1. Sta220 - Statistics
Mr. Smith
Room 310
Class #17

2. Section 5-3 and 5-5 Notes

3. 5-3: Confident Interval for a Population Mean: Student’s t-Statistic

4. 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.

5. The use of a small sample in making an inference about 𝜇 presents two immediate problems when we attempt to use the standard normal z as a test statistic.
We can no longer assume that the sampling distribution of 𝑥 is approximately normal, the CLT ensures normality only for samples that are sufficiently large. (n > 30).
Though still true that 𝜎 𝑥 =𝜎/ 𝑛 , the sample standard deviation s may provide a poor approximation for 𝜎 when the sample size is small.

6. T-statistic
t-statistic is used when the sample size is small. 𝑡 = 𝑥 –𝜇 (𝑠/ 𝑛 ) t-statistic has (n – 1) degrees of freedom (df)

7. Standard normal (z) distribution and t-distributions
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8. Table VI
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9. The t.025 value in a t-distribution with 4 df, and the corresponding z.025 value
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10. 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.
80% confidence interval
95% confidence interval

11. Solution on the Board

12. Example 5.3.2
Let 𝑡 0 be a specific value of t. Use Table IV to find 𝑡 0 values such that the following statements are true:
𝑃 𝑡≥ 𝑡 0 = .01, where df = 17
𝑃 𝑡≤ 𝑡 0 = .005, where df = 6

13. Solution on the Board

14. Procedure
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15. Example
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 𝜇, the mean increase in blood pressure associated with the new drug for all patients in the population.

16. 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

17. Solution On the Board
We can be 95% confident that the mean increase in blood pressure associated with taking this new drug is between 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 𝜇, assuming that the probability distribution of changes in blood pressure from which our sample was selected is normal.

18. 5-5: Determining the Sample Size

19. 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   𝑥 was within .78 days of the true mean length of stay, 𝜇, for all the hospital’s patients at the 95% confidence level for 𝜇 was 2(.78) = 1.56 days wide when 100 accounts were sampled. Suppose we want to estimate 𝜇 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.

20. Relationship between sample size and width of confidence interval: hospital-stay example
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21. How much will the sample size have to be increased to accomplish this
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.

22. Specifying the sampling error SE as the half-width of a confidence interval
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23. Procedure
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24. Example
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?

25. Solution on the Board
We round this up to n = 107. Realizing that 𝜎 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.

26. Assignment 5-2, 5-3, 5-5 all Due Friday Friday 6/1-6/2