Lecture 1.2 Confidence Interval Estimation 2. Introduction to

Transcription

CAS MA 116 • Statistics II • Summer 2, 2012 • Lecture 1.2
Lecture 1.2
“… The greater part of our happiness or misery depends upon our disposition, and not upon
our circumstances.” … Martha Washington (1732-1802)
1.
2.
3.
4.
5.
Point Estimation and Confidence Interval Estimation
Introduction to Hypothesis Testing
One-Sample Test for Population Proportion (large n)
Two-Sample Test for Difference in Population Proportions
Binomial Distribution. One-Sample Test for Population Proportion (large n)
Introduction to Point Estimation and Confidence Interval Estimation
Recall that population of interest represents the entire group of items (individuals), that we would
like to make an inference about. Statistical inference is the process of drawing conclusions about a
population parameter based on data, or statistic – an estimate or a summary computed from the
observations.
Population versus Sample
Parameter versus Statistic
Population
Parameter(s)
Inference
Parameter Estimation
Statistic(s)
Sample
Definitions:
There are two types of estimates for population parameter: confidence interval estimates and point
estimates. A confidence interval estimate is a range of values for the population parameter with a predefined level of confidence (e.g., 95% confidence interval). A point estimate for a population
parameter is the statistic and can be considered the “best” (available) single-number estimate of that
parameter.
1
Examples:
1) Based on a random sample of the 100 patients visiting ER this weekend, the average waiting
time is estimated to be 37.85 minutes.
37.85 minutes is a point estimate of the average waiting time
Question: How far is the sample mean of 37.85 minutes is from true (population) average waiting
time at ER (means for all weekends for all patients)?
2) Based on a random sample of the 100 patients visiting ER this weekend, the average waiting
time is estimated to be between 32.65 minutes and 42.10 minutes.
(37.65, 42.10) minutes is a confidence interval estimate of the average waiting time
Confidence Interval Estimates provide a researcher with the range of possible, or acceptable, values
of the true, unknown population parameter.
The basic structure for any confidence interval is:
point estimate ± multiplier * standard error,
where “multiplier * standard error” = the margin of error.
Note that:
•
The margin of error is a build in component that addresses how close (or how far) the point
estimates are from the true, unknown parameter.
•
The (estimated) variance in point estimates (e.g., 𝜎! , 𝑜𝑟 𝑠! ), is called the standard error.
•
The standard error depends on the sample size and the true population standard deviation (i.e.,
standard error goes down as the sample size goes up).
•
Standard error interpretation: If repeated samples of (sample size) are obtained from this same
population, we would estimate the resulting sample (statistic) to be about (value of standard error)
away from the true (population parameter) on average.
•
The multiplier depends on the confidence level and the population parameter, but not the
sample size deviation (i.e., multiplier is higher for higher values of confidence level).
2
Confidence Intervals for Decision Making
Principle 1: A value not in a CI can be rejected as possible value of the population parameter. A
value in a CI is an “acceptable” or “reasonable” possibility for the value of a population parameter.
Principle 2: When the CIs for parameters for two different population parameters do not overlap,
it is reasonable to conclude that the parameters for the two populations are different.
1. The probability that the true parameter falls in a particular, already computed, confidence
interval is either 0 or 1. The interval is now fixed and the parameter is NOT random, so the
parameter is either in that particular interval or it is not.
2. A 95% Confidence Interval Interpretation: We are 95% confident that the true parameter
value lies inside the confidence interval. The interval provides a range of reasonable values for
the population parameter.
3. The 95% Confidence Level Interpretation: If the procedure were repeated many times (that
is, if we repeatedly took a random sample of the same size and computed the 95% confidence
interval for each sample), we would expect 95% of the resulting confidence intervals to contain
the true population parameter.
Confidence Intervals for Population Proportion
The assumptions required for CI for a population proportion to be valid: (1) the sample size n is large
enough (check: npˆ ≥ 5 and n(1 − pˆ ) ≥ 5), and (2) the data are a random sample from that population.
