Download Sec. 7.2b Notes Critical Values and Rejection Regions

What is the P-value for a two-tailed z-test with
z = 1.91?
0.0561
Review 7.2a and collect assignment
Sec. 7.2b Notes
Critical Values and Rejection Regions
Mr. Ricks
Madison High School
A rejection region (critical region) is the range of
values for which H0 is not probable
If the test statistic falls in the rejection region, H0
is rejected
A critical value (z0) separates the rejection
region from the nonrejection region
Finding Critical Values for a Normal
Distribution
1. Specify the level of significance α
2. Decide if test should be left, right, or two-tailed
3. Find z0 for the following areas
a) Left-tailed: find z-score for area α
b) Right-tailed: find z-score for area 1 – α
c) Two-tailed: find z-score for
1
𝛼
2
and state ± cases
4. Sketch the standard normal distribution, showing
the critical value(s) and shading the rejection
region(s)
Find the critical value and rejection region for a
left-tailed test with α = 0.10.
z0 = -1.28
Find the critical values and rejection regions for
a two-tailed test with α = 0.08.
z0 = ±𝟏. 𝟕𝟓
Decisions Based on Rejection Regions
1. Find the standardized test statistic z
2. If z is in rejection region, reject H0
3. If z is not in the rejection region, fail to reject H0
Using Rejection Regions for a z-test of μ
1.
2.
3.
4.
5.
6.
State H0 and HA
Identify α
Find critical value(s)
Determine and sketch rejection regions
Find standardized test statistic z
Determine whether z is in the rejection region
and reject or fail to reject H0
7. Interpret decision in context of original claim
The CEO of a firm claims that the mean work
day of the firm’s accountants is less than 8.5
hours. A random sample of 35 of the firm’s
accountants has a mean work day of 8.2 hours
with a standard deviation of 0.5 hours. At α =
0.01, test the CEO’s claim.
z0 = -2.33, z = -3.55, reject H0 as there is enough
evidence to support the claim
The U.S. Department of Agriculture reports that
the mean cost of raising a child from birth to age
2 in a rural area is $8390. You believe this value
is incorrect, so you select a random sample of
900 children (age 2) and find that the mean cost
is $8275 with a standard deviation of $1540. At
α = 0.05, is there enough evidence to conclude
that the mean cost is different from $8390?
z0 = ±𝟏. 𝟗𝟔, z = -2.24, reject H0 as there is
enough evidence to support the claim
Sec. 7.2b
Pg. 344 #13-26,37-41,43
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