1. Sumukh Deshpande n Lecturer College of Applied Medical Sciences Statistics = Skills for life. BIOSTATISTICS (BST 211) Lecture 7

2. Confidence Intervals About 

3. Revision Sample parameters are ESTIMATES of population parameters Sample parameters → Population parameters… (INFERENTIAL STATS) Parameter Mean  SD S  SamplePopulation

4. Example pediatric paracetamol syrup claims to contain 120ml. Measurement of 49 bottles revealed an average of 116ml. Does this mean the bottle claim is wrong? How sure can you be when you say the bottle contains 120ml? 116ml? X ml?

5. Summary Sample mean is an ESTIMATE of population mean Point estimate means ONE single value.. One number! What is written on the bottle should be , the TRUE POPULATION MEAN What you measured is…. Sample mean, xbar

6. Point Estimate A point estimate of a parameter is the value of a statistic that estimates the value of the parameter. s is the best POINT ESTIMATE of  But, what if the xbar misses  ? or s misses  ?

7. Confidence Interval Estimate A confidence interval estimate of a parameter consists of an interval of numbers along with a probability that the interval contains the unknown parameter.

8. Confidence Level Confidence Interval Confidence LEVEL Interval width (Error) Higher confidence level → wider interval → less precise estimate

9. Confidence Intervals About   known

10. Back to Example1 Measurement of 49 bottles of syrup revealed an average of 116ml. If the bottles are normally distributed and SD is known to be 7ml, estimate the true mean with 95% confidence?.

11. What do we know/Don’t know?  Sample:  n = 49  Xbar = 116ml  Population:   = ?   = 7ml We want to calculate the LOWER and UPPER limits

12. Solution page1 Population SD,  = 7ml Sample mean SD is SEM = So, SEM = 7/  49 = 1ml We want a 95% CI, so error is only 5%,  = 0.05 Normal Distribution = symmetry… 0.025 on each side Find z corresponding to P(z) = 0.025 or (1 - 0.025)

13. Solution page2 

14. Solution Page3 We are 95%confident that  is between 114.04 and 117.96 ml Given this data, we can say that the claim of 120ml on syrup bottle is 95% FALSE!!!

16. Summary of Process  If the data follows N and  is known:  Have a sample of size n and mean xbar  Have level of confidence CI%, then  = (100 – CI) 

17. Interpretation of Confidence Interval (1 –  )100% Confidence Interval: if you take many random samples of size (n≥30), from a normally distributed population of known SD = , then (1 –  )100% of the intervals will contain .

18. Example 2 Weight of watermelons from Bagaa follows a normal distribution with SD = 0.5 kg. 36 melons were weighed and averaged 8 kg. Estimate the true population mean weight with 90% CI? Also 99% CI?

19. What do we know/Don’t know?  Sample:  n = 36  Xbar = 8 kg  Population:   = ?   = 0.5 kg Required levels of confidence 90% and 99%. then  = (100 – CI)= 0.1 and 0.01respectively.

20. Summary of Process  If the data follows N and  is known:  Have a sample of size n and mean xbar  Have level of confidence CI%, then  = (100 – CI) 

21. Solution page4 Population SD,  = 0.5 kg Sample mean SD is SEM = So, SEM = 0.5/  36 = 0.083 kg We want a 90% CI, so error is 10%,  = 0.1 Normal Distribution = symmetry… 0.05 on each side Find z corresponding to P(z) = 0.05 or (1 - 0.05)

22. Solution page5 

23. Solution Page3 We are 90%confident that  is between 8.14 and 7.86 kg Now you work out the 99% confidence interval? We are 99%confident that  is between 8.19 and 7.81 kg

24. Summary of Lecture 1  Point Estimate (single value)  Interval Estimate  SEM =   = (100 – CL)

25. Summary of Lecture 2 True mean sample mean How confident? Population SD sample size Margin of Error Estimate