Inferences for a Single Population Mean ( )



9 - Statistical Inference

9.5 - Comparing Two Population Means Using Dependent

Samples or Matched Pairs ([pic]) (Daniel, Section 7.4)

When using dependent samples each observation from population 1 has a one-to-one correspondence with an observation from population 2. One of the most common cases where this arises is when we measure the response on the same subjects before and after treatment. This is commonly called a “pre-test/post-test” situation. However, sometimes we have pairs of subjects in the two populations meaningfully matched on some pre-specified criteria. For example, we might match individuals who are the same race, gender, socio-economic status, height, weight, etc... to control for the influence these characteristics might have on the response of interest. When this is done we say that we are “controlling for the effects of race, gender, etc...”. By using matched-pairs of subjects we are in effect removing the effect of potential confounding factors, thus giving us a clearer picture of the difference between the two populations being studied.

DATA FORMAT

|Matched Pair |[pic] |[pic] |[pic] |

|1 |[pic] |[pic] |[pic] |

|2 |[pic] |[pic] |[pic] |

|3 |[pic] |[pic] |[pic] |

|... |... |... |... |

|n |[pic] |[pic] |[pic] |

The hypotheses are

[pic]

Test Statistic for a Paired t-test

[pic]t-distribution with df = n - 1

Note: [pic]the hypothesized value for the mean paired difference

100(1-[pic])% CI for [pic]

[pic] where [pic] comes from the appropriate quantile of t-distribution df = n – 1.

This interval has a 100(1-[pic])% chance of covering the true mean paired difference.

Example: Effect of Captopril on Blood Pressure

In order to estimate the effect of the drug Captopril on blood pressure (both systolic and diastolic) the drug is administered to a random sample n = 15 subjects. Each subjects blood pressure was recorded before taking the drug and then 30 minutes after taking the drug. The data are shown below.

[pic]

Research Questions:

• Is there evidence to suggest that Captopril results in a systolic blood pressure decrease of at least 10 mmHg on average in patients 30 minutes after taking it?

• Is there evidence to suggest that Captopril results in a diastolic blood pressure decrease of at least 5 mmHg on average in patients 30 minutes after taking it?

For each blood pressure we need to consider paired differences of the form [pic]. For paired differences defined this way, positive values correspond to a reduction in their blood pressure ½ hour after taking Captopril. To answer research questions above we need to conduct the following hypothesis tests:

[pic] and [pic]

Below are the relevant statistical summaries of the paired differences for both blood pressure measurements.

[pic]

The t-statistics for both tests are given below:

Systolic BP Diastolic BP

We can use the t-Probability Calculator in JMP to find the associated p-values or better yet use JMP to conduct the entire t-test.

Systolic Blood Pressure Diastolic Blood Pressure

[pic]

Both tests result in rejection of the null hypotheses. This we have sufficient evidence to suggest that taking Captopril will result in mean decrease in systolic blood pressure exceeding 10 mmHg (p = _______) and a mean decrease in diastolic blood pressure exceeding 5 mmHg (p = _______). Furthermore we estimate that the mean change in systolic blood pressure will be somewhere between _______ mmHg and ______ mmHg, and that the mean change in diastolic blood pressure could be as large as ______ mmHg.

9.6 – Comparing Two Population Means Using Independent

Samples (Daniel, Sections 6.4 & 7.3)

Example 1: Effect of Cadmium Oxide on Hemoglobin Levels in Dogs

An experiment was conducted to determine the examine the potential effect cadmium oxide might have on the hemoglobin levels of dogs. It is thought that cadmium oxide exposure would lead to decreased hemoglobin levels. 10 dogs were randomly assigned to the control group and 15 were randomly assigned to the cadmium oxide exposure group.

Research Question: Is there evidence to suggest the cadmium oxide exposure lowers the hemoglobin level found in dogs?

[pic]

To answer the question of interest we need tools for comparing the population mean hemoglobin level for dogs not exposed to cadmium oxide vs. that for dogs that have had cadmium oxide exposure, i.e. how does [pic] compare to [pic].

