Question

Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. How many times would we have to flip the coin in order to obtain a 90% confidence interval of width of at most .18 for the probability of flipping a head?a) 83b) 84c) 87d) 50e) 51f) None of the above

Answer 1

Answer:  b) 84

Step-by-step explanation:

Let p be the prior estimate of the required proportion.

As per given , we have

p =0.5   (The probability of getting heads on a fair coin is 0.5)

Significance level : $\alpha: 1-0.90=0.10$

Critical z-value (using z-value table ) : $z_{\alpha/2}=1.645$

Confidence interval width : w= 0.18

Thus , the margin of error : $E=\dfrac{w}{2}=0.09$

Formula to find the sample size ( if prior estimate of proportion is known.):-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

Substitute the values , we get

$n=0.5(1-0.5)(\dfrac{1.645}{0.09})^2$

Simplify ,

$n=83.5192901235\approx84$  [Round of to the next whole number.]

Hence, the number of times we would have to flip the coin =84

hence, the correct answer is b) 84