Hypothesis Testing
z-test
The sampling distributions of a mean (SDM)
describes the behavior of a sampling mean
x ~ N  ,  x 
where  x 

n
Sampling distribution of a
mean is based on:
1. the central limit theorem,
2. the law of large numbers
(unbiased nature of the
sample mean)


We compare means of two samples taken
from specific sub-groups of the population.
We analyze the difference between two
means
The question under consideration:
“Is the difference between the samples large
enough to allow us to conclude (with a
known probability of error) that the
populations represented by the samples are
different?”
Example 1:
A quality engineer would like to determine
whether the production process he is charged of
monitoring is still producing products whose
mean response value is supposed to be 0
(process is in-control), or whether it is producing
products whose mean response value is now
different from the required value of 0 (process
is out-of-control).


Statement 1 (Null):  = 0 (process in-control)
Statement 2 (Alternative):   0 (process outof-control)
Example 2:
Suppose we want to estimate the difference
between the mean weights of all participants
before (μ1) and after (μ2) a weight loss program.
To accomplish this, suppose we take a sample of
40 participants and measure their weights before
and after the completion of this program.


Statement 1 (Null): 1 = 2 (No weight loss)
Statement 2 (Alternative):  2 < 1 (Significant weight
loss)
Consider the following situation:
Previous records state that the birth weights of
babies in England are normally distributed with
a mean of 3000g and a standard deviation of
500g.

We think that maybe babies in Australia have a
mean birth weight greater than 3000g and we
would like to test this hypothesis.




Convert the research question to null and
alternative hypotheses
The null hypothesis (H0) is a claim of “no
difference in the population”
The alternative hypothesis (Ha) claims “H0 is
false”
Collect data and seek evidence against H0 as a
way of bolstering Ha (deduction)
Null hypothesis, denoted H0
–Assuming that H0 is true, there is no difference
between the parameters of the two populations
 The mean birth weight is equal to 3000g.
H0 : μ= 3000g

Alternative hypothesis, denoted H1,
We reject H0 and say there is a difference between
the populations (If the difference between the sample
statistics is large enough).


The mean birth weight of Australian babies is
greater than 3000g.
H1 :μ >3000g



Once we have set up our null and alternative
hypothesis we can collect a sample of data.
Imagine that we take a sample of 44 babies from
Australia, measure their birth weights and we
observe that the sample mean of these 44
weights is 𝑋= 3275.955g.
Now, we want to calculate the probability of
obtaining a sample with mean as large as
3275.955 by chance under the assumption of the
null hypothesis H0.
We know from the previous lecture that, If
X1,X2,...Xn are independent and identically
distributed random variables from a N(μ,σ)
Then, x ∼N(μ,  )

n

Now, we can calculate the probability of
obtaining a sample with a mean as large as
3275.955 using standardization.
z stat 
x  0
x
where  0  population mean assuming H 0 is true
and  x 

n
75.378





This probability is called the p-value of the test.
In this case the p-value is very low.
But how low does this probability has to be
before we can conclude that the null hypothesis
is false?
Convention: choose a level of significance α, as
an indicator of a significant difference (α =0.01
or α =0.05).
We conclude that there is significant evidence
against the null hypothesis if the p-value is less
than or equal to 0.01 (or 0.05).


We have to look at the tail of the Standard
Normal distribution beyond the zstat.
Convert z statistics to P-value:
For Ha: μ > μ0  p= P(Z > zstat) = right-tail beyond zstat
For Ha: μ < μ0  p= P(Z < zstat) = left tail beyond zstat
For Ha: μ μ0  p= 2 × one-tailed p-value

Use the Table to find these probabilities.



In the baby weight example, the p-value is
0.00015 which is lower than α/2
(0.01/2=0.005).
In this case, we conclude that “there is
significant evidence against the null
hypothesis at the 0.01 level.”
Another way of saying this is that “we reject
the null hypothesis at the 0.01 level”.