Type of the Formula
Formula for Approximate Confidence Interval
for the Population Proportion p
Formula for Conservative Confidence Interval
for the Population Proportion p
Note: here 𝒑=0.5 is used to compute the
standard error:
Formula
pˆ ± Z1−(α / 2)
pˆ ± Z1−(α / 2)
SE ( pˆ ) =
pˆ (1 − pˆ )
n
1
2 n
pˆ (1 − pˆ )
0.5(1 − 0.5)
1
=
=
n
n
2 n
Formula for the Sample Size required to
produce an estimate for the population
proportion p
Formula for the Approximate Sample Size. If
the population parameter p is unknown, the
sample estimate 𝒑 can be used instead
⎛ Z
⎞
n = p(1 − p)⎜⎜ 1−(α / 2) ⎟⎟
⎝ E ⎠
2
⎛ Z
⎞
n = pˆ (1 − pˆ )⎜⎜ 1−(α / 2) ⎟⎟
⎝ E ⎠
2
Formula for the Conservative Sample Size (p=0.5) ⎛ Z
⎞
n = 0.25⎜⎜ 1−(α / 2) ⎟⎟
⎝ E ⎠
3
2
Exercise (D’Agostino, Example 7.1, 7.2): Testing Population Proportion of Patients with
Osteoarthritis
The medical records of a random sample of 200 patients are reviewed for the diagnostics of
arthritis. Suppose that 38 patients are observed with diagnosed osteoarthritis. Sample proportion pˆ (a
point estimate for the population proportion p) was computes as: pˆ = 38 / 200 = 0.19 , or 19%.
a) Compute a 95% confidence interval for the population proportion p of all patients in the physician’s
practice with diagnosed osteoarthritis.
First, we have to check if the main conditions for using CI formula are satisfied. It is given that
the sample of 200 patients is a random sample, so we only need to check if the sample size is large
enough. A large sample is defined as one that satisfies the following: npˆ ≥ 5 and n(1 − pˆ ) ≥ 5
In our case, n=200, pˆ = 0.19 and both conditions are satisfied
npˆ = 200 * 0.19 = 38 > 5 and n(1 − pˆ ) = 200 * 0.81 = 162 > 5
The standard error of the sample proportion is computed as:
ˆ =
s.e.( p)
ˆ
ˆ
p(1−
p)
=
n
For 95% confidence interval Z 1−(α / 2 ) = Z 1−( 0.05 / 2 ) = 1.96 and the appropriate formula is the following:
pˆ ± Z1−(α /2)
ˆ
ˆ
p(1−
p)
=
n
b) Based on the computed interval, could you conclude that the proportion of all patents with
osteoarthritis in the clinic is significantly lower than 50%? Explain.
c) Compute the number patients required to estimate the population proportion p of all patients in the
physician’s practice with diagnosed osteoarthritis with precision of 1% (E=0.01) at 95% confidence
level.
Since the true population proportion p is unknown, we will the formula for approximate sample
size n with Z 1−(α / 2 ) = Z 1−( 0.05 / 2 ) = 1.96
2
⎛ Z1−(α / 2) ⎞
⎟⎟ =
n = pˆ (1 − pˆ )⎜⎜
E
⎝
⎠
Answer: 5913
4
Introduction to Hypothesis Testing
Recall that Hypothesis Testing uses the point estimate to attempt to reject/accept a hypothesis about
the population. Usually researchers want to reject the notion that chance alone can explain the sample
results.
ü Hypothesis testing is applied to population parameters by specifying a null hypothesis (H0) that
contains a null value for the population parameter—a value that would indicate a baseline, or that
nothing of interest is happening: “old news”, “no difference”, etc. In most cases, the researchers are
trying to show that the null value is not correct.
ü Hypothesis testing proceeds by obtaining a sample, computing a point estimate (sample statistic), and
assessing how unlikely to obtain this sample statistic if the null parameter value were correct.
Basic Steps in Hypothesis Test
Step 1: Determine the null (H0) and alternative (Ha) hypotheses.
Note: Hypotheses are statements ABOUT population parameters NOT ABOUT sample statistics.
Step 2: Verify necessary data conditions (assumptions), and if met, summarize the data into an
appropriate test statistic (using appropriate data summary, or sample statistic).
