Hypothesis Testing:
the test statistic and the P-value.
The P-value of the test, assuming H0 is true, is the probability
that the test statistic would take a value as extreme or more
extreme than that actually observed.
Text:
“After assessing the consequences of Type I and Type II
errors, identify the largest α that is tolerable for the
problem. Then employ a test procedure that uses this
maximum acceptable value -- rather than anything smaller - as the level of significance (because using a smaller α
increases β). In other words don’t make α smaller than it
needs to be.”
α = Probability (Type I error)
1) Sarah claims that she makes 80% of her basketball free
throws. You don’t believe that she is that good.
H0 : π = 0.80
Ha : π < 0.80
Choose a value for α, say 0.05.
Suppose Sarah shoots 60 free throws and makes 42. Is
this convincing evidence that Sarah is not an 80% free
throw shooter? In other words, is this convincing
evidence to reject the null-hypothesis? We will calculate
the probability that Sarah would make 42 or less free
throws out of 60 if indeed she was an “80% free throw
shooter.”
Assume that H0 is true, i.e. π = 0.80.
Use µp = 0.80, and σp = (0.80*0.20 / 60)0.5 = 0.05164
P-Value
Draw a picture
= Pr(p ≤ 42/60)
- one-tailed.
= Pr(p ≤ 0.70)
= Pr( (p – 0.80)/0.05164 ≤ (0.70 – 0.80)/0.05164) )
= Pr( z < -1.936)
= 0.0262
Since the P-Value < α (0.0262 < 0.05) there is convincing
evidence to reject the null-hypothesis at the 0.05
significance level.
2) Workers at a manufacturing plant have two working
modes, machine-paced and self-paced. Management
believes that there is no difference in job satisfaction with
these two modes.
H0 : µ = 0
Ha : µ ≠ 0
Choose a value for α, say 0.05.
Suppose that 25 workers, a random sample, were asked
their opinion and their scores gave = +17.5 and from
past experience with these questionnaires they know σ =
50. Is this convincing evidence that there is no difference
in job satisfaction with these two modes? In other words,
is this convincing evidence to reject the null-hypothesis?
We will calculate the Pr( ≥ +17.5).
Assume that H0 is true, i.e. µ = 0.
Use µ = 0, and σ = 50, and = 50/
= 10.
P-Value
Draw a picture
= Pr( ≥ +17.5)
-two-tailed.
= Pr( ( – 0)/10 ≥ (17.5 – 0)/10)
= Pr( z ≥ +1.75)
= 1 – 0.9599
= 0.0401
Since the P-Value > α/2 (0.0401 > 0.025) there is not
convincing evidence to reject the null-hypothesis at the
0.05 significance level.
3) A labor specialist thinks that the mean travel time to work
for all workers in Michigan is 33 minutes. A state congress
person believes that this is incorrect.
H0 : µ = 33
Ha : µ ≠ 33
Choose a value for α, say 0.01.
Suppose that 20 commuters, a random sample, measured
their time to commute to work and = 28.2 and s = 6. Is
this convincing evidence that the average commute time
is not 33 minutes. In other words, is this convincing
evidence to reject the null-hypothesis? We will calculate
the Pr( ≤ 28.2). Since  is not known we will use the tscore.
Assume that H0 is true, i.e. µ = 33.
Use µ = 33, and s = 6, and = 6/
= 1.34164.
Draw a picture
-two-tailed.
P-Value
= Pr( ≤ 28.2)
= Pr( ( – 33) / 1.34164 ≤ (28.2 – 33)/1.34164)
= Pr( t ≤ -3.58)
df = 20 – 1 = 19
= (1 – 0.998)/2
= 0.001
Since the P-Value < α/2 (0.001 < 0.005) there is convincing
evidence to reject the null-hypothesis at the 0.01
significance level.
4) A local state representative is willing to support casino
gambling if more than two-third of his constituents
support it.
H0 :  = 0.6667
Ha :  > 0.6667
Choose a value for α, say 0.05.
Suppose that a random sample of 500 people were
surveyed and 350 of them were in favor of casino
gambling.
Assume that H0 is true, i.e. π = 2/3
Use µp = 2/3, and σp = ( * / 500)0.5 = 0.02108185
P-Value
Draw a picture
= Pr(p > 350/500)
- one-tailed.
= Pr(p > 0.70)
= Pr( (p – 2/3)/0.02108185) > (0.70 – 2/3)/0.02108185) )
= Pr( z > 1.58)
= 1 – Pr(z <1.58)
= 1 – 0.9429
= 0.0571
Since the P-Value > α (0.0571 > 0.05) there is not
convincing evidence to reject the null-hypothesis at the
0.05 significance level.