Introduction to Hypothesis Testing. Chapter 11. 11.1 Introduction. The purpose of hypothesis testing is to determine whether there is enough statistical evidence in favor of a certain belief about a parameter. Examples
Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.
Introduction to Hypothesis Testing
Chapter 11
This is what you want to prove
m = 350
m = 350
Critical value
m = 350
H1 : m > 170
H0 : m = 170
The rejection region is a range of values such that if the test statistic falls into that range, the null hypothesis is rejected in favor of the alternative hypothesis.
Reject H0 here
Critical value of the
sample mean
a
= P( given that H0 is true)
we have:
Example 11.1 – solution continued
P(commit a Type I error) = P(reject H0 given that H0 is true)
… is allowed to be a.
Example 11.1 – solution continued
a
= 0.05
Conclusion
Since the sample mean (178) is greater than the critical value of 175.34, there is sufficient evidence to infer that the mean monthly balance is greater than $170 at the 5% significance level.
One tail test
Recall:H0: m = 170
H1: m > 170
The probability of observing a test statistic at least as extreme as 178,
given that m = 170 is…
The p-value
Note how the event
is rare under H0
when but...
…it becomes more
probable under H1,
when
Because the probability that the sample mean will assume a value of more than 178 when m = 170 is so small (.0069), there are reasons to believe that m > 170.
We can conclude that the smaller the p-value
the more statistical evidence exists to support the alternative hypothesis.
The p-value
a
= 0.05
The alternative hypothesis
is the more important
one. It represents what
we are investigating.
Left-tail test
Define the rejection region
H0: m = 17.09
a/2 = 0.025
a/2 = 0.025
If H0 is true (m =17.09), can still fall far above or far below 17.09, in which case we erroneously reject H0 in favor of H1
Solution - continued
17.09
We want this erroneous rejection of H0 to be a rare event, say 5% chance.
a/2 = 0.025
a/2 = 0.025
17.55
a/2 = 0.025
a/2 = 0.025
From the sample we have:
0
za/2= 1.96
-za/2= -1.96
Rejection region
Solution - continued
17.09
0
za/2= 1.96
-za/2= -1.96
Two-tail test
There is insufficient evidence to infer that there is a difference between the bills of AT&T and the competitor.
Also, by the p value approach:
The p-value = P(Z< -1.19)+P(Z >1.19)
= 2(.1173) = .2346 > .05
a/2 = 0.025
a/2 = 0.025
-1.19
1.19
a=.05
H0: m = 170
H1: m = 180
m= 170
m=180
Express the rejection region directly, not in standardized terms
Specify the alternative value under H1.
Do not reject H0
a=.05
H0: m = 170
H1: m = 180
m= 170
m=180
A false H0…
…is not rejected
H0: m = 170
H1: m = 180
m=180
m= 170
a2 >
b2 <
a1
b1
m= 170
m=180
By increasing the sample size the standard deviation of the sampling distribution of the mean decreases. Thus, decreases.
m= 170
m=180
Note what happens when n increases:
a does not change,
but b becomes smaller