In the 1970s, 20–29 year old men in the U.S.
had a mean μ body weight of 170 pounds.
Standard deviation σ was 40 pounds. We test
whether mean body weight in the population
now differs.
Null hypothesis H0: μ = 170 (“no difference”)
The alternative hypothesis can be either
Ha: μ ≠ 170 (two-sided test)
Sampling distribution of xbar
under H0: µ = 170 for n = 64 
x ~ N 170 ,5



For the illustrative example, μ0 = 170
We know σ = 40
Take an random sample of n = 64.
Therefore


40
x 

5
n
64
If we found a sample mean of 173, then:
zstat 
x  0
x
173  170

 0.60
5

If we found a sample mean of 185, then
x   0 185  170
zstat 

 3.00
x
5

One-sided p-value
AUC in tail beyond zstat

Two-sided p-value
consider potential
deviations in both
directions
 double the onesided p-value
•
If one-sided p = 0.0010,
then two-sided p = 2 ×
0.0010 = 0.0020.
•
If one-sided P = 0.2743,
then two-sided p = 2 ×
0.2743 = 0.5486.
If we choose α =0.05
•
For zstat = 3.0, two-sided p = 0.002,
•
•
•
•
•
In this case, 2 sided p-value is less than 0.05
Then, we reject the null hypothesis and say that the
mean weight of old men has increased significantly
since 1970s.
For zstat = 0.6, two-sided p = 0.5486.
In this case, 2 sided p-value is greater than 0.05
Then, we accept the null hypothesis and say that
there is no statistical difference between the mean
weight of old men from the 1970s until now.
1. Make assumptions and meet test
requirements
2. Define the null and alternative hypothesis
3. Select the sampling distribution and
establish the critical region
4. Compute the test statistic
5. Make a decision and interpret the test
results

Let X represent Weschler Adult Intelligence
scores (WAIS)
◦ Typically, X ~ N(100, 15)
◦ Take SRS of n = 9 from Lake Wobegon population
◦ Data  {116, 128, 125, 119, 89, 99, 105, 116, 118}
Calculate: x-bar = 112.8
 Does sample mean provide
strong evidence that population
mean μ > 100?

“where all the women are strong, all
the men are good-looking, and all the
children are above average”
A.
B.
Hypotheses:
H0: µ = 100
Ha: µ > 100 (one-sided)
Ha: µ ≠ 100 (two-sided)
Test statistic:

zstat
15
x 

5
n
9
x   0 112.8  100


 2.56
x
5
p-value: p = P(Z ≥ 2.56) = 0.0052
p =0.0052  For α/2 = 0.025, it is unlikely the
sample came from this null distribution  We reject
the null Hypothesis H0



Ha: µ ≠100
Ha considers random
deviations “up” and
“down” from μ0 tails
above and below ±zstat
Thus, two-sided P
= 2 × 0.0052
= 0.0104
2-sided p < α
(0.0104 <0.05)
 We reject the null
Hypothesis H0



In general, comparing two population means
is the way used to prove that one population
is different or better than another
Examples
◦ Competing Companies / Products
◦ Treatment vs. No treatment
◦ New method vs. Old method
1. Make assumptions and meet test
requirements
2. Define the null and alternative hypothesis
3. Select the sampling distribution and
establish the critical region
4. Compute the test statistic
5. Make a decision and interpret the test
results
Case study_ Safety of drinking water (Arizona
Republic, May 27, 2001)
 Water sampled from 10 communities in
Pheonix and 10 communities from rural
Arizona
 Determine if Arsenic concentration (As) is
different between these two areas?

