Question

A coin has a true probability \( \mu \) of turning up Heads. This coin is tossed 100 times and shows up Heads 60 times. The following hypothesis is tested: 
\[ H_0: \mu = 0.5 \quad ({Null Hypothesis}), \quad H_1: \mu>0.5 \quad ({Alternative Hypothesis}) \] 
Using the Central Limit Theorem, the \( p \)-value of the above test is ________ (round off to three decimal places). 
Hint: If Z is a random variable that follows a standard normal distribution, then P (Z ≤ 2) = 0.977.

Hint:

The \( p \)-value in hypothesis testing helps determine the strength of evidence against the null hypothesis. A small \( p \)-value (usually less than 0.05) suggests strong evidence against the null hypothesis.

Answer 1

Step 1: Define the problem setup.
The number of heads, \( X \), follows a binomial distribution: \[ X \sim {Binomial}(n = 100, \, p = 0.5) \] The sample proportion of heads is: \[ \hat{\mu} = \frac{60}{100} = 0.6 \] Step 2: Apply the Central Limit Theorem.
For large \( n \), the binomial distribution can be approximated by a normal distribution with mean \( \mu = 0.5 \) and standard deviation \( \sigma = \sqrt{\frac{p(1 - p)}{n}} \). Here, \( p = 0.5 \) and \( n = 100 \), so: \[ \sigma = \sqrt{\frac{0.5(1 - 0.5)}{100}} = \sqrt{\frac{0.25}{100}} = \frac{0.5}{\sqrt{100}} = 0.05 \] Step 3: Compute the \( Z \)-score.
The \( Z \)-score is given by: \[ Z = \frac{\hat{\mu} - \mu}{\sigma} = \frac{0.6 - 0.5}{0.05} = \frac{0.1}{0.05} = 2 \] Step 4: Find the \( p \)-value.
The \( p \)-value corresponds to the probability of observing a \( Z \)-score greater than or equal to 2. From the standard normal distribution table, \( P(Z \leq 2) = 0.977 \). Since the test is one-tailed (right tail), the \( p \)-value is: \[ p = 1 - P(Z \leq 2) = 1 - 0.977 = 0.023 \] So, the \( p \)-value is \( 0.023 \).