Download 8.3 Estimating a Population Mean Notes

8.3 ESTIMATING A POPULATION MEAN
When σ Is Known: Recall the Mystery Mean Activity where x­
bar = 240.79 and we have an SRS of size 16 Task was to estimate the mean when we know that the situation is Normal and σ=20. To Calculate a 95% conf. interval for μ we use
statistic ± (critical value)x(standard deviation of statistic)
x­bar ± z*(σ/√n) = 240.79 ± 9.8 = (230.99, 250.59)
We call this a one­sample z interval for a population mean.
The One­Sample z Interval for a Population Mean
Draw an SRS of size n from population having unknown mean μ and known standard deviation σ. As long as the Normal and Independent conditions are met, a level C confidence interval for μ is x­bar ± z*(σ/√n)
The critical value z* is found from the standard Normal distribution.
1
Example: For a data set obtained from a sample, n = 20 and a sample mean equals 24.5. It is known that the population standard deviation is 3.1. The population is Normally distributed.
Make a 99% CI for the population mean.
n = 20
x­bar = 24.5
σ =3.1 Normal
99% implies alpha = z* = 2.576
x­bar ± z*(σ/√n) = 24.5 ± 2.576(3.1/√20) = 24.5 ± 1.7856
so (22.7144, 26.2856 ) is CI
We would say " With 99% confidence the population mean is between (22.7144, 26.2856 )
2
Ex: Assume the population standard deviation is 2.8 inches for the height of adult males. A sample size of 16 is selected and the sample mean is 70 inches. The population is normally distributed.
Make a 95% CI for the population mean.
σ = 2.8
n = 16
x­bar = 70
Normal
z* = 1.96 chart
x­bar ± z*(σ/√n) = 70 ± 1.96(2.8/√16) = 70 ± 1.225
(68.775, 71.225)
Remember when n ≥ 30, any distribution the x­bar is approx. Normal so the conf. interval will be the same.
3
Ex: The standard deviation for a population is 15.3. A sample of 36 observations selected from this population gave a mean of 74.8.
Make a 95% CI for the pop. mean.
σ = 15.3
n = 36
x­bar = 74.8
Normal (N > 30)
z* = 1.96
x­bar ± z*(σ/√n) = 74.8 ± 1.96(15.3/√36) = 74.8 ± 4.998
(69.802, 79.798)
4
Ex: The standard deviation for a population is 15.3. A sample of 36 observations selected from this population gave a mean of 74.8.
Make a 90% CI for the pop. mean.
σ = 15.3
n = 36
x­bar = 74.8
Normal (N > 30)
z* = 1.645
x­bar ± z*(σ/√n) = 74.8 ± 1.645(15.3/√36) = 74.8 ± 4.19475
(70.60525, 78.99475)
So CI got smaller when we lowered the Conf. Level.
and If raise C. level then the CI gets wider.
5
More Examples p141 Black Book
6
CHOOSING THE SAMPLE SIZE
Choosing Sample Size for a Desired Margin of Error When Estimating μ
To determine the sample size n that will yield a level C confidence interval for a population mean with a specified margin of error ME:
1. Get a reasonable value for the population standard deviation σ from an earlier or pilot study.
2. Find the critical value z* from a standard Normal curve for confidence level C.
3. Set the expression for the margin of error to be less than or equal to ME and solve for n:
z*(σ/√n) ≤ ME
In other words solve for n and get
n =(z*)2σ2/E2
7
Example: How Many Monkeys? p500
Always round up to the next whole
number when finding n.
8
Ex: We want to determine the mean weight of all boxes of cereal labeled 400 grams. We need to be 98% confident that our sample mean is within 3 grams of the population mean and a pilot study suggests that the pop. standard deviation is 10 grams.
How large must our sample be?
E = 3
σ = 10
z* = 2.326 chart
n = (z*)2(σ2)/(E)2 = (2.326)2(102)/(3)2 = 60.11418 = round up
sample size = 61 boxes
9
Ex: A psychologist has developed a new test of spatial perception and she wants to establish the mean score achieved by adult male pilots. How many people must she test if she wants the sample mean to be in error by no more than 2.0 points, with 95 % confidence? A pilot study suggests that the pop. standard deviation is 21.2.
E = 2
σ = 21.2
z* = 1.96
n = (z*)2(σ2)/(E)2 = (1.96)2(21.22)/(2)2 = 431.6422 = round up 432 people
10
More Examples p144 Black Book
11
WHEN σ IS UNKNOWN:
The t Distributions
Activity: Calculator Bingo p502
12
The t Distributions; Degrees of Freedom
Draw an SRS of size n from a large population that has a Normal distribution with mean μ and standard deviation of σ, The statistic we will notate t*
t = (xbar ­ μ)/(sx/√n)
has the t distribution with degrees of freedom df = n­1. This statistic will have approximately a tn­1 distribution as long as the sampling distribution of x­bar is close to Normal.
(William S. Gosset)
1. Bell Shaped Distribution
2. Lower Height and wider spread than a standard Normal distribution.
3. As sample size grows larger, the t­distribution approaches standard Normal distribution. [n approaches infinity implies t approaches z]
4. Mean of the t ­ distribution is Zero
5. Standard Deviation of the t­ distribution is Nu
√[df/(df ­ 2)] or √[ν/(ν­2)]
a) df or v = degrees of freedom (i.e. the # of observations that can be chosen "freely".)
b) v = n­1 or sample size ­1 13
Example: 1. We know area in the right tail is .05 and the v = 12 Find t
See Table B
t = 1.782
2. We know area in the Left tail is .025 and n = 66 Find t
n = 66 so v = 66 ­ 1 = 65
see chart t = 1.997 so t = ­1.997 b/c left of curve.
