test statistics - LICH

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Hypothesis testing
• Null hypothesis Ho - this hypothesis holds
that if the data deviate from the norm in any
way, that deviation is due strictly to chance.
• Alternative hypothesis Ha - the data show
something important.
• Doing decision = accept/reject Ho (the
decision centers around null hypothesis)
Errors in hypothesis testing
• Type I – False Positive
• Type II – False Negative
• The probability of Type I error: α
• The probability of Type II error: β
• Test involving sample from a normally
distributed population.
• Because it’s a normal distribution, you use zscores in the hypothesis test.
z
x

n
• The z-score here is called test statistics.
• The test statistics constructed according the
above formula holds only for the mean.
• Tests for other statistics (e.g. variance) use
different formulas.
• Suppose you think that people living in a
particular zip code have higher-than-average
IQs. Your data are given in sheet ZIP, test this
hypothesis.
• n = 16, μZIP = 107.75, α = 0.05
• We know about IQ scores: μ = 100, σ = 16
• What will be the Ho and Ha?
– Ha: μZIP > 100
– Ho: μZIP ≤ 100
• Can you reject Ho?
z
x

n

107.75  100
16

7.75
4
16
 1.94
What is the value of z that
cuts off 5% of the area in a
standard normal
distribution?
It’s exactly 1.645.
So what’s the decision?
The calculated value, 1.94,
exceeds 1.645, so it’s in the
rejection region. The
decision is to reject Ho.
• This hypothesis test is called one tailed (one sided).
• The rejection region is in one tail of the sampling
distribution.
• A hypothesis test can
be one tailed in the
other direction.
– Ha: μZIP < 100
– Ho: μZIP ≥ 100
• What is the critical
value?
-1.645
http://www.emathzone.com/tutorials/basic-statistics
• Test can be also two-tailed.
• The rejection region is in both tails of the Ho
sampling distribution.
– Ho: μZIP = 100
– Ha: μZIP ≠ 100
• What is the critical value now?
– Find z-score that cuts 2.5% from right (=1.96) and
from left (=-1.96).
– 1.94 does not exceed 1.96, we do not reject Ho.
http://www.emathzone.com/tutorials/basic-statistics
• Using one tailed test we rejected Ho, while
using two tailed test we did not!!
• A two tailed test indicates that you’re looking
for a difference between the sample mean
and the null-hypothesis mean, but you don’t
know in which direction.
• A one tailed test shows that you have a pretty
good idea of how the difference should come
out.
• For practical purposes, this means you should
try to have enough knowledge to be able to
specify a one tailed test.
z-test in Excel
• ZTEST
– Do now: examples2.xlsx | ZIP
– provide sample IQ data, null hypothesis
value, σ (if omitted, s is used)
– p-value is returned
– If p-value < α, reject Ho.
– Will you reject Ho or not?
This is the result of ZTEST
• For one tailed test you reject Ho.
• What if you do two tailed test?
Critical value for
one tailed test
α = 0.05
Our actual value (red line, pvalue = 0.026) is in the rejection
region of one tailed test (0.026 <
0.05).
However, it is outside rejection
region for two tailed test. To see
this, you must compare 0.026 >
0.025. Or you can 2x multiply this
equation → 0.052 > 0.05.
So if you have α set to 0.05, and
you get p-value for one sided
test, you get p-value for two
sided test doubling the one sided
p-value.
t for one
• In the real world you typically don’t have the
luxury of working with such well-defined
populations as results of IQ test.
• Real world:
– small samples
– you often don’t know the population parameters
• When that’s the case, you
– use the sample data to estimate the population
standard deviation
– you treat the sampling distribution of the mean as a tdistribution
– You use t as a test statistic
• The formula for the test statistic
t
x
s
n
with DF = n – 1. The higher the DF, the more
closely the t-distribution resembles the
normal distribution.
• Company claims their vacuum cleaner averages
four defects per unit. A consumer group believes
this average is higher. The consumer group takes
a sample of 9 cleaners and finds an average of 7
defects, with a standard deviation of 3.16.
• Is companie’s claim correct or not?
• Ho, Ha?
– Ho: μ ≤ 4
– Ha: μ > 4
• And what else is missing in defining the
hypothesis?
– α = 0.05
• Now calculate t test statistic
t
x
s

