Transcript Hypothesis Test

Chapter 12
Tests of a Single
Mean When σ is
Unknown
A Research Question
Children’s growth is stunted by a
number of chemicals (lead, arsenic,
mercury)
 The tap water in the local community
contains a bit of each of these
chemicals
 Are children in this town smaller than
other children their age?

A Research Project
16 (n = 16) 6-yr old children are
randomly selected from around town
 Each child’s height is measured
 In the US the average height of 6-yr
olds is 42” (μ = 42)
 The variance of 6-yr-old’s height,
however is not known

The Data

The 16 kid’s heights were:
44, 38, 42, 37, 35, 41, 46, 39,
40, 42, 34, 39, 41, 42, 45, 35
Hypothesis Test
1.State and Check Assumptions
Heights normally distributed ? - probably
(n = 16 large enough)
Interval level data
Random Sample
Population variance unknown
Hypothesis Test
2.
Null and Alternative Hypotheses
HO : μ = 42 (6-yr old’s height is 42”)
HA : μ < 42 (6 yr-old’s height is less than 42”)
Hypothesis Test
3.Choose Test Statistic
Parameter of interest - μ
Number of Groups - 1
Independent Sample
Normally distributed - probably
Variance - unknown
What do we do?
z-test requires that we know the
population standard deviation (σ)
 Can we use s as a substitute for σ?
 Not with a z statistic, but…
 We can use s with a t statistic
(Student’s t) and a t sampling
distribution

Single Sample t statistic
M-m
t=
sM
where :
sX
sM =
n
Back to the Hypothesis Test
3.Choose the test statistic
Parameter of interest - μ
Number of Groups - 1
Independent Samples
Normally distributed - probably
Variance - unknown
One Sample t-test
Hypothesis Test
4.Set significance level
α = .05
critical value is found in table C
What’s a df?
Degrees of Freedom (df)

Degrees of Freedom (df) - the number
of components in a statistic’s
calculation that are free to vary
df Explained


If you have a M = 10 obtained from 5 scores,
what are the scores?
Let’s say the first four are 15, 10, 11, and 5
– in this case the last score has to be 9, in order to
have a mean of 10

As a second example, let’s say the first four
are 8, 14, 3, and 11
– the last score has to be 14 in order to have a
mean of 10
df Explained



Therefore, the first 4 scores can vary, the
fifth score is not free to vary - it must take on
some value (in order to maintain the mean
of 10)
In our example, there are 4 degrees of
freedom
The first four scores can take on any value
(they are free to vary), but that last one is
fixed in order to maintain the mean
One Sample t test

In a one sample t test the degrees of
freedom are always equal to n - 1
– df = n -1
Back to the Hypothesis test
4. Set significance level and make decision rule
α = .05
df = n -1 = 16 - 1 = 15
critical value at .05 of t(15) = 1.753
(read: “critical value at .05 of a t test with 15 degrees of freedom is 1.753”)
But, since we have a directional hypothesis (<
42), then the critical value is -1.753
Thus, if our computed t ≤ -1.753, we reject HO
Or…

If we compute the p-value associated
with our t, with 15 df, we can state the
decision rule as:
– If p ≤ α, Reject the HO
Hypothesis Test
5.Compute test statistic
M-m
t=
sM
need :
M, m,s X ,sM ,n,df
Hypothesis Test
6. Draw conclusions
Since our obtained t (-2.236) is less
than the critical t (-1.753) we,
Reject HO, and conclude
That our town’s 6-yr olds are smaller,
on average, than 6-yr olds in the US
Careful…a warning
We have rejected the HO and
concluded that our town’s 6-yr-olds are
smaller, on average, than 6-yr-olds in
the US
 But, we are not allowed, in this case,
to conclude that it is because of
chemicals in the water, or any other
cause

Alternative Explanations

There are likely many causes for children’s
small stature, not limited to:
–
–
–
–
–

Genetics
Diet
Environmental contaminants
Chemicals in ground water
Etc.
The hypothesis test allows us to conclude that
these children are smaller, on average, but
does not allow us to say why
Before we move on…
Although we already rejected the null
hypothesis,
 We can determine the actual
probability of our results if the null
hypothesis were true (p-value)
 We know that it is less than .05, but
how much less?

Ugghh!!!
Excel recognizes only
positive values for
a t distribution, but
because the t is
symmetrical, use the
absoute value
function (ABS) to
find the p-value