Transcript Chapter 9, part D

Chapter 9, part D
More Hypothesis Testing
V. Tests about : Small sample case
• If you have a small (n<30) sample, it’s appropriate
to use the t-distribution for hypothesis testing.
• Your test statistic,
x
t
 s 


 n
has a t-distribution with (n-1) degrees of freedom.
A. One-Tailed Test
Historically the population of Hanover students has a
cumulative G.P.A. of no more than 2.77. Suppose
you want to test the hypothesis that (due to
smarter students) the mean GPA is higher.
Ho:   2.77
Ha:  > 2.77
The Test
• In a sample of 25 students, you calculate a mean
GPA of 2.8491 and a standard deviation of .4796.
• Test the hypothesis at the 95% confidence.
• Your critical t.05 with 24 degrees of freedom is
1.711.
2.8491  2.77
t

• The test statistic:
=.8246
 .4796 



25 
• You can’t reject Ho and must conclude that the
GPA has not increased.
B. Two-Tailed Test
A golf ball producer tests driving distance of a new
golf ball. Due to PGA specifications, the distance
needs to average 280 yards.
Ho:  = 280 yds.
Ha:   280 yds.
The Test
• You sample n=10 golf balls, calculate a sample
mean of 282.2 yards and a standard deviation of
13.53.
Can you do
• With 9 degrees of freedom, the critical
this?
t.025=±2.262
• Your test statistic is .514, so you cannot reject Ho.
• Your conclusion is that the golf balls are o.k.
VI. Tests about a population
proportion
• The techniques are virtually identical to those
testing a population mean.
• “p” is the population proportion, and p0 is the
hypothesized value of the population proportion.
• One and two-tailed tests are structured in the same
way as before.
A. Testing a proportion hypothesis
The main difference when dealing with a proportion
problem is in the calculation of the standard error.
Recall from chapter 7:
p 
p(1  p)
n
We also have to test to see if we can safely apply the
Central Limit Theorem.
If np>=5 and n(1-p)>=5, our sample is “large
enough”.
B. An example
A restaurant believes that 30% of their customers do
not drink water with their meal. In a recent
sample of 480 customers, 128 did not drink their
water. Use this sample to test their belief at a 95%
confidence level.
Ho:  = .30.
Ha:   .30
p-bar = 128/480 =.2667.
p 
p(1  p)
=
n
.0209
The critical value is  1.96 and the test statistic is Z
= -1.5933 so we fail to reject the null.