74
 2.85
3.16
n
9
• Can you reject Ho?
– Get critical value from tables or TINV.
• Use Excel TDIST
– returns p-value
• reject Ho
Testing a variance
• The family of distributions for the test is called
chi-square - χ2
• The formula
for test statistics
 
2
 n  1 s 2

2
With this test, you have to assume that what you’re measuring has a normal distribution.
• Solve the following example using CHIDIST.
• You produce a part of some machine that has to be a
certain length with at most a standard deviation of
1.5 cm.
• After measuring a sample of 26 parts, you find a
standard deviation of 1.8 cm.
• Is your process producing these parts OK?
– Ho: σ2 ≤ 2.25 (remember to square the “at-most” standard
deviation of 1.5 cm)
– Ha: σ2 > 2.25
– α = 0.05
 
2
 n  1 s

2
2

25 1.8 
1.5 
2
2
 36
p-value = 0.0716. Do not
reject Ho.
Two sample hypothesis testing
• Compare one sample with another.
• Usually, this involves tests of hypotheses about
population means. You can also test hypotheses
about population variances.
• Here’s an example. Imagine a new training
technique designed to increase IQ. Take a sample
of 25 people and train them under the new
technique. Take another sample of 25 people and
give them no special training. Suppose that the
sample mean for the new technique is 107, and
for the no-training sample it’s 101.2.
– Did the technique really increased IQ?
• Same principles: Ho (no difference between
means), Ha, α
– one-tailed test
• Ho: μ1 – μ2 = 0, Ha: μ1 – μ2 > 0
• Ho: μ1 – μ2 = 0, Ha: μ1 – μ2 < 0
– two-tailed test
• Ho: μ1 – μ2 = 0, Ha: μ1 – μ2 ≠ 0
– The zero is typical case, but it’s possible to test for
any value.
The first sample in the pair always
has the same size, and the second
sample in the pair always has the
same size. The two sample sizes
are not necessarily equal.
CLT strikes again
• If the samples are large, the sampling distribution of the
difference between means is approximately a normal
distribution.
• If the populations are normally distributed, the sampling
distribution is a normal distribution even if the samples are
small.
• The mean of the sampling distribution
 x  x  1   2
1
2
• The standard deviation of the sampling distribution (standard
error of the difference between means)
1
2
 x x 
1
2
n1
2
2

n2
• Because CLT says that the sampling
distribution is approximately normal for large
samples (or for small samples from normally
distributed populations), you use the z-score
as your test statistic.
• i.e. you perform a z-test.
• The z test statistics:
z
 x1  x 2     1   2 
x
1
x2
• Solve the following.
• Imagine a new training technique designed to
increase IQ. Take a sample of 25 people and
train them under the new technique. Take
another sample of 25 people and give them
no special training. Suppose that the sample
mean for the new technique sample is 107,
and for the no-training sample it’s 101.2.
• Did the technique really increased IQ?
• Ho: μ1 – μ2 = 0, Ha: μ1 – μ2 > 0, α = 0.05
• The IQ is known to have a standard deviation of
16, and I assume that standard deviation would
be the same in the population of people trained
on the new technique.
z
 x1  x 2     1   2 
 x x
1
2

