Section 5-3 and 5

Report
Sta220 - Statistics
Mr. Smith
Room 310
Class #17
Section 5-3 and 5-5 Notes
5-3: Confident Interval for a
Population Mean: Student’s t-Statistic
Federal legislation requires pharmaceutical
companies to perform extensive test on new
drugs before they can marketed. First phase, a
new drug is tested on animals. If the drug is
deemed safe, then they are permitted to begin
human testing on a limited basis. During phase,
inferences must be made about the safety of the
drug on the basis of information obtained from
a very small samples.
The use of a small sample in making an inference about
 presents two immediate problems when we attempt to
use the standard normal z as a test statistic.
1. We can no longer assume that the sampling
distribution of  is approximately normal, the CLT
ensures normality only for samples that are
sufficiently large. (n > 30).
2. Though still true that  = / , the sample
standard deviation s may provide a poor
approximation for  when the sample size is small.
T-statistic
t-statistic is used when the sample size is small.
 –
 =
(/ )
t-statistic has (n – 1) degrees of freedom (df)
Standard normal (z) distribution and tdistributions
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Table VI
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The t.025 value in a t-distribution with 4 df, and
the corresponding z.025 value
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Example 5.3.1
Suppose you have selected a random sample of
n = 7 measurements from a normal distribution.
Compare the standard normal z-values with the
corresponding t-values if you were forming the
following confidence intervals.
A. 80% confidence interval
B. 95% confidence interval
Solution on the Board
Example 5.3.2
Let 0 be a specific value of t. Use Table IV to
find 0 values such that the following statements
are true:
A.   ≥ 0 = .01, where df = 17
B.   ≤ 0 = .005, where df = 6
Solution on the Board
Procedure
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Example 5.3.3.
Consider the pharmaceutical company that
desires an estimate of the mean increase in
blood pressure of patients who take a new drug.
The blood pressure increases (measured in
points) for the n = 6 patients in the human
testing phase are shown below. Use this
information to construct a 95% confidence
interval for , the mean increase in blood
pressure associated with the new drug for all
patients in the population.
Blood Pressure Increase (Points) for Six Patients
1.7
3.0
.8
3.4
2.7
2.1
We need to calculate the mean and standard
deviation.
TI-84
Insert data:
Stat-> Edit -> L1
Mean and Standard deviation:
Stat-> Go to CALC->1-Var Stats -> Enter Twice
Solution On the Board
We can be 95% confident that the mean
increase in blood pressure associated with
taking this new drug is between 1.286 and 3.28
points. We know that if we were to repeatedly
use this estimation procedure, 95% of the
confidence intervals produced would contain
the true mean , assuming that the probability
distribution of changes in blood pressure from
which our sample was selected is normal.
5-5: Determining the Sample Size
Estimating a Population Mean
Go back to the hospital example, which we estimated
the mean length of stay for patients in a large
hospital. The sample was 100 patients’ records
produced the 95% confidence interval 4.5 ± .78.
Our estimate  was within .78 days of the true mean
length of stay, , for all the hospital’s patients at the
95% confidence level for  was 2(.78) = 1.56 days
wide when 100 accounts were sampled. Suppose we
want to estimate  to within .25 days with 95%
confidence is, we want to narrow the width of the
confidence interval from 1.56 days to .50 days.
Relationship between sample size and width of
confidence interval: hospital-stay example
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How much will the sample size have to be
increased to accomplish this?
Over 983 patients’ records will have to be
sampled to estimate the mean of stay to within
.25 days (approximately) 95% confidence.
Specifying the sampling error SE as the half-width
of a confidence interval
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Education, Inc.. All rights
reserved.
Procedure
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Education, Inc.. All rights
reserved.
Example 5.5.1.
Suppose the manufacturer of official NFL footballs
uses a machine to inflate the new balls to pressure
of 13.5 pounds. When the machine is properly
calibrated, the mean inflation pressure is 13.5
pounds, but uncontrollable factors cause the
pressures of individual foots to vary randomly from
13.3 to 13.7 pounds. For quality control purposes,
the manufacturer wished to estimate the mean
inflation pressure to with .025 pounds of its true
value with a 99% confidence interval. What sample
size should be specified for the experiment?
Solution on the Board
We round this up to n = 107. Realizing that
 was approximated by R/4, we might even
advise that the sample size be specified as n =
110 to be more certain of attaining the objective
of 99% confidence interval with sampling error
of .025 pound or less.
Assignment
5-2, 5-3, 5-5 all Due Friday
Friday
6/1-6/2

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