Title: Hypothesis Testing (Decisions on Means, Proportions and Variances)
1Hypothesis Testing(Decisions on Means,
Proportions and Variances)
- QSCI 381 Lecture 28
- (Larson and Farber, Sect 7.3 - 7.5)
2Overview
- Yesterday, we dealt with the case in which the
sample size was large. Today we address the
issue of how to handle cases in which the sample
size is small. - By small we mean less than 30 samples note
that the population needs to be normal (or at
least approximately normal).
3The Test Procedure
- State the claim mathematically and verbally.
Identify the null and alternative hypotheses. - H0 ? Ha ?
- Specify the level of significance.
- ??
- Identify the degrees of freedom (d.f n-1) and
hence the critical value(s) and the rejection
region. - Determine the standardized test statistic
- Check whether t is in the rejection region.
4Example-I(Comparison of Treatments)
- Under normal feeding conditions, a given fish is
2 kg after two years of growth. - We develop an alternative feeding program (which
is cheaper). - We feed a sample of 20 fish and, after 2 years,
the fish are 1.9 kg on average (s.d. 0.4). - Can someone reject the claim that fish fed on our
diet are greater or equal to those fed using the
usual diet.
5Example-II
Note that We multiply by 2
- H0 ??2 Ha ?lt2.
- ?0.05.
- d.f 19 tc-1.73 (TINV(20.05,20-1))
- t (1.9-2)/(0.4/?20)-1.11
- t is not in the rejection region (even though the
sample mean is less than the value we are
comparing it with).
6Example-III
Rejection region
Do not reject
What would have happened had our standard
deviation been 0.2 rather than 0.4? Just because
we cannot reject the null hypothesis does not
mean that it is not false!
7Hypothesis Tests for Proportions-I
- Recall that we can approximate a binomial
distribution, B(n,p) with the normal
distribution, as long as np ? 5,
nq ? 5. - We will make use of this to define a hypothesis
test for proportions.
8Hypothesis Tests for Proportions-II
- State the claim mathematically and verbally.
Identify the null and alternative hypotheses. - H0 ? Ha ?
- Specify the level of significance, ?, and hence
the rejection region. - Determine the standardized test statistic
- Check whether z is in the rejection region
9Example-I
- Under the Sustainable Fisheries Act, stocks
should be managed so that they are at a target
level. The expectation is therefore that 50 of
stocks are above their target levels. - We review the status of 300 stocks and find that
47 (a slight exaggeration of reality) are below
their target levels. Can we reject the claim that
management is achieving its goal? Use a
significance level of 0.01.
10Example-II
- H0 p??0.5 Ha pgt0.5.
- ?0.01. The rejection region is z gt 2.33.
- The standardized test statistic is
- We fail to reject H0.
-
- The U.S. government supplies statistics on its
performance managing fisheries. See if - its performance is really as good as this??
http//www.nmfs.noaa.gov/sfa/statusoffisheries/S
OSmain.htm
11Hypothesis Tests for Variances-I(the ?2 test)
- State the claim mathematically and verbally.
Identify the null and alternative hypotheses. - H0 ? Ha ?
- Specify the level of significance, ?, and
determine the degrees of freedom (d.f. n-1) - Find the rejection region (note that the
chi-square distribution is not symmetric) - Determine the standardized test statistic
- Check whether ?2 is in the rejection region.
- Note that the population must be normally
distributed to use this test.
12Hypothesis Tests for Variances-II(Finding the
critical values and the rejection region)
- We need
- The level of significance.
- The degrees of freedom
- Whether this is a
- right-tailed test (?2 corresponding to ?)
- left-tailed test (?2 corresponding to 1-?)
- two-tailed test (?2 corresponding to ½? and 1- ½?)
13Hypothesis Tests for Variances-III(Finding the
critical values and the rejection region)
- Given n26 (i.e. d.f25). Find the rejection
regions for ?0.1
CHIINV(0.9,25)
CHIINV(0.1,25)
CHIINV(0.05,25) CHIINV(0.95,25)
14Example-I
- A restaurant claims that the standard deviation
of the length of its serving times is less than
2.9 minutes. A random sample of 23 serving times
has a standard deviation of 2.1 minutes. Is there
enough evidence to support the claim at the 0.1
level of significance?
15Example-II
- The null hypothesis is ??2.9 (not ?lt2.9 why).
- The level of significance is 0.1 and degrees of
freedom is 22 (23-1). - This is left-tailed test so we determine the
critical value of ?2 to be 14.042
(CHIINV(0.9,22)). - We now calculate the standardized test statistic
- Now 11.54 lt 14.042 so we reject H0.
16Caveats About Hypothesis Testing
- Check the assumptions of the test
- Random sampling
- Must the population be normal?
- Not rejecting the null hypothesis is not the same
as proving it to be true. We should never say
that the null hypothesis is proven if the p-value
? ?. - Hypothesis tests focus on Type I error, but never
forget about type II error.