107  101.2 
16
2
25

16
2

5.8
 1.28
4.53
25
• Use either NORMSDIST (supply 1.28, you get pvalue = 1-0.899=0.101) or NORMSINV (probability
= 0.95, you get critical value equaling to 1.645).
• Accept Ho.
• Excel provides a tool z-Test: Two Sample for
Means (Data | Data Analysis)
• Do now
– IQ_Test sheet
Variable variance is 162 = 256 (16 is population standard
deviation of IQ test distribution)
t for Two
• The previous example involves a situation you
rarely encounter - known population
variances.
• Not knowing the variances takes the CLT out
of play. This means that you can’t use the
normal distribution as an approximation of
the sampling distribution of the difference
between means.
• Instead, you use the t-distribution. You
perform a t-test.
• Unknown variances lead to two possibilities
for hypothesis testing:
– although the variances are unknown, you have
reason to assume they’re equal
– you cannot assume they’re equal
t for Two – equal variances
• Put sample variances together to estimate a
population variance – pooling
sp 
2
 N 1  1  s12   N 2  1  s 22
 N 1  1   N 2  1
DF
t 
 x1  x 2     1   2 
sp
1
N1

1
N2
• FarKlempt Robotics is trying to choose between two machines
to produce a component for its new microrobot. Speed is of
the essence, so they have each machine produce ten copies of
the component, and time each production run.
• Which machine should they choose? Do now using Data
Analysis, Mechine_speed sheet.
s 
2
p
 N 1  1 s   N 2  1 s
 N 1  1   N 2  1
2
1
2
2
t 
 x1  x 2     1   2 
sp
1
N1

1
N2
• Ho: μ1 - μ2 = 0, Ha: μ1 - μ2 ≠ 0, α = 0.05
• This is a two-tailed test, because we don’t
know in advance which machine might be
faster.
Get critical value using TINV (+-2.10)
or
p-value using TDIST (0.0252).
Result: reject Ho.
• The worksheet function TTEST eliminates the muss, fuss,
and bother of working through the formulas for the t-test.
• Do now – Machines example in examples2.xlsx |
Machine_speed
It’s more desirable to use the equal variances t-test, which typically
provides more degrees of freedom than the unequal variances t-test.
• Do now – Data|Data Analysis, use t-Test: TwoSample Assuming Equal Variances
t for Two – unequal variances
• In the case of unequal variances, the t
distribution with (N1-1) + (N2-1) DF is not as
close an approximation to the sampling
distribution.
• DF must be reduced, fairly involved formulas
are used to do this.
• A pooled estimate is not appropriate. t-test is
calculated as
x  x      
t
1
2
1
2
s1
N1
2

s2
N2
2
Testing two variances
• classic: Ho: σ12 = σ22, Ha: σ12 ≠≥≤ σ22, α=0.05
• When you test two variances, you don’t
subtract one from the other. Instead, you
divide one by the other to calculate the test
statistic.
• This statistics is called F-ratio, and you’re
doing F-test.
F 
larger s
smaller
2
s
2
• The family of distributions for the test is called the Fdistribution.
• Each member of the family is associated with two
values of DF (each DF is n - 1)!
• And it makes a difference which DF is in the
numerator and which DF is in the denominator.
• One use of the F-distribution is in conjunction
with the t-test for independent samples.
• Before you do the t-test, you use F to help decide
whether to assume equal variances or unequal
variances in the samples.
• Excel: FTEST, FDIST, FINV, F-Test: Two-Sample for
Variances
• Do now
– FarKlempt Robotics produces 10 parts with Machine 1
and finds a sample variance of .60 cm2. They produce
15 parts with Machine 2 and find a sample variance of
.44 cm2. Are these variances same?
– Data are in the examples2.xlsx | Machine_var
F 
0 . 60
 1 . 36
0 . 44
• Estimate variances from data using VAR.
• FDIST value is exactly ½ of FTEST. (FDIST is
one-tailed, FTEST is two-tailed)
• It finds a critical value.
• Probability is 0.025, because two tailed test is
with α = 0.05.
For future use
http://www.intuitor.com/statistics/CurveApplet.html
β
power
α
• Theoretically, when you test a null hypothesis
versus an alternative hypothesis, each
hypothesis corresponds to a separate
sampling distribution.
• When you do a hypothesis test, you never
know which distribution produces the results.
You work with a sample mean - a point on the
horizontal axis. It’s your job to decide which
distribution the sample mean is part of. You
set up a critical value - a decision criterion. If
the sample mean is on one side of the critical
value, you reject Ho. If not, you don’t.

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