number, calculate and record the new 95% CI to confirm that it does not include the value of 50%. (Use 4-1.9600.)

Answer 1

Solution

Back-up Theory

100(1 - α) % Confidence Interval for the population proportion, p is: p_cap ± MoE, ………………………. (1)

where

MoE = Z_α/2[√{p_cap(1 –p_cap)/n}] ……………………………………………………………………………. (2)

with

Z_α/2 is the upper (α/2)% point of N(0, 1),

pcap = sample proportion, and

n = sample size.

Now, to work out the solution,

As already hinted in the question itself, we want: [vide (2)]

E = MoE = Z_α/2[√{p_cap(1 –p_cap)/n}] < 0.048077, where also given are: p_cap = 0.548077 and Z_α/2 = 1.96.

Thus, we must have: 1.96√(0.548077 x 0.451923/n) < 0.048077, Or

n > (1.96/0.048077)²(0.548077 x 0.451923) , Or

n > 439.

Thus, n = 440 Answer 1

With the above n = 440, vide (2),

E = MoE = 1.96√(0.548077 x 0.451923/440)

= 0.0465

And hence, vide (1), 95% CI = 0.548077 ± 0.0465

= [0.5016, 0.5946] Answer 2

DONE