Basic Idea:

Case 1 ~ Equal Populations Variances/Standard Deviations ([pic]=[pic]( common variance to both populations)

Assumptions:

For this case we make the following assumptions

1. The samples from the two populations were drawn independently.

2. The population variances/standard deviations are equal.

3. The populations are both normally distributed. This assumption can be relaxed

when the samples from both populations are “large”.

100(1 - [pic])% Confidence Interval for ([pic])

[pic]

where

[pic]

[pic] is called the “pooled estimate of the common variance [pic]”. The degrees of freedom for the t-distribution is [pic].

CI Example: Cadmium Exposure and Hemoglobin Levels

[pic]

Hypothesis Testing ([pic])

The general null hypothesis says that the two population means are equal, or equivalently there difference is zero. The alternative or research hypothesis can be any one of the three usual choices (upper-tail, lower-tail, or two-tailed). For the two-tailed case we can perform the test by using a confidence interval for the difference in the population means discussed above.

[pic]

Test Statistic

[pic]

where the [pic] is as defined in the confidence interval section above.

Testing Example: Cadmium Exposure and Hemoglobin Levels

In JMP

EXAMPLE 2: Normal Human Body Temperatures Females vs. Males

Do men and women have the same normal body temperature? Putting this into a statement involving parameters that can be tested:

[pic] or [pic]

[pic]or [pic]

Intuitive Decision

In order to determine whether or not the null or alternative hypothesis is true, you could review the summary statistics for the variable you are interested in testing across the two groups.  Remember, these summary statistics and/or graphs are for the observations you sampled, and to make decisions about all observations of interest, we must apply some inferential technique (i.e. hypothesis tests or confidence intervals)

One of the best graphical displays for this situation is the side-by-side boxplots.  To get side-by-side boxplots, select Analyze > Fit Y by X.  Place Gender in the X box and Temperature in the Y box.  Place the mean diamonds on the boxplots and jitter the points.  The more separation there is in the mean diamonds, the more likely we are to reject the null hypothesis (i.e data tends to support the alternative hypothesis).

[pic]

[pic]

Assumptions

1. The two groups must be independent of each other.

2. The observation from each group should be normally distributed.

3. Decide whether or not we wish to assume the population variances are equal.

Assessing Normality of the Two Sampled Populations

To assess normality we select Normal Quantile Plot from the Oneway Analysis pull-down menu as shown below.

[pic]

Checking the Equality of the Population Variances

To test the equality of the population variances select Unequal Variances from the Oneway Analysis pull-down menu.

[pic]

The test is:

[pic]

JMP gives four different tests for examining the equality of population variances.  To use the results of these tests simply examine the resulting p-values. If any/all are less than .10 or .05 then worry about the assumption of equal variances and use the unequal variance t-Test instead of the pooled t-Test.

[pic]

Performing the Test

To perform the two-sample t-test for independent samples:

• assuming equal population variances select the Means/Anova/Pooled t option from Oneway-Analysis pull-down menu.

• assuming unequal population variances select t-Test from the Oneway-Analysis pull-down menu.

[pic]

Several new boxes of output will appear below the graph once the appropriate option has been selected, some of which we will not concern ourselves with. The relevant box for us will be labeled t Test as shown below for the mean body temperature comparison.

[pic]

• What is the test statistic for this test?

 

• What is the p-value?

 

• What is your decision for the test? 

 

• Write a conclusion for your findings. 

 

Construct and Interpret a 95% CI for the Difference in the

Mean Body Temperatures [pic]

For body temperature and gender example we have:

Interpretation of the CI for [pic]

Case 2 ~ Unequal Populations Variances/Standard Deviations ([pic])

Assumptions:

For this case we make the following assumptions

1. The samples from the two populations were drawn independently.

2. The population variances/standard deviations are NOT equal.

(This can be formally tested or use rule o’thumb)

3. The populations are both normally distributed. This assumption can be relaxed

when the samples from both populations are “large”.

100(1 - [pic])% Confidence Interval for ([pic])

[pic]

where

[pic]

The t-quantiles are the same as those we have seen previously.

