For H₀：p=p₀,
Z = (p̂ − p₀)/√(p₀(1−p₀)/n)
Example ① A coin is tossed 100 times and lands heads 60 times. Can we say at 5% that it is “not fair”?
Solution:H₀: p=0.5, H₁: p≠0.5. p̂=0.6.
σ_{p̂} = √(0.5·0.5/100) = 0.05.
Z = (0.6−0.5)/0.05 = 2.0
Since |2.0| > 1.96, reject H₀.
Solution:H₀: p=0.5, H₁: p≠0.5. p̂=0.6.
σ_{p̂} = √(0.5·0.5/100) = 0.05.
Z = (0.6−0.5)/0.05 = 2.0
Since |2.0| > 1.96, reject H₀.
Key Point
- Calculate the denominator using p₀ (this differs slightly from the estimation formula)
Practice Problems
- A die is rolled 600 times and 1 appears 80 times. Can we say at 5% in a one-tailed test that “1 is unlikely to appear”?
Solutions and Explanations
- Solution:p̂=80/600≈0.133、p₀=1/6≈0.167。Z=(0.133−0.167)/√(0.167·0.833/600)≈−2.21。Since Z<−1.65, reject H₀.