Transcript Chapter 18 Notes MINE

Chapter 18
MATH 2400
Z-distribution  T-distribution
When σ is known, we base confidence intervals and tests for μ on the
𝑥−𝜇
𝜎
on-sample z-statistic 𝑧 =
where 𝜎𝑥 = .
𝜎𝑥
𝑛
But,
in
practice,
we
don’t
know
σ,
so
we
substitute
the
standard
error
𝑠
𝜎
for 𝑥 for its standard deviation . The statistic that results does
𝑛
𝑛
NOT have a normal distribution. It has a new distribution called the
“t distribution”. The one-sample t statistic
𝑥−𝜇
𝑡=𝑠
𝑛
has the t distribution with n-1 degrees of freedom.
t-statistic
The t-statistic has the same
interpretation as a z-statistic: It
says how far 𝑥 is from its mean μ
in standard deviation units.
There is a different t-distribution
for each sample size n. We
specify a particular t distribution
by giving its degrees of freedom.
The degrees of freedom is based
on the sample size.
d.f. = n - 1
Example 1
t Confidence Interval
𝑆
𝑥
𝑛
where 𝑡 ∗ is the critical value for the confidence interval.
± 𝑡∗
Practice Finding t*
Example 2
Psychologists studied the size of the tip in the restaurant when a
message indicating that the next day’s weather would be good was
written on the bill. Here are tips from 20 patrons, measured in percent
of the total bill. Determine a 95% confidence interval for μ.
Example 3
A manufacturer of high-resolution video terminals must control the tension on the
mesh of fine wires that lies behind the surface of the viewing screen. The tension is
measured by an electrical device with output readings in millivolts (mV). A random
sample of 20 screens has the following mean and standard deviation:
x = 306.32 mV
and
sx = 36.21 mV
We want to estimate the true mean tension µ of all the video terminals produced
this day at a 90% confidence level.
t Test (Example 4 – By Hand)
Very similar to a z Test.
The level of dissolved oxygen (DO) in a stream or river is an important indicator of
the water’s ability to support aquatic life. A researcher measures the DO level at
15 randomly chosen locations along a stream. Here are the results in milligrams
per liter:
4.53
5.04
3.29
5.23
4.13
5.50
4.83
4.40
5.42
6.38
4.01
4.66
2.87
5.73
5.55
A dissolved oxygen level below 5 mg/l puts aquatic life at risk.
We want to perform a test at the α = 0.05 significance level of
H0: µ = 5
H a: µ < 5
where µ is the actual mean dissolved oxygen level in this stream.
t Test (Example 5 – Using the Calculator)
A random sample of 22 fifth grade pupils have a grade point average of
5.0 in maths with a standard deviation of 0.452, whereas marks range
from 1 (worst) to 6 (excellent). The grade point average (GPA) of all fifth
grade pupils of the last five years is 4.7. Is the GPA of the 22 pupils
different from the populations’ GPA?
5
5.5 4.5 5
5
6
5
5
4.5 5
5
4.5 4.5 5.5 4
5
5
5.5 4.5 5.5 5
5.5
Mean: 5.0
Variance: 0.2045
t Test (Example 6 – Using the Calculator)
Bon Air Elementary School has 300 students. The principal of the
school thinks that the average IQ of students at Bon Air is at least 110.
To prove her point, she administers an IQ test to 20 randomly selected
students. Among the sampled students, the average IQ is 108 with a
standard deviation of 10. Based on these results, should the principal
accept or reject her original hypothesis? Assume a significance level of
0.01.
t Test -OR- z Test?
(Example 7 – Using the Calculator)
An inventor has developed a new, energy-efficient lawn mower engine.
He claims that the engine will run continuously for 5 hours (300
minutes) on a single gallon of regular gasoline. Suppose a simple
random sample of 50 engines is tested. The engines run for an average
of 295 minutes, with a standard deviation of 20 minutes. Test the null
hypothesis that the mean run time is 300 minutes against the
alternative hypothesis that the mean run time is not 300 minutes. Use
a 0.05 level of significance. (Assume that run times for the population
of engines are normally distributed.)