Hypothesis Testing

Test Statistic

[pic]t-distribution with df = (see formula above)

where the [pic] is as defined in the confidence interval section above.

Example: Cell Radii of Malignant vs. Benign Breast Tumors

In your previous work with these data you noticed that the radii of malignant breast tumor cells were generally larger than the radii of benign breast tumor cells. Assuming the researchers initially hypothesized that cancerous breast tumor cells have larger radii than non-cancerous cells, conduct a test to see if this is supported by these data.

[pic]

The cell radii of the malignant tumors certainly appear to be larger than the cell radii of the benign tumors. The summary statistics support this with sample means/medians of rough 17 and 12 units respectively. The 95% CI’s for the mean cell radius for the two tumor groups do not overlap, which further supports a significant difference in the cell radii exists.

Formally Testing the Equality of Population Variances (see Section 7.8)

In JMP

[pic]

Because we conclude that the population variances are unequal we should use the non-pooled version to the two-sample t-test. No one does this by hand, so we will use JMP.

[pic]

Conclusion:

9.7 - Comparing Two Population Proportions Using

Independent Samples [pic] (Daniel, Section 7.6)

100(1 - [pic])% Confidence Interval for ([pic])

[pic] ( (provided [pic]are “large”)

where,

[pic]

and

|Confidence Level | z |

|95 % ([pic] | 1.96 |

|90 % ([pic]) | 1.645 |

|99 % ([pic]) | 2.576 |

Hypothesis Testing

Test Statistic

[pic]standard normal dist. provided [pic] are “large” (see above)

where the [pic] is as defined in the confidence interval section above

[pic]

Example: In a study conducted to investigate the non-clinical factors associated with the method of surgical treatment received for early-stage breast cancer, some patients underwent a modified radical mastectomy while others had a partial mastectomy accompanied by radiation therapy. We are interested in determining whether the age of the patient affects the type of treatment she receives. In particular, we want to know whether the proportions of women under 55 are identical in the two treatment groups.

In a sample of n = 658 women who underwent a partial mastectomy and subsequent radiation therapy contains 292 women under 55, which is a sample percentage of 44.4%.

In another independently drawn sample of n = 1580 women who received a modified radical mastectomy 397 women were under 55, which is a sample percentage of 25.1%.

Conduct a test comparing the proportion of women each group under the age of 55 and construct a 95% confidence interval for the difference in these two proportions.

Fisher’s Exact Test for Comparing Two Proportions (in JMP)

Enter these data as you would for setting up a 2 X 2 contingency table.

[pic]

In JMP, select Analyze > Fit Y by X and place Surgery in the X box and Age in the Y.

The following output from JMP is obtained.

[pic] [pic]

The results of Fisher's Exact Test are always included in the JMP output whenever we are working with a

2 X 2contingency table.

[pic]

The three p-values given are for testing the following:

(1) Left,  p-value = 1.000 is for testing if the proportion of women under age 55 is larger for the modified radical mastectomy group than the partial mastectomy group. This is clearly not supported as the p-value >> .05. Obviously we would not conclude this when only 25% of women in the mod. rad. Group were under age 55 compared to 44.4% in the partial mastectomy group.

(2) Right, p-value < .0001 is for testing if the proportion of women under age 55 is larger for the partial mastectomy group than the modified radical mastectomy group.  The fact this p-value is highly significant suggests that the proportion of women under age 55 in partial mastectomy group is indeed greater than the proportion under 55 in the modified radical mastectomy group.  This was the research hypothesis.

(3) 2-Tail, p-value < .0001 is for testing if the proportion of women under age 55 differs between the two surgery groups.  The fact this p-values is highly significant suggests that these proportion of women under age 55 is not the same for both surgery groups. 

