Transcript No Slide Title

Today’s lesson
• Confidence intervals for the expected value
of a random variable.
• Determining the sample size needed to have
a specified probability of a Type II error and
probability of a Type I error.
Confidence Interval for the Mean
of a Normally Distributed
Random Variable
• ASS-U-ME that Y is normally distributed
with unknown mean μ and known standard
deviation (say 100).
• Confidence interval for the mean is the set
of “reasonable values” based on the data
observed.
Example Problem 1
• The random variable Y is normally
distributed with unknown mean and
standard deviation 100. A random sample of
25 observations is taken from Y and has
mean value 515. What is the 99 percent
confidence interval for the mean of Y?
Solution to Problem 1
• Find the standard error of the mean:
– standard deviation of Y/square root of sample
size=100/square root of 25=20
• Find the factor for the size of the confidence
interval:
– use 1.960 for 95 percent CI
– use 2.576 for 99 percent CI
Solution to Problem 1
• Multiply standard error and factor:
– 2.576*20=51.52
• Add and subtract this product to the mean:
– Left end point is mean-51.52=463.48
– Right end point is mean+51.52=566.52
• In real life, round off CI to “non-obsessive”
numbers.
Statement of Solution
• The 99 percent confidence interval for the
expected value of Y (mean of Y) is the
interval between 463.48 and 566.52.
• The difference between the left and right
end points of the CI is a measure of how
much sampling variability is present in the
experimental results.
Example Problem 2
• Test the following two situations and select
the answer that describes your conclusions:
• I. Test H0: E(Y)=500, α=0.01 against H1:
E(Y) not equal to 500.
• II. Test H0: E(Y)=600, α=0.01 against H1:
E(Y) not equal to 600.
Example Problem 2 Options
• A) Reject null hypothesis I and reject null
hypothesis II.
• B) Reject null hypothesis I and accept null
hypothesis II.
• C) Accept null hypothesis I and reject null
hypothesis II.
• D) Accept null hypothesis I and accept null
hypothesis II.
Solution to Problem 2
• Use the confidence interval calculated in
problem 1, the interval between 463.48 and
566.52.
• Null mean in I is 500, which is in 99 percent
CI; hence accept in I.
• Null mean in II is 600, which is not in 99
percent CI; hence reject in II.
• Answer is C.
Hints and Reminders
• READ YOUR COMPUTER OUTPUT.
– Use the numbers that the computer calculates.
• MAKE SURE THAT THE PARAMETER
IN THE QUESTION AND THE
PARAMETER YOU ARE CALCULATING
ARE THE SAME!
Example Question 3
• The random variable Y is normally
distributed with unknown mean and
unknown standard deviation. A random
sample of 4 observations is taken from Y
The mean is 515, and the unbiased estimate
of the variance is 8100. What is the 99
percent confidence interval for the mean of
Y?
Solution to Problem 3
• Recognize that this problem requires the use
of Student’s t (the standard deviation is not
known).
• Find the estimated standard error of the
mean:
– square root of the unbiased estimate of the
variance/square root of sample size=90/square
root of 4=45
Solution to Problem 3 Continued
• Find the degrees of freedom for the estimate
of the unknown standard deviation:
– size of sample-1=4-1=3.
• Find the factor for the size of the confidence
interval:
– stretch 1.960 for 95 percent CI; for 3 df, 3.182
– stretch 2.576 for 99 percent CI; for 3 df, 5.841
Solution to Problem 3 Continued
• Multiply standard error and factor:
– 5.841*45=262.8
• Add and subtract this product to the mean:
– Left end point is mean-262.8=252.2
– Right end point is mean+262.8=777.8
• In real life, round off CI to “non-obsessive”
numbers.
How to Use Student t Confidence
Interval for Mean
• Exactly the same as the use of the normal
confidence interval for the mean.
Example Problem 4
• Test the following two situations and select
the answer that describes your conclusions:
• I. Test H0: E(Y)=500, α=0.01 against H1:
E(Y) not equal to 500.
