Chapter 10. AP Statistics St. Francis High School Fr. Chris. Introduction to Inference. 1. Estimating with Confidence. 2. Tests of Significance. 3. Using Significance Tests. 4. Inference as a Decision. Confidence Interval. Two parts: the estimate and the confidence level
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1. Estimating with Confidence
2. Tests of Significance
3. Using Significance Tests
4. Inference as a Decision.
To establish the concentration of an active ingredient of a drug, the same specimen of a particular batch is measured three times: 0.8403, 0.8363, and 0.8447. The standard deviation is known to be 0.0068 grams per liter. What is the 95% confidence interval?
So .8404 is a good estimate of the actual amount, but how big a margin of error do we need to be 95% confident the actual measurement is in our interval?
On a standard normal curve (N[0,1]), 95% is between -1.96 and +1.96 standard deviations.
Since we know sigma, we convert the drug’s distribution to the standard normal so the interval is
95%
+1.96
-1.96
Where z = 1.96 (95% Confidence)
So we can be 95% sure that the actual measurement of the sample is between .8303 and .8505
We would have a more narrow interval if we are willing to drop our confidence interval to 90%. On a standard normal curve, 90% of the area is between -1.645 and +1.65...
Or (.8339, .8469)
Males age 33-44 have a systolic blood pressure of 128 with a =15.
A researcher looks at the records of 72 male ATC’s in this age
group and found their average systolic blood pressure to be
131.07. Obviously this is higher than the average. It is likely thatthis is due to chance variation, or that ATC’s have a significantly higher blood pressure? It is SIGNIFICANT if the result is unlikelyto be caused by chance.
Males age 33-44 have a systolic blood pressure of 128 with a =15.
72 of our ATC’s average at over 131. The null Hypothesis is to say
There is no difference between ATC’s and males in general.
H0 : = 0
One way to state the alternative is to say that the average of ATC’s is different, so the alternative hypothesis is
Ha : ≠ 0
This is a “Two-tail” test, since the alternative includes two possibilities: Either the ATC’s have a higher blood pressure or they have a lower blood pressure.
A two tailed test means that the sample mean falling on either extreme will be considered grounds to reject the H0
Knowing what we know about the Normal Distribution, we know the critical z score that partitions the inner 95% is where the cumulative area is at 2.5% on both ends (thus our =.05)
If they belong to the same distribution as anyone else, we would expect the of the sample mean of 72 randomly selected males to be
So to convert to the Standard Normal Curve...
Since 1.7366 is less than 1.96, we are in the 95% of sample means, and we cannot reject the H0 …
But what if we only consider the alternative that ATC’s, due to the stressful nature of their work, would have a HIGHER blood pressure than other males...
The one tailed test has the same H0 : =0but a different alternative hypothesis, viz., Ha : >0
The critical value is different, since 95% is on the left, and 5% is on the upper end.
Since our critical value is now 1.645, and the sample mean of 131.07 has a z value of 1.7366, it is in the extreme upper 5%, so we can reject the null hypothesis in favor of the alternative hypothesis: The mean of ATC’s systolic blood pressure is higher than other males of this age group.
Notice our is still .05, but only when we consider the one-tailed test (which is appropriate in this case), are we able to reject the null hypothesis. Computers and statistical calculators can compute a p-value (probability) which allows to to see how unlikely the average is if the null hypothesis is true.
First let’s select the two tailed alternative hypothesis then press Calculate
Using your calculatorSelect STAT, then TESTS
Select “1:Z-Test”
Select “Stats” since we don’t have the raw data (72 sbp’s)
Notice that it shows the z-score, and the p-value (.08 is bigger than .05). So we cannot reject the null hypothesis.
Go back and instead of selecting “Calculate”, select “Draw”
Select STAT, then TESTS
Select “1:Z-Test”
Select “Stats” since we don’t have the raw data (72 sbp’s)
First let’s select the two tailed alternative hypothesis >0 then press Calculate
Notice that it shows the z-score, and the p-value (.04 is less than .05). So we can reject the null hypothesis.
Go back and instead of selecting “Calculate”, select “Draw”
Hawthorne Effect: Will background music increase productivity? Any short term change can effect, regard less of what it is (See Example 10.19, p 590)
A researcher looking for evidence of ESP tests 500 subjects. Four do significantly better (P<.01) than random guessing.
Can we conclude these 4 have ESP? What should the researcher do to test whether these 4 have ESP? (10.61 p. 592)
Patients with kidney failure are often treated by dialysis, and can also cause the retention of phosphorous that must be corrected by changes in the diet. One patient in a study had these 6 measurements over time: 5.6, 5.3, 4.6, 4.8, 5.7, 6.4 (These are separated over time and can be considered an SRS of the patient’s blood phosphorous level). If this level varies normally with = 0.9 mg/dl, give a 90% confidence interval for the mean blood phosphorus level.
(4.7956, 6.0044)
The normal range of phosphorous in the blood is considered to be 2.6 to 4.8 mg/dl. Is there strong evidence that the patient has a mean phosphorous level that exceeds 4.8?
What are the hypotheses?
Conclusion?
H0: =4.8 Ha: >4.8
There is evidence that the patient’s mean phosphorous level is higher than 4.8 mg/dl, but it is not strong. Whether this is sufficient evidence is a judgement call. .05 is not sacred
What if the inference is use to determine a change in diet? Surgury?
What is the z-score/P-value?
Z=1.633
P-value is .051235