Example 2: Low Birth Weight and Smoking

These data come from a study looking at the effects of smoking during pregnancy on birth weight. Amongst the 381 non-smokers in the study, 13 had babies with low birth weight, while amongst the 299 mothers who smoked during pregnancy, 28 had babies with low birth weight. Is there evidence to suggest that the proportion of babies born with low birth weight is greater for mothers who smoked during pregnancy?

| | | | |

| | | | |

| |Normal Birth | | |

|Smoking |Weight |Low Birth |Row |

|Status | |Weight |Totals |

| | | | |

|Non-smoker |368 |13 |381 |

| |96.59% |3.41% | |

| | | | |

|Smoker |271 |28 |299 |

| |90.64% |9.36% | |

| | | | |

|Column |639 |41 |680 |

|Totals | | | |

Hypothesis Test:

1)

2)

3)

4)

5)

Construct and interpret a 95% CI for ([pic]

Fisher’s Exact Test Results from JMP

[pic]

Conclusion:

Find and interpret the RR and OR for low birth weight associated with smoking during pregnancy (Note: this was on Assignment 3).

9.8 - Confidence Intervals for the RR and OR

(Daniel, Section 12.7)

1) OR = ________

RR = _________________

95% CI for OR:

1) Take natural logarithm of OR, to obtain [pic].

2) Compute SE(ln(OR)) = [pic]

3) Find [pic] to obtain (LCL, UCL)

4) 95% CI for OR is then given by [pic]

Smoking and Birthweight Example:

CI for RR (not in text):

1) Take natural logarithm of RR to obtain [pic].

2) Compute SE(ln(RR)) = [pic]

3) Find [pic] to obtain (LCL, UCL)

4) 95% CI for RR is then given by [pic]

Smoking and Birthweight example:

-----------------------

For the sample paired differences ([pic]) find the sample mean [pic] and standard deviation ([pic].

We actually can hypothesize any size difference for the mean of the paired differences that we want. For example if wanted to show a certain diet resulted in at least a 10 lb. decrease in weight then we could test if the paired differences: d = Initial weight – After diet weight had mean greater than 10 ([pic])

Syspre – initial systolic blood pressure

Syspost – systolic blood pressure 30 minutes after taking the drug

Diapre – initial diastolic blood pressure

Diapost – diastolic blood pressure 30 minutes after taking the drug

Rule of Thumb for Checking Variance Equality

If the larger sample variance is more than twice the smaller sample variance do not assume the variances are equal.

[pic] if [pic]

[pic]

[pic] mean body temperature for females.

[pic] mean body temperature for males.

Summary Statistics

[pic]

Normality appears to be satisfied here.

p-values for testing variances

Because we have no evidence against the equality of the population variances assumption we will use a pooled t-Test to compare the population means.

[pic]

Because we have concluded that the equality of variance assumption is reasonable for these data we can refer to the output for the t-Test assuming equal variances.

[pic]

Summary Statistics

[pic]

[pic]

CI for [pic]

[pic]

[pic]

[pic] or equivalently [pic]

Test Statistic

[pic] which has an F-distribution with numerator df = [pic]and denominator df = [pic] if [pic] and are reversed if [pic].

If F is large then one variance is several times larger than the other and we should reject the null in favor of the alternative. There is separate F-table for each level of significance. If our test statistic value exceeds the value in the table for appropriate level of significance and degrees of freedom we reject the null hypothesis. BETTER TO JUST USE JMP!!!

“Large” sample sizes

Both samples should be larger than 25 and both samples should have more than 5 “successes” and more than 5 “failures”

| |Disease |Disease |

| |Present |Absent |

|Risk factor present |a |b |

|Risk factor absent |c |d |

| | | | |

| | | | |

| |Normal Birth | | |

|Smoking |Weight |Low Birth |Row |

|Status | |Weight |Totals |

| | | | |

|Non-smoker |368 |13 |381 |

| |96.59% |3.41% | |

| | | | |

|Smoker |271 |28 |299 |

| |90.64% |9.36% | |

| | | | |

|Column |639 |41 |680 |

|Totals | | | |

| | | | |

| | | | |

| |Normal Birth | | |

|Smoking |Weight |Low Birth |Row |

|Status | |Weight |Totals |

| | | | |

|Non-smoker |368 |13 |381 |

| |96.59% |3.41% | |

| | | | |

|Smoker |271 |28 |299 |

| |90.64% |9.36% | |

| | | | |

|Column |639 |41 |680 |

|Totals | | | |

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