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Unformatted text preview: late the relevant test statistic:
t= 30 − 29.5
0.5
x − µ0
¯
=
= 1.515
√=
s/ n
10/30
0.33 Step 3: Find the critical value: Let’s pick α = 0.05. Then tα/2,n−1 = t0.025,899 ≈ 1.96 (Table 2
in Appendix B) (n is too large, use inﬁnity as df, note that this is the same critical value as z).
Step 4: Use the decision rule: Reject H0 if t  > tα/2,n−1 . t = 1.515 is outside the rejection
region (−1.96, 1.96). We failed to reject the null.
As a ﬁnal step, state your conclusion: We do NOT have enough evidence to reject the claim that
µ = 29.5.
Big picture note: this does not mean that we know that µ = 29.5. We would not have rejected
the null with 29.6 or 30 either! Utku Suleymanoglu (UMich) Hypothesis Testing 34 / 39 σ not Known MIDTERM TOPICS END HERE! This means the hypothesis tests that you will need to perform in the second midterm II will have
normal distribution as the distribution of the population values. The remainder of the chapter will be included in the last midterm. Utku Suleymanoglu (UMich) Hypothesis Testing 35 / 39 σ not Known Case 3: σ not Known, Population Not Normal The ttest relies on the assumption that population values are distributed normally.
If not, we can still test hypothesis thanks to CLT, but these are going to be approximations which
are close to true values in large samples.
To make the procudure work: you switch back to standard normal distribution even though σ is
not known.
¯ −µ0
So the test statistic is zt = x /√n , and ztest rules apply. I gave this test statistic a slightly
s
diﬀerent name so that you don’t confuse it with the case 1 statistic. PLEASE BE CAREFUL: We have the same formula for test statistic for cases 2 and 3. The
diﬀerence is
Case 2: Population values are normally distributed, so test statistic has tdistribution. Hence
called “t”. And you use the ttable.
Case 3: Population values are from unknown distribution, so test statistic is approximately
normal. So it is called “z” and ztable is the relevant one.
If one knows that population has normal distribution, “ttest” must be performed (especially with
small n). This yields more accurate test results. Utku Suleymanoglu (UMich) Hypothesis Testing 36 / 39 Hypothesis about Population Proportions Hypothesis Tests about Population Proportion We have learned
Sampling distribution of p
¯
Building conﬁdence intervals for p using p
¯
These worked only in large samples (CLT). Using the same ideas, we can perform hypothesis
testing for population proportions.
The test statistic you will need will be a zstatistic:
z= p − p0
¯
p0 (1−p0 )
n where p0 is the value in the null hypothesis. Note that denominator uses p0 , not p .
¯
And critical values looked up from the standard normal table. Utku Suleymanoglu (UMich) Hypothesis Testing 37 / 39 Hypothesis about Population Proportions Example Suppose you work at a polling company and you are interested the proportion of Americans who
support Obamacare. In particular, you are interested in whether the majority of people support it.
Your company interviewed 10, 000 people, 5, 200 of which said they support Obamacare. Perform
a test to support the Democratic claim that the majority of Americans support Obamacare with
0.01 signiﬁcance level. [this is ﬁctitious data]
We have a large enough sample, so we can do hypothesis testing.
The null and the alternative:
H0 :p ≤ 0.50
H1 :p > 0.50 Utku Suleymanoglu (UMich) Hypothesis Testing 38 / 39 Hypothesis about Population Proportions p = 0.52 so the test statistic is:
¯
z= 0.52 − 0.50
0.5(1−0.5)
10000 = 0.02
= 40
0.005 Find the critical value zα : z0.01 = F (2.33) = 0.99
Because this is righttailed test, decision rule says “reject H0 if z > zα ”.
So we reject the null hypothesis that says a minority supports Obamacare. The ﬁctitious data
supports the Democratic claim. Utku Suleymanoglu (UMich) Hypothesis Testing 39 / 39...
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