OMS 201
Hypothesis testing
Determining whether a statement about a population parameter is correct
Hypothesis testing can take on three forms:
Remember, we can conclude to:
HO:  = 0
HA:   0
Reject the null hypothesis (Reject HO)
Do not reject the null hypothesis (Do Not Reject HO)
HO:   0
HA:  < 0
HO:   0
HA:  > 0
5 steps in Hypothesis testing:
1.
2.
3.
4.
5.
State the hypotheses (both the null and alternative) and set up the rejection region on the normal curve
State the level of significance and the Z critical value (good time to also indicate sample: mean, size, &
standard deviation)
State the test statistic and the P-value
State the decision rule (when do you reject HO ?)
State the conclusion
Some formulas to live by:
x 
z

x 
n
x
/ n
x  z / 2

x  t / 2
z
s
n
x
s/ n
x  z / 2
n
s
n
Hypothesis Testing Notes
p 
s
n
t  x 
/ n
0
z
p(1  p)
n
x
z

p  z / 2
p  p0
p
p(1  p )
n
t  x 
s/ n
0
1
Case 1. Sample size is large.
Hypotheses:
Upper tail test!
Rejection
region is
always based
on HA:
HO:   
HA:  > 
Level of Significance & sample data:
X-bar = 16.5
n = 40
=7
 = .02 (98% level of significance)
Z2.05 (look up .48 in normal table)
Test Statistic:
z
x
/ n
Z = (16.5 – 15)/(7 / 6.32)
Z = 1.35
P-Value: probability that we use to determine if we can reject the null hypothesis. It is the probability of
observing a result (such as X-bar) at least as unlikely as what is actually observed. To calculate
the p-value, calculate the probability of the area of the Z test statistic in rejection region. In this
example, the p-value would be the probability of observing a sample mean of greater than or
equal to 16.5.
Rejection
Region
P-Value: Calculate the area
for the Z test statistic score of
1.35: P(Z > 1.35) =
.5 - .4115 = .0885
1.35 2.05
Once you have the p-value, compare it to . This will determine if the null hypothesis
can be rejected.
Decision Rule:
Reject HO: if Z Test Statistic >ZOR:
Reject HO: if the p-value is < 
Conclusion:
Z= 2.05
Z test statistic = 1.35. Do NOT reject HO:
p-value = .0885  = .02. Do Not Reject HO: since p-value is greater than .
Hypothesis Testing Notes
2
Hypotheses:
H O:   
HA:  < 
Lower tail
test: rejection
region is
always based
on HA:
Level of Significance & sample data:
X-bar = 2.92
s = .18
n = 36
 = .05
Z = -1.645 (look up .45 in normal table) (negative value because it is in lower/left tail)
z
Test Statistic:
x
s/ n
Z = (2.92 – 3) / (.18 / 6)
Z=
-2.67
P(Z < -2.67) = .5 – .4962 = .0038
p-value:
Decision Rule: Reject HO: if Z Test Statistic < Z
OR: Reject HO: if the p-value is < 
Conclusion:
Since –2.67 < -1.645, Reject Ho:
Since p-value .0038 < .05, Reject Ho:
Rejection
region is set
up in both
tails
Hypotheses: (Two Tail Test)
HO:  = 
HA:   
Level of Significance & sample data:
X-bar = 278.5
 = .12
n = 36
 = .05
Z =  1.96 (look up .475 in table .05/2 = .025 and .5-.025 = .475)
Test Statistic:
p-value:
Decision
Rule:
z
x
/ n
Z = (278.5 – 280) / (.12 / 6)
Z=
-.75
P(Z < -.75) = .5 – .2734 = ..2266 – only one tail so: (.2266 * 2) = .4532
Reject HO: if Z Test Statistic < -ZOR
Reject HO: if Z Test Statistic is > Z
OR:
Reject HO: (p-value*2) < 
Hypothesis Testing Notes
Conclusion:
Since –.75 is greater than -1.96, Do
Not Reject Ho:
Since p-value .4532 greater than .05,
Do Not Reject Ho:
3
A Two-tail test can also calculate a confidence level to determine if the null hypothesis can be rejected:
Hypotheses: (Two Tail Test)
HO:  = 
HA:   
Level of Significance & sample data:
X-bar = 278.5
 = .12
n = 36
 = .05
Z =  1.96 (look up .475 in table .05/2 = .025 and .5-.025 = .475)
Confidence Intervals:
x  z / 2

x  z / 2
n
278.5  1.96 (12/6)=
s
n
278.5  3.92 =
274.58 to 282.42
Decision Rule:
Do NOT Reject Ho: if the population parameter is within the interval.
Conclusion:
Since  = 280 falls within the interval, we can NOT reject Ho:
Case 2. n < 30
Normal distribution is assumed and s is used for . Additionally, we will use the t-distribution.
Level of significance
t with
Test Statistics:
t  x 
/ n
n-1 degrees of freedom
0
t  x 
s/ n
0
Decision Rules:
HO:  = 0
HA:   0
HO:   0
HA:  < 0
HO:   0
HA:  > 0
Reject:
t < -t critical value or
t > t critical value or
p-value < 
Reject:
t < -t critical value
or
p-value < 
Reject:
t > t critical value
or
p-value < 
p-values: in the t-distribution, we do not have exact values, so we will use a range. t-distribution gives the
area in the upper tail.
Hypothesis Testing Notes
4
Hypotheses:
H O:   
HA:  > 
Level of Significance & sample data:
X-bar = 7.75
s = 1.215
n = 12
 = .05
t = 1.796 (.05 with 11 degrees of freedom)
Test Statistic:
t  x 
s/ n
t = (7.75 – 7) / (1.215 / 3.46)
0
t = 2.14
p-value: the following areas for 11 (n-1) degrees of freedom
Area
t-value
.1
1.363
.05
1.796
.025
2.201
.01
2.718
.005
3.106
2.14 is between .025 and .05
Conclusion:
Decision Rule: Reject HO: if t Test Statistic > t
OR: Reject HO: if the p-value is < 
Since 2.14 > 1.796, Reject Ho:
Since p-value is < .05, Reject Ho:
In a two tail test: make sure to multiply the p-value by 2!
Case 3. Population Proportion
State the Hypotheses:
HO: p = P0
HA: p  P0
HO: p  P0
HA: p < P0
HO: p  P0
HA: p > P0
State the Test Statistic and p-value:
z
p  p0
p
(p-values are calculated the same as before)
Decision Rules:
HO: p = P0
HA: p  P0
HO: p  P0
HA: p < P0
HO: p  P0
HA: p > P0
Reject:
z < -z critical value or
z > z critical value or
p-value < 
Reject:
z < -z critical value
or
p-value < 
Reject:
z > z critical value
or
p-value < 
Hypothesis Testing Notes
5
Hypothesis Testing Notes
6