Single sample z test - explain (final)

Single-Sample Z Test
Theoretical Explanation

A one-sample Z-test for proportions is a test that helps
us compare a population proportion with a sample
proportion.

proportion.
Sample Population
30% 32%

proportion.
Here is our question: Are the population and the
sample proportions (which supposedly have the same
general characteristics as the population) statistically
significantly the same or different?
Sample Population
30% 32%

Consider the following example:

A survey claims that 9 out of 10 doctors recommend
aspirin for their patients with headaches. To test this
claim, a random sample of 100 doctors is obtained. Of
these 100 doctors, 82 indicate that they recommend
aspirin. Is this claim accurate? Use alpha = 0.05

Which is the sample?

What is the population?

The sample proportion is .82 – 82 out of doctors
recommend aspirin.

The sample proportion is .82 – 82 out of doctors
recommend aspirin.
The population proportion .90 (the claim that 9 out of
10 doctors recommend aspirin).

We begin by stating the null hypothesis:

The alternative hypothesis would be:
The proportion of a sample of 100 medical doctors who
recommend aspirin for their patients with headaches IS NOT
statistically significantly different from the claim that 9 out of
10 doctors recommend aspirin for their patients with
headaches.

The alternative hypothesis would be:
recommend aspirin for their patients with headaches IS NOT
headaches.
recommend aspirin for their patients with headaches IS
headaches.

State the decision rule: We will calculate what is called
the z statistic which will make it possible to determine
the likelihood that the sample proportion (.82) is a rare
or common occurrence with reference to the
population proportion (.90).

State the decision rule: We will calculate what is called
the z statistic which will make it possible to determine
the likelihood that the sample proportion (.82) is a rare
or common occurrence with reference to the
population proportion (.90).
If the z-statistic falls outside of the 95% common
occurrences and into the 5% rare occurrences then we
will conclude that it is a rare event and that the sample
is different from the population and therefore reject
the null hypothesis.

Before we calculate this z-statistic, we must locate the
z critical values.

Before we calculate this z-statistic, we must locate the
z critical values.
What are the z critical values? These are the values
that demarcate what is the rare and the common
occurrence.

Let’s look at the normal distribution:
It has some important properties that make it possible
for us to locate the z statistic and compare them to the
z criticals.

Here is the mean and the median of a normal
distribution.

50% of the values are above and below the orange
line.
50% - 50% +

68% of the values fall between +1 and -1 standard
deviations from the mean.
34% - 34% +

34% - 34% +
68%
mean-1σ +1σ

34% - 34% +
95%
mean-1σ +1σ-2σ +2σ

Since our decision rule is .05 alpha, this means that if
the z value falls outside of the 95% common
occurrences we will consider it a rare occurrence.
34% - 34% +
95%

Since are decision rule is .05 alpha we will see if the z
statistic is rare using this visual
95%
2.5% 2.5%
rarerare

Or common
95%
Common

Before we can calculate the z – statistic to see if it is
rare or common we first must determine the z critical
values that are associated with -2σ and +2σ.
95%
Common

We look these up in the Z table and find that they are -
1.96 and +1.96
95%
Common
-1.96 +1.96Z values

So if the z statistic we calculate is less than -1.96 or
greater than +1.96 then we will consider this to be rare
and reject the null hypothesis and state that there is
statistically significant difference between .9
(population) and .82 (the sample).

Let’s calculate the z statistic and see where if falls!

We do this by using the following equation:

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 =
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛

Zstatistic is what we are trying to find to see if it is
outside or inside the z critical values (-1.96 and +1.96).
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛

𝒑 is the proportion from the sample that
recommended aspirin to their patients (. 𝟖𝟐)
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛
A survey claims that 9 out of 10 doctors recommend aspirin for
their patients with headaches. To test this claim, a random
sample of 100 doctors is obtained. Of these 100 doctors, 82
indicate that they recommend aspirin. Is this claim accurate?
Use alpha = 0.05

𝐩 is the proportion from the population that
recommended aspirin to their patients (.90)
𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛
Use alpha = 0.05

𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛
Use alpha = 0.05
𝒏 is the size of the sample (100)

𝑝 − 𝑝
𝑝(1 − 𝑝)
𝑛
Use alpha = 0.05
Let’s plug in the numbers

.82 − 𝑝
𝑝(1 − 𝑝)
𝑛
Use alpha = 0.05
Sample Proportion

.82 − .90
.90(1 − .90)
𝑛
Use alpha = 0.05
Population Proportion

−.08
.90(1 − .90)
𝑛
Use alpha = 0.05
The difference

−.08
.90(1 − .90)
𝑛
Use alpha = 0.05
Now for the denominator which is the estimated
standard error. This value will help us know how many
standard error units .82 and .90 are apart from one
another.

−.08
.90(1 − .90)
𝑛
Use alpha = 0.05
If the standard error is small then the z statistic will be
larger. If it exceeds the -1.96 or +1.96 boundaries, then
we will reject the null hypothesis. If it is smaller than
we will not.

−.08
.90(1 − .90)
𝑛
Use alpha = 0.05
Let’s continue our calculations and find out:

−.08
.90(.10)
𝑛
Use alpha = 0.05

−.08
.09
𝑛
Use alpha = 0.05

−.08
.09
100
Use alpha = 0.05
Sample Size:

−.08
.0009
Use alpha = 0.05
Let‘s continue our calculations:

−.08
.03
Use alpha = 0.05

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 = −2.67
Use alpha = 0.05

𝒛 𝒔𝒕𝒂𝒕𝒊𝒔𝒕𝒊𝒄 = −2.67
Use alpha = 0.05
Now we have our z statistic.

Let’s go back to our distribution:
rarerare
95%
Common
-1.96 +1.96

Let’s go back to our distribution: So, is this result
rare or common?
rarerare
95%
Common
-1.96 +1.96-2.67

Looks like it is a rare event therefore we will reject the
null hypothesis in favor of the alternative hypothesis:

Looks like it is a rare event therefore we will reject the
null hypothesis in favor of the alternative hypothesis:
The proportion of a sample of 100 medical doctors
who recommend aspirin for their patients with
headaches IS statistically significantly different from
the claim that 9 out of 10 doctors recommend aspirin
for their patients with headaches.

Single sample z test - explain (final)

Recomendados

Recomendados

Mais conteúdo relacionado

Mais procurados

Mais procurados (20)

Destaque

Destaque (7)

Semelhante a Single sample z test - explain (final)

Semelhante a Single sample z test - explain (final) (20)

Mais de CTLTLA

Mais de CTLTLA (8)

Último

Último (20)

Single sample z test - explain (final)