ERRORS: TYPE 1 (a) TYPE 2 (b) & POWER
• Federal funds are to be allocated for financial aid to
students for the cost of books. It is believe the mean
cost of books to a college student is $300. We are
concerned that it is really higher. So what do we do?
We perform a hypothesis test.
• Ho: μ = 300 the mean cost of books is $300
• Ha: μ > 300 the mean cost of books is > $300
m = 300
If the true mean is really higher then we will not be giving
the students enough money to purchase books. So we take
a random sample of 36 students who purchased books and
the sample test data we collected has x = 325 s = 72 this
Let a = .01
m = 300
Is the p-value < .01 ?
x
=325
Lets assume the null is correct µ = 300 and the sample test
data we collected of x = 325 s = 12 gives a p-value =
.0186 Let a = .01. So what do we do?
m = 300
p-value = .0186
x
=325
Since the p- value = .0186 not < a = .01, we do not
reject the null.
m = 300
p-value = .0186
x
= 325
If null is wrong and the true mean is really µ = $335
then the real p-value for
= 325 is .2023.
x
m = 335
p-value = .2023
x
=325
If null is wrong and the true mean is really µ = $335
then. So what is the probability to make a wrong
decision? The red region below is known as β the
probability of a type 2 error.
m = 335
b
p-value =
x
=
So by overlapping the two graphs we can see what
to do. We need to find the data value that
corresponds to the alpha= .01
m = 335
Ho:m = 300
p-value = .01
p-value = ?
b
x
=?
The z score for alpha of .01 is 2.33
2.33 =
x  300
Solving this equation for
get a value of 328.
x we
72
36
m = 335
Ho:m = 300
p-value = ?
b
p-value = .01
x
= 328
We now need to find the p-value for 328 with the true
b 2 error.
μ = 335. This value is known as type
x


328

335
 = .2810 .28
P z 


72
36


So the p-value for 328
with a true mean of
335 is .28
m = 335
Ho:m = 300
p-value = .28
b
p-value = .01
x
= 328
Below we see Alpha and Beta together.
Ho:m = 300
m = 335
b
p-value = .28
α = .01
x
= 328
Now when the null is wrong then the area to the
right of 328 is the probability of making the correct
decision. This is called power.
Ho:m = 300
m = 335
p-value = .72
b
POWER
p-value = .28
1b
x
=328
We really do not know what the power is until
you know the true value for m.
m = 300
m = 335
p-value = .72
POWER
1b
x
=328
Now notice if m is even farther to the right of
300 then power gets even bigger.
m = 300
m = 350
p-value = .97
POWER
1b
x
=328
m = 335
m = 300
m = 350
p-value = .97
POWER
1b
x
= 328
Now if a = .05 then we would have rejected a false
null
Since the p- value = .0186 < a = .05, reject the null.
m = 300
p-value = .0186
a =
.05
x
= 325
When a is small then b is big, if the null is false.
So the probability of a type1 error is small,
rejecting a truthful null. But the probability of
making a type 2 error is big, failing to reject a
false null.
b
a
When a is big then b is smaller.
So the probability of a type1 error is bigger,
rejecting a truthful null. But we reduce the
probability of making a type 2 error. This
area is smaller, failing to reject a false null.
Truth
b
a
Not knowing whether the null is true or false is what
makes picking a correct alpha challenging.
So you need to determine which type of error you
do not want to make an then choose accordingly.
a=?
If you do not want to reject a truthful null pick a
small a. (Type 1 error) But you stand the chance
of making a type two error. Failing to reject the
null when it is false.
m
a=?
If you want to reject a false null, which decreases the
type 2 error pick a big a. But you increase the
chance of making a type one error. Rejecting the null
when it is true.
m
a=?
So looking at our problem what is the worse error?
Type 1 reject a truthful null. So we would give the students
more money when we did not need to. Big alpha will do this.
Type 2 error. Failing to reject a false null. Not giving the
students enough money. Small alpha will do this.
So you need to determine which type of error you do not
want to make an then choose accordingly.
a=?
• A doctor thinks that a new diet will significantly increases the birth
weight of babies. In 2002, the birth weight of full term babies were
normally distributed with a mean of 7.53 pounds and a standard
deviation of 1.15 pounds. The doctor randomly selects 50 recently
pregnant mothers and persuades them to partake in the new diet.
The mean weigh from these 50 babies 7.79 pounds. Is there
sufficient evidence to support the claim the new diet will increase
the birth weights of newborns? The significance level is 0.05.
• Write the null and alternative:
• Describe a type 1 and type 2 error.
• What is the probability for a type 1 error.
• We can not determine the probability of a type 2 error with out
knowing the truth.
We are done
x
x
ERRORS: TYPE 1 (a) TYPE 2 () & POWER
m = 1.5
a
1.578
m = 1.6
b
1.578
m = 1.5
m = 1.6
1.578
m = 1.5
m = 1.6
Power
1.578
m = 1.5
Power
1.578
m = 1.5
m = 1.6
Power
1.578
m = 1.5
p-value = .01
1.578
m = 1.6
b
p-value = .255
1.578
m = 1.5
b
p-value = .255
m = 1.6
p-value = .01
1.578
m = 1.5
m = 1.6
p-value = .745
b
POWER
p-value = .255
1b
1.578
m = 1.5
m = 1.6
p-value = .745
POWER
1b
1.578
m = 1.5
m = 1.65
p-value = .985
POWER
1b
1.578
m = 1.6
m = 1.5
m = 1.65
p-value = .985
POWER
1b
1.578