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 95.8% confidence interval of width of at most .14 for the probability of flipping a head? (note that the z-score was rounded to three decimal places in the calculation)
a) 217
b) 153
c) 212
d) 209
e) 150
f) None of the above

Answer 1

Answer:

153 times

Step-by-step explanation:

We have to flip the coin in order to obtain a 95.8% confidence interval of width of at most .14

Width = 0.14

ME = $\frac{width}{2}$

ME = $\frac{0.14}{2}$

ME = $0.07$

$ME\geq z \times \sqrt{\frac{\widecap{p}(1-\widecap{p})}{n}}$

use p = 0.5

z at 95.8% is 1.727(using calculator)

$0.07 \geq 1.727 \times \sqrt{\frac{0.5(1-0.5)}{n}}$

$\frac{0.07}{1.727}\geq sqrt{\frac{0.5(1-0.5)}{n}}$

$(\frac{0.07}{1.727})^2 \geq \frac{0.5(1-0.5)}{n}$

$n \geq \frac{0.5(1-0.5)}{(\frac{0.07}{1.727})^2}$

$n \geq 152.169$

So, Option B is true

Hence  we have to flip 153 times the coin in order to obtain a 95.8% confidence interval of width of at most .14 for the probability of flipping a head