• II. Test H0: E(Y)=600, α=0.01 against H1:
E(Y) not equal to 600.
Example Problem 4 Options
• A) Reject null hypothesis I and reject null
hypothesis II.
• B) Reject null hypothesis I and accept null
hypothesis II.
• C) Accept null hypothesis I and reject null
hypothesis II.
• D) Accept null hypothesis I and accept null
hypothesis II.
Solution to Example Problem 4
• Use the confidence interval calculated in
problem 3, the interval between 252.2 and
777.8.
• Null mean in I is 500, which is in 99 percent
CI; hence accept in I.
• Null mean in II is 600, which is also in 99
percent CI; hence reject in II.
• Answer is D.
Determining Sample Size
• Design in a statistical study is crucial for
success.
• Key issue is how large does the sample size
have to be.
• There is a key formula for determining the
sample size.
One-sample test sample size
parameters
• ASS-U-ME sampling for Y, a normally
distributed random variable.
• Null hypothesis values
– E0, expected value of Y under the null
– σ0, standard deviation of Y (a SINGLE value
drawn from Y) under the null
– α, the level of significance
– |zα|, the percentile from the standard normal
corresponding to α.
One-sample test sample size
parameters (continued)
• Alternative hypothesis values
– E1, expected value of Y under the alternative
– σ1, standard deviation of Y (a SINGLE value
drawn from Y) under the alternative
– β, the probability of a Type II error.
– |zβ|, the percentile from the standard normal
corresponding to β.
Sample Size Formula
• Use a sample size n that is as large or larger
than:
n
| z |  0  | z  |  1
| E1  E 0 |
Example Problem Scenario
• A research team will test the null hypothesis
that E(Y)=1000 at the 0.05 level of
significance against the alternative that
E(Y)<1000. When the null hypothesis is
true, Y has a normal distribution with
standard deviation 600.
Standard Warm-up Problem
• What is the standard deviation of the mean
of 900 observations under the null
hypothesis?
• Solution:
– The standard error is the standard deviation of
Y under H0 divided by the square root of the
sample size.
– 600/square root of 900=600/30=20.
A Sometime Warm-up Problem
• What is the critical value of the null
hypothesis in the scenario when using the
average of 900 observations as the test
statistic.
• Solution
– E0 sign |zα|*standard error of test statistic
– left sided test, hence use – 1000 - 1.645*20=967.1
Example Problem 5
• What is the probability of a Type II error for
an alternative in which Y is normally
distributed, E(Y)=950, and its standard
deviation is 600 using the average of a
random sample of 900 as the test statistic of
the null hypothesis in the scenario?
Solution to Example Problem 5
• The probability of a Type II error β is equal
to Pr1{Accept H0}= Pr1{Test
statistic>critical value}= Pr1{Sample
mean>967.1}.
• Under alternative, sample mean is
– normal
– with mean 950
– with standard error 20
Solution to Example Problem 5
Continued
• The problem now becomes: what is the
probability that a normally distributed
random variable with mean 950 and
standard deviation 20 is larger than 967.1?
–
–
–
–
Find standard units value of 967.1.
(967.1-950)/20=0.855
What is Pr{Z>0.855}?
This is about 0.197.
Example Problem 6
• What is the smallest sample size so that the
probability of a Type II error is 0.05 when
the (alternative) distribution of Y is
normally, E(Y)=950, and its standard
deviation is 600. The test statistic is the
average of a random sample of n as the test
statistic of the null hypothesis in the
scenario.
Solution
• Use formula:
• For null,
– E0=1000, σ0=600, α=0.05, |zα|=1.645
• For alternative,
– E1=950, σ1=600, β=0.05, |zβ|=1.645
Solution Continued
• Plug and chug:
• square root of sample size required is 39.48
• Required sample size is the square of
39.48,which is 1558.6.
• Round up to 1559.
• This is optimistic. How do you account for
nonresponse?
Review of Today’s lecture
• One sample procedures (today confidence
intervals).
• Determining Sample Size
• When you solve problem, think about the
meaning of the answer.