To answer: “Whether or not there is a
difference between the means of Arsenic
levels (μ1 and μ2) in these two areas?” is
equivalent to test: whether μ1-μ2 is different
from 0.
Two independent populations


Question: Do men and women significantly differ
for their support of gun control?
For men (Sample 1)
– Mean 𝑋1 = 6.2
– Standard deviation s1 = 1.3
– Sample size N1= 324

For women (Sample 2)
– Mean 𝑋2= 6.5
– Standard deviation s2 = 1.4
– Sample size N2= 317
Three conditions:
The two samples are independent
The standard deviations σ1 and σ2 of the two
populations are known
3. At least one of the following conditions is
fulfilled:
i. Both samples are large (i.e., n1 ≥ 30 and n2 ≥
30)
ii. If either one or both sample sizes are small,
then both populations from which the samples
are drawn are normally distributed
So, the Central Limit Theorem applies and we can
assume a standard normal distribution (Z)
1.
2.
Null hypothesis, H0 : μ1=μ2
– The null hypothesis asserts there is no
difference between the 2 populations

Alternative hypothesis, H1: μ1≠ μ2
– The research hypothesis contradicts the
H0 and asserts there is a difference
between the populations

x1  x 2

When the 3 conditions are satisfied, the sampling
distribution of x1  x 2 is (approx.) normal with its
mean and standard deviation as follows:
 x  x  1  2
1
2
and
 x x 
1
2
 12
n1

 22
n2
Significance level
– Alpha (α) = 0.05 (two-tailed)
– The decision to reject the null hypothesis has
only a 0.05 probability of being incorrect

Z(critical)= ±1.96
– If the probability (p-value) is less than 0.05
– Z(obtained) will be beyond Z(critical)


Sample outcomes

Pooled standard error
1

2
0.1067
Z(obtained) = –2.80
–This is beyond z(critical) = ± 1.96
–The obtained z score falls in the critical region
–Therefore, the H0 is false and must be rejected

The difference between men and women is
statistically significant
- that we can conclude (at 95% confidence) that a
difference exists between men and women for
their support to gun control.




According to Kaiser Family Foundation survey in
2011 and 2010, the average premium for health
insurance for family coverage was $15,073 in 2011
and $13,770 in 2010 and the standard deviations
were$2,160 and $1,990, respectively.
Suppose that these averages were based on
random samples of 250 and 200 employees who
had such health insurance plans for 2011 and
2010, respectively.
Test at 1% significance level whether the population
means for the two years are different.
Step 1:

Population standard deviations, σ1 and σ2, are known

Both samples are large; n1 > 30 and n2 > 30

Therefore, we use the normal distribution to perform the
hypothesis test.
Step 2:

H0: μ1 – μ2 = 0 (The two population means are not different.)

H1: μ1 – μ2 ≠ 0 (The two population means are different)
Prem Mann, Introductory Statistics, 8/E
Copyright © 2013 John Wiley & Sons. All rights reserved.
Step 3:
α = 0.01.
The ≠ sign in the alternative hypothesis indicates that the
test is two-tailed
Area in each tail = α / 2 = 0.01 / 2 = 0.005
The critical values of z are 2.58 and -2.58
Prem Mann, Introductory Statistics, 8/E
Copyright © 2013 John Wiley & Sons. All rights reserved.
Prem Mann, Introductory Statistics, 8/E
Copyright © 2013 John Wiley & Sons. All rights reserved.
Step 4:
𝜎𝑥1 −𝑥2 =
=
𝑍=
𝜎12
𝑛1
+
𝜎22
𝑛2
2160 2
250
+
1990 2
200
= $196.1196
𝑥1 −𝑥2 −(𝜇1 −𝜇2 )
𝜎𝑥1 −𝑥2
$15,073 − $13,770 − 0
=
= 6.64
196.1196
From H0
Prem Mann, Introductory Statistics, 8/E
Copyright © 2013 John Wiley & Sons. All rights reserved.
Step 5:
Because the value of the test statistic z = 6.64 falls in the
rejection region, we reject the null hypothesis H0.
Therefore, we conclude that the average annual premiums
for employer-sponsored health insurance for family
coverage were different for 2011 and 2010.
Prem Mann, Introductory Statistics, 8/E
Copyright © 2013 John Wiley & Sons. All rights reserved.