This preview shows page 1. Sign up to view the full content.
Unformatted text preview: red to have a weekly labor supply (hours
worked) of 30 hours. The standard deviation is calculated to be s = 10. You believe that
in the population mean hours worked per week is less than 32 hours/week. Perform a
test to support your idea.
Step 1: The null and alternative hypotheses are:
H0 :µ ≥ 32
H1 :µ < 32 Utku Suleymanoglu (UMich) Hypothesis Testing 29 / 37 σ not Known Step 2: Calculate the relevant test statistic:
t= x − µ0
¯
30 − 32
−2
√=
=
= −6.06
10/30
0.33
s/ n Step 3: Find the critical value: Let’s pick α = 0.05. Then tα,n−1 = t0.05,899 ≈ 1.645
(Table 2 in Appendix B) (n is too large, use inﬁnity as df).
Step 4: Use the decision rule: Reject H0 if t < −tα,n−1 .
t = −6.06 < −1.645 ≈ −t0.05,899 , so reject the null.
Finally, state your conclusion: We have enough evidence to reject the claim that µ ≥ 32.
So our claim that it is less than 32 is supported by the data. Utku Suleymanoglu (UMich) Hypothesis Testing 30 / 37 σ not Known TwoTailed Tests: σ Unknown We will again use tdistribution instead of standard normal. First state the null and
alternative hypotheses.
H 0 : µ = µ0
H 1 : µ = µ0
We reject the null the hypothesis if test statistic (same as onetailed tests):
t= x − µ0
¯
√
s/ n is such that
t  > tα/2,n−1
where tα/2 is the critical value with probability α/2 in the upper tail.
So if test statistic falls outside the interval (−tα/2 , tα/2 ), the null is rejected. Utku Suleymanoglu (UMich) Hypothesis Testing 31 / 37 σ not Known Example Remember the previous data set that consists of 900 people aged 65. They are measured
to have a weekly labor supply (hours worked) of 30 hours. The standard deviation is
calculated to be s = 10. Now, somebody claims that mean hours worked is exactly 29.5.
Test this claim.
Step 1: The null and alternative hypotheses are:
H0 :µ = 29.5
H1 :µ = 29.5 Utku Suleymanoglu (UMich) Hypothesis Testing 32 / 37 σ not Known Step 2: Calculate the relevant test statistic:
t= x − µ0
¯
0.5
30 − 29.5
√=
=
= 1.515
10/30
0.33
s/ n 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).
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.
Note that 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 (...
View
Full
Document
This note was uploaded on 03/17/2014 for the course ECON 404 taught by Professor Staff during the Spring '08 term at University of Michigan.
 Spring '08
 STAFF

Click to edit the document details