STATISTICS - CH 10, LECTURE
10.1 Tests
of Hypotheses
Be able to identify a null hypothesis and an alternate hypothesis
Be able to identify a type 1 error and a type 2 error
Note that the probability of a type 1 error is the significance level α
In hypothesis testing, we should minimize the probability of a type 1 error; α is usually set at .01 or .05
10.2 Significance Tests
One-sided tests vs. two-sided tests
α is the total tail area; so for a two-sided test, each tail has an area of α/2
10.3-4 Tests Concerning Means
5-step process
1. Identify hypotheses:
H0: μ = claimed value
HA: μ > or < or ≠ claimed value
> or <: one-tailed tests
≠: two-tailed test
2. Set the level of significance
α = .01 corresponds to 99% confidence
α = .05 corresponds to 95% confidence
3. Determine the criterion for rejection of the null hypothesis
Find critical value of the test statistic (z for n≥30, t for n<30)
t or z: based on α
One-tailed test with z: table area = .5−α; z could be pos or neg.
One-tailed test with t: subscript for t is α
Two-tailed test with z: table area = .5−α/2; z could be pos or neg
Two-tailed test with t: subscript for t is α/2
p: critical value is α
Write an inequality statement, such as: Reject H0 if F>1.96
For p-value test, "reject H0 if p < α
***You should do all this before collecting your data.
It is fundamentally dishonest to decide on your criteria after you know survey results.
4. Calculations: calculate the test statistic based on the sample data
t or z: use formula for central limit theorem
p: Extra step: translate t or z to the total tail area (i.e. both tails for a 2-tail test)
5. Decision: Compare the test statistic based on the data (step 4) with the critical value (step 3)
Do we reject the null hypothesis, or do we "reserve judgment?"
Note: It does not make sense, based on hypotheses testing, to accept the null hypothesis. By setting the significance level to minimize the probability of a type I error, we have a relatively high probability of a type II error. Remember that hypothesis testing is done to challenge an existing claim about a mean.
If what we really want to do is validate the mean, we should not be hypothesis testing. Instead, we should find a 95% or 99% confidence interval for the mean.
CAN YOU USE A COMPUTER (without buying a special statistics package)?
It is possible, if not very user-friendly, to use Excel for hypothesis testing
The example below includes two solved problems from the text, a 2-tail test and a 1-tail test; and it examines them both as large samples and as small samples, and it also shows the use of the p value.
You have to enter the formula for the test statistic yourself, but functions do exist in Excel for calculating the critical value of t or z, as well as the p-value of the data.
Notice, in the formulas for the critical values (in the Criteria column), that:
The inverse standard normal formula assumes a 1-tail test; divide α by 2 for a 2-tail test.
The inverse t formula assumes a 2-tail test; multiply α by 2 for a 1-tail test.
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p333, X10.11 |
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Criteria |
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Data: |
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Concl. |
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Large |
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n=60 |
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n |
x-bar |
s |
z or t |
p |
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H0: m = |
HA: m<>12.8 |
LS 0.01 |
z<-? Or >-? |
p<LS of .01 |
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12.8 |
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0.01 |
-2.575829304 |
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60 |
11.2 |
3.5 |
-3.541013345 |
0.000785564 |
reject |
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Formulas: |
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NORMSINV(.005) |
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(xbar - m)/(s/sqrt(n)) |
TDIST(abs(z),df,tails) |
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Small |
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n=20 |
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t<-? Or >? |
p<LS of .01 |
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12.8 |
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0.01 |
2.860934604 |
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20 |
11.2 |
3.5 |
-2.044405008 |
0.05502263 |
don't rej |
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Formulas: |
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TINV(.01,19) |
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(xbar - m)/(s/sqrt(n)) |
TDIST(abs(t),df,tails) |
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p335,X10.21 |
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Criteria |
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Data: |
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Concl. |
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Large |
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n=60 |
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n |
x-bar |
s |
z or t |
p |
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H0: m=9 |
HA: m<9 |
LS .05 |
z<-? |
p<LS of .05 |
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9 |
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0.05 |
-1.644853627 |
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60 |
7.2 |
1.8 |
-7.745966692 |
7.36365E-11 |
reject |
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Formulas: |
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NORMSINV(.05) |
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(xbar - m)/(s/sqrt(n)) |
TDIST(abs(z),df,tails) |
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Small |
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n=12 |
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t<-? |
p<LS of .05 |
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9 |
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0.05 |
1.795884814 |
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12 |
7.2 |
1.8 |
-3.464101615 |
0.002647366 |
reject |
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Formulas: |
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TINV(0.1,11) |
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(xbar - m)/(s/sqrt(n)) |
TDIST(abs(t),df,tails) |
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10.5 Differences Between Means (Large Samples)
Uses the statistic 
10.6 Differences Between Means (Small Samples)
Omit: We will use 10.9 or 13.6 for this
10.7
Use for before-and -after comparisons.
Use a t-or z-statistic based on the signed differences between the before and after values
10.8 Differences
Among Several Means
Concept: use the F-statistic, which is (Variation among the samples) / (Variation within the samples)
Definition: (n times variance of
means) / (mean of variances) 
10.9 Analysis
of Variance (ANOVA)
Collect, for each sample, statistics for n, x, and x2. Summarize as follows:
= N
ΣΣx
ΣΣx2
Σ [(Σx)2/n]
Summary statistics:
Based on the totals, calculate:
Calculations: 
Where
k = number of treatments (or data sets)
n = sample size of each data set
N = Σn = total 1
SST = ΣΣx2 - (ΣΣx)2 / N, = total 3 - (total 2)2 / total 1
SS(Tr) = Σ[(Σx)2/n] - (ΣΣx)2 / N = total 4 - (total 2)2 / total 1
SSE = SST -SS(Tr)
MS(Tr) = SS(Tr) / (k-1)
MSE = SSE / (N-k)
F = MS(Tr) / MSE
When calculating the critical value of F,
DF (numerator) = DF (Treatments) = k - 1, and
DF (denominator) = DF (Error) = N - k.
Minitab example for ANOVA (One-way analysis of variance), based on the data on pg 346:
Tests whether the difference among several means is significant
All data goes into column 1; column 2 tells which data set the entry comes from.
In this example, the first 4 items are from set 1, the next 4 from set 2, and the last 4 from set 3.
Here is the same calculation, with Excel. The data from each sample is in a different column.
To run this test, call up the Analysis Tool-Pack, and ask to do a Single-Factor ANOVA:
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65 |
80 |
72 |
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Anova: Single Factor |
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69 |
84 |
76 |
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71 |
86 |
77 |
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SUMMARY |
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75 |
90 |
79 |
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Groups |
Count |
Sum |
Average |
Variance |
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Column 1 |
4 |
280 |
70 |
17.33333333 |
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