1. Confidence intervals for the mean - continued
Population Mean - μ
Sample mean: X

2. Reminder Point estimator for μ: Limitations of point estimators
Interval estimation for μ

3. A (1-α)% confidence interval for μ
A (1-α)% confidence interval for μ is:
(1-α)%
α/2%
α/2%
Z α/2
Z 1-α/2

4. Example
A test for the level of potassium in the blood is not perfectly precise. Moreover, the actual level of potassium in a person’s blood varies slightly from day to day. Suppose that repeated measurements for the same person on different days vary normally with σ=0.2.

5. (a) Julie’s potassium level is measured three times and the mean result is Give a 90% confidence interval for Julie’s mean blood potassium level.
90% confidence interval for :
=[3.21,3.59]

6. (b) A confidence interval of 95% level would be:
(i) wider than a confidence interval of 90% level
(ii) narrower than a confidence interval of 90% level
(c) Give a 95% confidence interval for Julie’s mean blood
potassium level.
95% confidence interval for :
=[3.174,3.626]

7. (d) Julie wants a 99% confidence interval of [3. 3,3. 5]
(d) Julie wants a 99% confidence interval of [3.3,3.5]. What sample size should she take to achieve this (=how many times should she measure her potassium blood level?)
[3.3,3.5]=3.4±Z0.995(0.2/√n)
3.4-Z0.995(0.2/√n)=3.3
3.4+Z0.995(0.2/√n)=3.5
solve for n:
subtract the first equation from the second  2Z0.995(0.2/√n)=0. 2
Z0.995(0.2/√n)=0.1
2.575(0.2/√n)=0.1
(0.2/√n)=0.0388
√n=5.15
 n=
she needs 27 blood tests to achieve a 99% CI [3.3,3.5].

8. Finding n for a specified confidence interval
Suppose we want a specific interval with a confidence level 1-α. What sample size should be taken to obtain this CI?
Define m = the distance from the mean to the upper/lower limit of the CI
For the blood potassium example:
m= = (| |=0.1)

9. Example
An agricultural researcher plants 25 plots with a new variety of corn. The average yield for these plots is bushels per Acre. Assume that the yield per acre for the new variety of corn follows a normal distribution with unknown μ and standard deviation σ=10 bushels per acre. A 90% confidence interval for μ is:
(a) 150±2.00
(b) 150±3.29
(c) 150±3.92
(d) 150±32.9

10. (a) Plant only 5 plots rather than 25
An agricultural researcher plants 25 plots with a new variety of corn. The average yield for these plots is bushels per Acre. Assume that the yield per acre for the new variety of corn follows a normal distribution with unknown μ and standard deviation σ=10 bushels per acre. Which of the following will produce a narrower confidence interval than the 90% confidence interval that you computed above?
(a) Plant only 5 plots rather than 25
(b) Plant 100 plots rather than 25
(c) Compute a 99% confidence interval rather than a 90% confidence interval.
(d) None of the above

11. Example (a) [59.61,60.39] (b) [59,61] (c) [58.04,61.96]
You measure the weight of a random sample of 25 male runners. The sample mean is kilograms (kg). Suppose that the weights of male runners follow a normal distribution with unknown mean μ and standard deviation σ=5 kg. A 95% confidence interval for μ is:
(a) [59.61,60.39]
(b) [59,61]
(c) [58.04,61.96]
(d) [50.02,69.8]

12. (a) The lengths of the confidence interval would increase
You measure the weight of a random sample of 25 male runners. The sample mean is kilograms (kg). Suppose that the weights of male runners follow a normal distribution with unknown mean μ and standard deviation σ=5 kg. Supposed I had measured the weights of a random sample of 100 runners rather than 25 runners. Which of the following statements is true?
(a) The lengths of the confidence interval would increase
(b) The lengths of the confidence interval would decrease
(c) The lengths of the confidence interval would stay the same
(d) σ would decrease

13. Example The number of observations required is closest to: 25 30 609
Suppose we wanted a 90% confidence interval of length $4 for the average amount spent on books by freshmen in their first year at a major university. The amount spent has a normal distribution with σ=$30.
The number of observations required is closest to: 25 30
609
865

14. m=half of the CI length (=marginal error)α=
4/2=2
0.1

15. Example
You plan to construct a confidence interval for the mean μ of a normal population with known standard deviation σ. Which of the following will reduce the size of the confidence interval?
use a lower level of confidence
Increase the sample size
Reduce σ
All the above

16. Example
A 95% confidence interval for the mean μ of a population is computed from a random sample and found to be 9±3. We may conclude that:
(a) There is a 95% probability that μ is between 6 and 12
(b) There is a 95% probability that the true mean is 9 and there is a 95% probability that the true mean is 3
(c) If we took many additional random samples and from each computed a 95% confidence interval for μ, approximately 95% of these intervals would contain μ.
(d) All of the above

17. Example
The heights of young American women, in inches, are normally distributed with mean μ and standard deviation σ=2.4. I select a simple random sample of four young American women and measure their heights. The four heights, in inches, are:
Based on these data, a 99% confidence interval for μ, in inches, is:
(a) 65±1.55
(b) 65±2.35
(c) 65±3.09
(d) 65±4.07

18. Mean of four heights:
CI=
CI=[61.91,68.09] = = = = = =

19. The heights of young American women, in inches, are normally distributed with mean μ and standard deviation σ=2.4. I select a simple random sample of four young American women and measure their heights. The four heights, in inches, are:
If I wanted the 99% confidence interval to be ± 1 inch from the mean, I should select a simple random sample of size: 2 7 16 39

20. m=half of the CI length (=marginal error)α= 1
0.01