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pickupchik [31]
3 years ago
15

Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. How many times would we have to f

lip 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
Mathematics
1 answer:
STALIN [3.7K]3 years ago
8 0

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 =<u>84</u>

hence, the correct answer is b) 84

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