Step 3: Assuming the null (H0) hypothesis is true, find either rejection region or the p-value.
Step 4: Decide whether or not the result is statistically significant based on rejection region or based
on the p-value, the probability of getting a test statistic as extreme or more extreme (in the direction of
Ha) than the observed value of the test statistic, assuming H0 is true.
Decision based on the rejection region:
--- If the test statistic falls into the rejection region, then we reject, then the result IS statistically
significant, the decision is to reject H0; otherwise the result IS NOT statistically significant, the
decision is to fail to reject H0.
Decision based on the p-value:
If the p-value ≤ α, then the result IS statistically significant, the decision is to reject H0.
If the p-value >α, then the result IS NOT statistically significant, the decision is to fail to reject H0.
Step 5: Report the conclusion in the context of the problem (question of interest).
5
One-Sample Hypothesis Test for Population Proportion (Large n)
Test
Scenario
Population
Proportion
Data
1 Sample
Population
Parameter
𝑝
Sample
Statistics
Response
𝑥
𝑝 =
𝑛
# 𝑠𝑢𝑐𝑒𝑠𝑠𝑒𝑠
=
𝑛
Explanatory
Variable
________
Categorical
(YES/NO,
Success/Failure,
True/False)
Is the true population proportion of adults who believe in life after death is more than 70%?
Large Sample Hypothesis Test for Population Proportion H0: p=p0
Required assumptions:
• Sample size n is large enough (check: 𝑛𝑝! ≥ 5 and 𝑛(1 − 𝑝! ) ≥ 5 )
• Data are a random sample from Binomial population.
pˆ − p0
!
# !"#$!!$!
Test Statistics: Z =
where 𝑝 = ! =
!
p0 (1 − p 0 )
n
Decision Rule: Ha: p>p0 𝒁 ≥ 𝒁𝟏!𝜶
Ha: p≠p0 𝒁 ≥ 𝒁𝟏!𝜶/𝟐
Ha: p<p0 𝒁 ≤ −𝒁𝟏!𝜶
Example (D’Agostino, Example 7.3): Testing Proportion of Cases with Abnormality Correctly
Detected Against a Referent. Suppose that a diagnostic test has been shown to be 80% effective in detecting a genetic abnormality in human cells. An investigator modifies the diagnostic testing protocol
and wishes to test if the new protocol has a detection rate that is significantly different from 80% in
specimens known to possess the abnormality. The new protocol is applied to 300 independent
specimens of human cells known to possess the abnormality. The abnormality is detected in 222
specimens. Run the appropriate test at a 5% level of significance.
Step1: Define parameter of interest and state the hypothesis.
Parameter: p = true detection rate
H0: p = 0.8
Ha: p ≠ 0.8 Significance level α = 0.05
Step2:
Verify necessary data conditions and if met, summarize the data into an appropriate test statistic:
pˆ =
x
=
n
n (1− p0 ) =
np 0 =
Both conditions are satisfied, so we will use Z as our test statistic:
Z=
pˆ − p0
=
p0 (1− p0 )
n
6
Step3: Assuming the H0 is true, define decision rule:
Reject Ho if
𝒁 ≥ 𝒁𝟏!𝜶/𝟐 , 𝒐𝒓 𝒁 ≥ 𝟏. 𝟗𝟔
Step 4: Decide whether or not the result is statistically significant based on rejection region:
Based on the sample of n =______ abnormal specimens, there _________ significant evidence,
at ______ level, to conclude that the modified protocol has a significantly different detection rate
than 80% (the rate for original diagnostic test).
Exercise: (D’Agostino, Example 7.3): Testing Proportion of Cases with Flu Following Vaccination
Against Referent.
Step 1: Define parameter of interest and state the hypothesis.
