Observed Significance Level (P-value)

The observed significance level, or P-value, for a specific statistical test is the probability (assuming the null hypothesis is true) of observing a value of the test statistic that is at least as contradictory to the null hypothesis, and supportive of the alternative hypothesis as the actual one computed from the sample data.

Decision Criterion for a Hypothesis Test Using the P-value:

If P-value is less than a, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

Examples:

Ha: µ  30 versus Ho: µ = 30
 

Assumptions: X is normally distributed with s = 8
Test Statistic: 

a = .05 RR: z < -1.96 or z >1.96 (P-value < .05)
 

Calculation: z = 1.54
 

P-value = 2P(z > |zcalculated|) = 2P(z > |1.54|) = 2P(z < -1.54)
= 2(.0618) = .1236
 

Decision: Fail to reject Ho.
 

Suppose s is not known and n = 31.

Test Statistic: 
 

a = .05     RR: t < -2.042 or t > 2.042 (P-value < .05)
 

Calculation: t = 1.54 df = 30
 

P-value = 2P(t > |tcalculated |) = 2P(t > 1.54)

    P(t > 1.310) = .10 and P(t > 1.697) = .05.

    Hence, .05 < P(t > 1.54) < .10; therefore .10 < P-value < .20.

Decision: Fail to reject Ho.
 
 

P-value Calculations


I. For each test of hypothesis, compute the p-value. Sketch a figure.
 

1) Ha: µ < 100 versus Ho: µ > 100

If n = 20 and t = -1.78, find the p-value of t = -1.78.
 

2) Ha: µ > 240 versus Ho: µ < 240
 

If z = 2.35, find the p-value of z = 2.35.
 

3) Ha: µ 75 versus Ho: µ = 75

If n = 22 and t = -2.236, find the p-value of t = -2.236.
 

4) Ha: µ > 80 versus Ho: µ < 80

If t = 2.35 and n = 24, find the p-value of t = 2.35.
 
 

5) Ha: s12 < s22 versus Ho: s12 > s22
 

If F = 4.63, n1 = 18 and n2 = 16, find the p-value of F = 4.63.
 

6) Ha: s2 < 144 versus Ho: s2 > 144
 

If cn-12 = 7.44 and n = 18, find the p-value of cn-12 = 7.44.