Question

p = 0.63 of obtaining a head. The experimenter tosses the coin 300 times in order to estimate the probability p. What is the probability that the experimenter’s point estimate of p will be within the interval (0.62, 0.64)

Answer 1

Answer:

The probability that the experimenter’s point estimate of p will be within the interval (0.62, 0.64) is 0.72

Step-by-step explanation:

To find a probability for the confidence interval we need to calculate z-statistic of the margin of error of the confidence interval.

margin of error for (0.62, 0.64) with p=0.63 is 0.01

Margin of error (ME) can be calculated using the equation

$ME=\frac{z*\sqrt{p*(1-p)}}{\sqrt{N} }$ where

• z is the corresponding statistic for the given confidence level

• p is the probability of obtaining a head(0.63)

• N is the sample size (300)

Then solving  $0.01=\frac{z*\sqrt{0.63*0.37}}{\sqrt{300} }$ gives z≈0.3587

two tailed p-value for z≈0.3587 is 0.72.

Thus, the probability that the experimenter’s point estimate of p will be within the interval (0.62, 0.64) is 0.72