Parameter: _________________________________________________________
H0: _____________
Ha: _________________ Significance level α =_______
Step 2: Verify necessary data conditions and if met, summarize the data into an appropriate test
statistic:
Check: 𝑛𝑝! ≥ 5 and 𝑛(1 − 𝑝! ) ≥ 5
Test statistic is:
Z=
pˆ − p 0
p 0 (1 − p 0 )
n
=
Step 3: Assuming the H0 is true, define decision rule: Reject Ho if _______________________________
Based on the sample of n=___________, there is ________ significant evidence, at level
α=________, to conclude that ________________________________________________
7
Two-Sample Hypothesis Test for Difference in Population Proportions
Assumptions
Large n1, n2
𝒏𝟏 𝒑𝟏 ≥ 𝟓 and
𝒏𝟏 (𝟏 − 𝒑𝟏 ) ≥ 𝟓
and 𝒏𝟐 𝒑𝟐 ≥ 𝟓 𝒏𝟐 (𝟏 − 𝒑𝟐 ) ≥ 𝟓
Test Statistic
pˆ 1 − pˆ 2
Z=
⎛ 1 1 ⎞
pˆ (1 − pˆ )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠
Confidence Interval
( pˆ 1 − pˆ 2 ) ± Z1−(α / 2)
where p
ˆ 1 = X 1 / n1 , pˆ 2 = X 2 / n2
pˆ =
X1 + X 2
n1 + n2
pˆ 1 (1 − pˆ 1 ) pˆ 2 (1 − pˆ 2 )
+
n1
n2
⎛ pq 2Z1−(α / 2) + p1 q1 + p 2 q 2 Z1− β
ni = ⎜
⎜
ES
⎝
⎞
⎟
⎟
⎠
2
where ES=|p2-p1| (under Ha)
p1 + p 2
,q = 1− p
2
Note: when test for (p1-p2) or construct confidence interval for (p1-p2) necessary conditions are the same.
p=
Example (D’Agostino, Chapter 7, Problem 20): Suppose an observational study is conducted to
investigate the smoking behaviors of male and female patients with a history of coronary heart disease.
Among 220 men surveyed, 80 were smokers. Among 190 women surveyed, 95 were smokers.
a. Construct a 95% confidence interval for the difference in the proportions of males and females
who smoke.
-What do we have? Categorical Variable: Yes = “smoke”, No = “does not smoke”
Two groups: group 1 - male and group 2 – female, n1=220 x1=80 and n2=190 x2=95
-
We can compute the corresponding sample proportions:
80
95
= 0.36 and pˆ2 =
= 0.5
pˆ1 =
220
190
-
The formula for a corresponding 95% CI for (p1 - p2) is
pˆ (1 - pˆ1) pˆ2 (1 - pˆ2)
(pˆ1 - pˆ2) ± Z1− α 1
+
2
n1
n2
In order to use it, we have to check if both samples are sufficiently large:
min(n1 pˆ 1,n1(1- pˆ 1))=min(220(0.36),220(0.64))=79 > 5
min(n2 pˆ 2,n2(1- pˆ 2))=min(190(0.50),190(0.50))=95 >5
-
Both conditions are satisfied, so we can use the formula and compute a 95% CI for (p1 - p2)
(recall that for 95% confidence Z1− α = 1.96 ):
2
(0.36 - 0.50) + (1.960)
0.36(1 - 0.36) 0.50(1 - 0.50)
=
+
220
190
8
b. Based on part a), would you reject H0:p1=p2 (equivalent to H0:p1-p2=0)?
______________________________________
c. Perform an appropriate test of hypothesis to test if the proportion of female who smoke is
significantly higher than the proportions of males who smoke at 5% significance level.
Step 1: Define parameter of interest and state the hypothesis.
Parameter: p1 – p2, true population difference in proportions of male and female smokers
H0:___ p1-p2=0 ____
Ha:___ p1-p2<0________ Significance level α =0.05
Step 2: Verify necessary data conditions and if met, summarize the data into an appropriate
test statistic:
Check: 𝑛! 𝑝! ≥ 5 and 𝑛! (1 − 𝑝! ) ≥ 5 and 𝑛! 𝑝! ≥ 5 and 𝑛! (1 − 𝑝! ) ≥ 5.
