The one-proportion z test and P-values

Three quick questions from earlier lessons.

A coin lands heads 60 times in 100 tosses. If it were fair, $\hat{p}$ would have SD 0.05. How many SDs above 0.5 is 0.6?

How surprising is the data if H₀ is true?

One-proportion z test:

z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}}

(the SE uses the null value

p_0

). The P-value is the probability, assuming H₀ is true, of a result at least as extreme as the one observed, in the direction(s) of Hₐ.

H₀: $p = 0.25$ , $n = 300$ , $\hat{p} = 0.30$ .

SE = 0.025

√(0.25 × 0.75/300).

z = 2

(0.30 − 0.25)/0.025.

P-value

\approx 0.023

Upper tail for Hₐ: p > 0.25.

Your turn: SE = 0.04, $\hat{p} - p_0 = 0.06$ . Find $z$ .

z = 1.5

0.06/0.04.

Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 52/100. Find the z statistic.
Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 58/100. Find the z statistic.

A test gives a P-value of 0.05. What does it mean?
1. If H₀ were true, a result at least as extreme as ours would happen with probability 0.05
2. There is a 0.05 probability that H₀ is true
3. The result is due to chance with probability 0.05

Boss problem. A coin is tossed 100 times and lands heads 52 times. Test H₀: p = 0.5 against Hₐ: p > 0.5.

Three questions to check you have got it.

Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 52/100. Find the z statistic.
Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 58/100. Find the z statistic.
Test H₀: p = 0.5 with a random sample of 100 in which p̂ = 56/100. Find the z statistic.

z test for p

$z = (\hat{p} - p_0)/\sqrt{p_0(1 - p_0)/n}$ ; P-value assumes H₀.

Watch out: use $p_0$ in the SE.

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