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Introduction to Inference
Tests of Significance
Wording of conclusion revisit
• If I believe the statistic is just too extreme and
unusual (P-value < a), I will reject the null
hypothesis.
• If I believe the statistic is just normal chance
variation (P-value > a), I will fail to reject the null
hypothesis.
reject
p-value<a, there is
We
Ho, since the
fail to reject
p-value>a, there is not
enough evidence to believe…(Ha in context…)
Example 3
test of significance
m = true mean distance
Ho: m = 340
Ha: m > 340
Given random sample
Given normally distributed.
Safe to infer a population of at least 100 missiles.
348  340
z
 1.26
p-value=.1038
20
10
let a = .05
We fail to reject Ho. Since p-value>a there is not
enough evidence to believe the mean distance
traveled is more than 340 miles.
Familiar transition
• What happened on day 2 of confidence
intervals involving mean and standard
deviation?
• Switch from using z-scores to using the tdistribution.
• What changes occur in the write up?
Example 3
test of significance
m = true mean distance
Ho: m = 340
Ha: m > 340
Given random sample.
Given normally distributed.
Safe to infer a population of at least 100 missiles.
348  340
z
 1.26
p-value=.1038
20
10
let a = .05
We fail to reject Ho. Since p-value>a there is not
enough evidence to believe the mean distance
traveled is more than 340 miles.
Example 3
t-test
m = true mean distance
Ho: m = 340
Ha: m > 340
Given random sample.
Given normally distributed.
Safe to infer a population of at least 100 missiles.
348  340
z
 1.26
p-value=.1038
20
10
let a = .05
We fail to reject Ho. Since p-value>a there is not
enough evidence to believe the mean distance
traveled is more than 340 miles.
Example 3
t-test
m = true mean distance
Ho: m = 340
Ha: m > 340
Given random sample
Given normally distributed.
Safe to infer a population of at least 100 missiles.
df  9
348  340
t
 1.26
p-value=.1038
20
10
let a = .05
We fail to reject Ho. Since p-value>a there is not
enough evidence to believe the mean distance
traveled is more than 340 miles.
Example 3
t-test
m = true mean distance
Ho: m = 340
Ha: m > 340
Given random sample.
Given normally distributed.
Safe to infer a population of at least 100 missiles.
df  9
348  340
t
 1.26
p-value=
20
10
let a = .05
We fail to reject Ho. Since p-value>a there is not
enough evidence to believe the mean distance
traveled is more than 340 miles.
t-chart
t  1.26
Example 3
t-test
m = true mean distance
Ho: m = 340
Ha: m > 340
Given random sample.
Given normally distributed.
Safe to infer a population of at least 100 missiles.
df  9
348  340
t
 1.26
.10<p-value<.15
20
10
let a = .05
We fail to reject Ho. Since p-value>a there is not
enough evidence to believe the mean distance
traveled is more than 340 miles.
Example 3
t-test
m = true mean distance
Ho: m = 340
Ha: m > 340
Given random sample.
Given normally distributed.
Safe to infer a population of at least 100 missiles.
df  9
348  340
t
 1.26
p-value=.1188
20
10
let a = .05
We fail to reject Ho. Since p-value>a there is not
enough evidence to believe the mean distance
traveled is more than 340 miles.
1 proportion z-test
p = true proportion pure short
Ho: p = .25
Ha: p = .25
Given a random sample.
np = 1064(.25) > 10
n(1–p) = 1064(1–.25) > 10
Sample size is large enough to use normality
Safe to infer a population of at least 10,640 plants.
.2603  .25
z
 .78
.25(1  .25)
1064
.p-value=.4361
let a = .05
We fail to reject Ho. Since p-value>a there is not
enough evidence to believe the proportion of pure
short is different than 25%.