min(n1 pˆ 1,n1(1- pˆ 1))=min(220(0.36),220(0.64))=79 > 5
min(n2 pˆ 2,n2(1- pˆ 2))=min(190(0.50),190(0.50))=95 >5
Test statistic is:
pˆ 1 − pˆ 2
Z=
⎛ 1 1
pˆ (1 − pˆ )⎜⎜ +
⎝ n1 n2
, where p
ˆ 1 = X 1 / n1 , pˆ 2 = X 2 / n2 and pˆ =
X1 + X 2
n1 + n2
⎞
⎟⎟
⎠
X + X2
80
95
80 + 95
= 0.36 , pˆ2 =
= 0.5 , and pˆ = 1
=
= 0.4268
pˆ1 =
220
190
n1 + n2
220 + 190
pˆ 1 − pˆ 2
0.36 − 0.5
− 0.14
Z=
=
=
= −2.8
0
.
05
0
.
43
(
1
−
0
.
43
)(
1
/
220
+
1
/
190
)
⎛ 1 1 ⎞
pˆ (1 − pˆ )⎜⎜ + ⎟⎟
⎝ n1 n2 ⎠
Step 3: Assuming the H0 is true, define decision rule:
____________________________________________________________________
_______________________________________________________________
Based on two samples of male n1=220 and female n2=195, there is a significant evidence,
at level α = 0.05,to conclude that the proportion of female who smoke is significantly higher
than the proportion of males who smoke.
Exercise: Go over examples in the D’Agostino’s book, Example 7.7 and Example 7.8.
9
Sample Size Determination in Tests for Differences in Proportions
Example (D’Agostino, Example 7.17, Chapter7): Suppose we wish to design a study to compare
two treatments with respect to the proportion of successes in each. A two-sided test is planned at a 5%
level of significance. Based on a review of the literature, p1=0.20. How many subjects would be
required per group to detect p2=0.10 with 90% power?
The formula to determine sample size is given by:
⎛ pq 2Z1−(α / 2) + p1 q1 + p 2 q 2 Z1− β
ni = ⎜
⎜
ES
⎝
2
⎞
⎟ , where ES=|p2-p1| and p = p1 + p 2 , q = 1 − p
⎟
2
⎠
The ES (the effect size) is: ES=|p2-p1| = |0.2-0.1| = 0.1
Since α = 0.05, Z1-α/2 = Z0.975 = 1.96.
Similarly, for the power = 0.90, β = 1 – power = 1-0.9 = 0.1, and Z1-β =Z0.90= 1.282
p1 + p 2 0.1 + 0.2
=
= 0.15,
2
2
q = 1 − p = 1 − 0.15 = 0.85
p=
Substituting,
⎛ pq 2 Z 1−(α / 2 ) + p1 q1 + p 2 q 2 Z 1− β
ni = ⎜
⎜
ES
⎝
= 265.9 ≈ 266
2
2
⎞
⎟ = ⎛⎜ 0.15 * 0.85 * 2 *1.96 + 0.2 * 0.8 + 0.1 * 0.9 *1.282 ⎞⎟ =
⎜
⎟
⎟
0.1
⎝
⎠
⎠
Thus, we need at least 266 subjects in each group.
10
Binomial Distribution
An important class of discrete variables is called the Binomial random variables. The
Binomial distribution model has specific attributes:
There are n trials (performances of the binomial experiment), where n is determined in
advance and is not random.
Each trial results in only one of the two possible outcomes, which we call either a
success or a failure.
The probability of a success on each trial is constant, denoted p, with 0≤p≤1.
The probability of a failure on each trial is also constant, denoted q=1-p, with 0≤q≤1.
The trials are independent.
X~ Binomial (n, p) = “number of successes in the n trials of a binomial experiment”
The Binomial Distribution X~ Binomial (n, p)
Probability of exactly k successes in trials:
𝑃 𝑋 = 𝑘 = 𝐶!! 𝑝! 1 − 𝑝
where
𝑛
!!
= !! !!! !.
𝑘
!!!
=
𝑛 !
𝑝 1−𝑝
𝑘
!!!