14
Example: p 506 Find t*
df
10
11
12
z*
Upper­tail probability p
.05
.025
.02
1.812
2.228
2.359
1.796
2.201
2.328
1.782
2.179
2.303
1.645
1.960
2.054
90%
95%
96%
.01
2.764
2.718
2.681
2.326
98%
Confidence Level C
15
Technology Corner Inverse t on calculator p506
16
CONSTRUCTING A CONFIDENCE INTERVAL FOR μ
When the conditions for inference are satisfied, the sampling distribution of x­bar has roughly a Normal distribution with mean μ and standard deviation σ/√n. Because we don't know σ, we estimate it by the sample standard deviation sx. We then estimate the standard deviation of the sampling distribution by sx/√n. This value is called the standard error of the sample mean x­bar, or just the standard error of the mean.
Definition: The standard error of the sample mean x­bar is
sx/√n, where sx is the sample standard deviation. It describes how far x­bar will be from μ, on average, in repeated SRS's of size n.
17
The One­Sample t Interval for a Population Mean
Choose an SRS of size n from a population having unknown mean μ. A level C confidence interval for μ is x­bar ± t*(sx/√n)
where t* is the critical value for the tn­1 distribution. Use this interval only when 1) The population distribution is Normal or the sample size is large (n ≥ 30).
2) The population is at least 10 times as large as the sample.
18
Conditions for Inference about a Population Mean
1. RANDOM: the data come from a random sample of size n from the population of interest or a randomized experiment. This condition is very important.
2. NORMAL: The population has a Normal distribution or the sample size is large (N ≥ 30)
3. INDEPENDENT: The method for calculating a confidence interval assumes that individual observations are independent. To keep the calculations reasonably accurate when we sample without replacement from a finite population, we should check the 10% condition: verify that the sample size is no more than 1/10 of the population size.
19
Example: Video Screen Tension p508
269.5 297.0 269.6 283.3 280.4 233.5 257.4 317.5
327.4 264.7 307.7 310.0 343.3 328.1 342.6 338.8
340.1 374.6 336.1
Note: When the sample size is small n<30, as in this example, the Normal condition is about the shape of the population distribution. We inspect the distribution to see if it's believable that these data came from a Normal population.
df
18
19
20
Upper­tail Probability p
.1
.05
.025
1.330
1.734
2.101
1.328
1.729
2.093
1.325
1.725
2.086
80%
90%
95%
Confidence Level C
calculator: invT(.05, 19) = ­1.729
20
Example: Auto Pollution p509
df
29
30
40
Upper­tail probability p
.05
.025
.02
1.699
2.045
2.150
1.697
2.042
2.147
1.684
2.021
2.123
90%
95%
96%
Confidence Level C
21
Using t Procedures Wisely
Definition: Robust procedures
An inference procedure is called robust if the probability calculations involved in that procedure remain fairly accurate when a condition for using the procedure is violated.
Example: More Auto Pollution p511
22
Using One­Sample t Procedures: The Normal Condition
1. Sample Size less than 15: Use t procedures is the data appear close to Normal (roughly symmetric, single peak, no outliers). If the data are clearly skewed or if outliers are present, do not use t.
2. Sample Size at least 15: The t procedure can be used except in the presence of outliers or strong skewness.
3. Large samples: The t procedures can be used even for clearly skewed distributions when the sample is large, roughly N ≥ 30.
23
Example: People, Trees, and Flowers p. 513
24
Technology Corner: p514
25
Ex: A random sample of 25 mid­sized cars gave a mean of 26.4 mpg and a standard deviation fo 2.3 mpg. Assuming that mpg's are Normally distributed, find 95% CI.
n =25
ν = 25­1 = 24
x­bar = 26.4
s = 2.3
x­bar ± t*24(s/√n) = 26.4 ± (2.064)(2.3/√25) = 26.4 ± .94944
(25.4505, 27.34944)
26
EX: Suppose we have only 10 scores representing the ages (yrs) of randomly selected US commercial aircraft. The mean of 10 planes is 14.25 yrs and standard deviation = 9.35 yrs and assuming that the ages of the aircraft are Normally distributed, find 95% CI for the mean age of all US commercial aircraft.
n =10
ν = 10­1 = 9
x­bar = 14.25
s = 9.35
x­bar ± t*9(s/√n) = 14.25 ± (2.262)(9.35/√10) = 14.25 ± 6.6881
(7.5619, 20.9381)
27
Example: A sample computer connection time (in hrs) for 18 selected computers
n =18
ν = 18­1 = 17
x­bar = 16.2
s = 3.4
x­bar ± t*17(s/√n) = 16.2 ± (2.898)(3.4/√18) = 16.2 ± 2.38975
(13.81125,15.58975)
28
Homework: p498 and p518
p518
49­52, 55­59odd, 63
65, 67, 71, 73, 75­78
Chapter Review
Unit 8 Practice Test
Frappy and 10 MC
29