Mean (Expected value) of X is 𝜇 = 𝑛𝑝
Standard Deviation of X is 𝜎 = 𝑛𝑝(1 − 𝑝)
If 𝑛𝑝 ≥ 5 and 𝑛 1 − 𝑝 ≥ 5 , X is approximately 𝑁(𝜇 = 𝑛𝑝, 𝜎 =
,
𝑛𝑝 1 − 𝑝 )
Exercise (Priority of Information)
According to a national survey, 25% of teenagers can correctly name the city where the US
Constitution was written. Assume a random sample of eight teenagers.
a)
Find the probability that none of the teenagers can name the city.
b)
Find the probability that at most two of the teenagers can name the city.
11
One-Sample Hypothesis Test for Population Proportion (small n)
Small Sample Hypothesis Test for Population Proportion H0: p=p0
Required assumption: Data are a random sample from Binomial population.
Test Statistics: X (observed # successes), X ~ Binomial(n, p0)
Decision rule is based on p-value:
Ha: p>p0 p-value = P(X≥x) = P(X=x)+P(X=x+1) +...+ P(X=n)
Ha: p<p0 p-value = P(X≤x) = P(X=0)+P(X=1) +...+ P(X=x)
Ha: p≠p0 p-value = P(X≠x) =1- P(X=x)
If the p-value ≤ α, then reject H0.
If the p-value >α, then fail to reject H0.
Recall Probability of exactly k successes in trials:
𝑛 !
𝑛
!!
𝑃 𝑋 = 𝑘 = 𝐶!! 𝑝!! 1 − 𝑝! !!! =
𝑝! 1 − 𝑝! !!! , where
= !! !!! !
𝑘
𝑘
Example: Kudzu Vine and Curbing Binge Drinking
The hardy, invasive kudzu vine (shown in the picture at the
right), introduced to this country decades ago to control soil
erosion, could have what it takes to curb binge drinking, new
research suggests.
A researcher recruited 10 people in their 20s to spend two 90minute sessions consuming beer and watching TV. Researchers
selected people who said they regularly consumed three to four drinks per day. After the first session,
subjects received capsules of kudzu. None of the subjects had any side effects from mixing kudzu with
beer. Findings showed that 80% of the subjects decreased their amount of beer intake (as compared
to the first session levels).
Question of interest: Can the researcher conclude that a majority of users of kudzu would decrease
their beer intake?
Step1. Parameter: p, true proportion of the subjects who decreased their beer intake
H0: p=0.5 Ha: p>0.5 Significance level α = 0.10 (choose and fix in advance)
Step 2: Verify necessary data conditions and compute an appropriate test statistic:
np 0 = 10 * 0.5 = 5
n(1 − p0 ) = 5
Both conditions are met on the boundary, so we may, but would better not use Z statistics, instead
we will use the observed number of successes x =
10*0.8 =8 as our test statistics.
Step 3: Instead of definition of the rejection
region, we will compute p-value and compare it to
10% significance level:
12
Step 4: Since p-value < .10 we reject H0.
Step 5: Conclusion.
Based on a study of n=10 people, there is significant evidence that a majority of users of kudzu
would decrease their beer intake.
Exercise: A multiple choice test has 10 questions each with 4 possible choices. A particular student
claims he has a ‘gift’ and without even looking at the questions can do better than just guessing. This
leads to the following hypotheses H0: p = 0.25 versus Ha: p > 0.25 where p = the probability of
correctly guessing any question; note that p > ¼ implies the student is better than random guessing,
supporting he has this ‘gift’. His test results come back and he got 8 out of
10 answers correct. Perform small sample hypothesis test for p.
Step1. Determine the null (H0) and alternative (Ha) hypotheses.
Parameter______________________________________________
H0:_______________
Ha:_________________ Significance level α =_________
Step 2: Verify necessary data conditions, and compute an appropriate test statistic
Test statistic is:
Step 3: Assuming H0 is true, define decision rule (based on rejection region or p-value)
The decision rule:
Reject Ho if ______________
Do not reject Ho if __________________________
Step 4: Decide whether or not the result is statistically significant (reject or fail to reject H0):
Based on the sample of n=____________________, there is________ significant
evidence, at level α = ________________,to conclude that ____________________
____________________________________________________________________
13
14

Lecture 1.2 Confidence Interval Estimation 2. Introduction to

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