Download 7.3 SAMPLE MEANS Activity:ааPenny for your Thoughts

7.3 SAMPLE MEANS
Activity: Penny for your Thoughts
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Example: Making Money p443
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THE SAMPLING DISTRIBUTION OF X: MEAN AND STANDARD DEVIATION
Mean and Standard Deviation of the Sampling Distribution of x.
Suppose that x(bar) is the mean of an SRS of size n drawn from a large population mean μ and standard deviation σ. Then
The mean of the sampling distribution of x(bar) is μx = μ
The standard deviation of the sampling distribution of x is σx = σ/√n
as long as the 10% condition is satisfied: n ≤ (1/10)N
AP Exam Tip: Notation matters. The symbols p, x, p, μp, σp, μx, and σx all have specific and different meanings. Either use notation correctly ­­­­­or don't use it at all. You can expect to lose credit if you use incorrect notation. 3
Example: This Wine Stinks p444
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SAMPLING FROM A NORMAL POPULATION
Activity p. 446
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Sampling Distribution of a Sample Mean from a Normal Population
Suppose that a population is Normally distributed with mean μ and standard deviation σ. Then the sampling distribution of x(bar) has the Normal distribution with mean μ and standard deviation σ/√n, provided that the 10% condition is met.
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Example: p447 Young Women's Heights
a) μ = 64.5 and σ = 2.5 inches
X = 66.5 standardize z = (66.5­64.5)/2.5 = .8
tableA P(X>66.5) = P(z>.8) = 1­.7881 = .2119
b) 10% condition met ­ Standardize: z = (66.5­64.5)/.79 = 2.53
P(z>2.53) = 1 ­ .9943 = .0057 very unlikely 7
THE CENTRAL LIMIT THEOREM
Activity: p449
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Definition: Central Limit Theorem (CLT)
Draw an SRS fo size n from any population with mean μ and finite standard deviation σ. The central limit theorem (CLT) says that when n is large, the sampling distribution of the sample mean x
(bar) is approximately Normal.
Example: A Strange Population Distribution
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Normal Condition for Sample Means
If the population distribution is Normal, then so is the sampling distribution of x(bar). This is true no matter what the sample size n is. If the population distribution is not Normal, the central limit theorem tells us that the sampling distribution of x will be approximately Normal in most cases if n ≥ 30.
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Example: Servicing Air Conditioners p451
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SHAPE OF THE SAMPLING DISTRIBUTION
1. If the sampling is from a Normal Distribution then the μx = μ
2. The standard deviation of the sample mean is σx = σ/√n
3. The Shape of the sampling distribution of the sample mean is Normal
So we can get Z ­ scores for our observations where
Z = (x ­ μ)/(σ/√n)
SHAPE OF THE SAMPLING DISTRIBUTION Continued
1. If the sampling is from a Non ­ Normal Distribution then the μx = μ
2. The standard deviation of the sample mean is σx = σ/√n
3. and By the Central Limit Theorem if n ≥ 30 the sampling distribution of the sample mean is Approximately Normal.
So we can use Z scores Z = (x ­ μ)/(σ/√n)
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Example: A highway construction zone has a speed limit of 40mph. The actual speed of the vehicles are normally distributed with a mean of 46mph and a standard deviation of 3 mph. Find the probability that the mean speed of a random sample of 20 cars going through the zone is more than 45mph.
Write out info that you know!
Normally Dist.
P(x > 45)
μ = 46
σ = 3
n = 20
*Must standardize into Z scores!! where Z = (x ­ μ)/(σ/√n)
P(Z > (45­46)/(3/√20)) P(Z > ­1.49) Go to the Table
Remember Z table gives area to the LEFT
total area chart(­1.49)
So P(Z > ­1.49) = 1 ­ .0681 = .9319
Implies there is a 93.19% chance that cars are going over 45mph in the construction zone.
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Example: Construction Zone Less than 45.5 mph
Normally Dist.
μ = 46
σ = 3
n = 20
P(x < 45.5) = P(Z < (45.5­46)/(3/√20)) = P(Z < ­.75)
Table (­.75) = .2266
So 22.66% chance that cars are traveling less than 45.5 mph through zone.
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Example: Speed zone Between 44.5 and 47 mph
Normally Dist.
μ = 46
σ = 3
n = 20
P( 44.5 < Xbar < 47) convert to Z
P((44.5­46)/(3/√20)) < Z < (47­46)/3/√20))
P (­2.24 < Z < 1.49) go to chart .9319 ­ .0125 = .09194 = 9.194% that a car will be traveling between 44.5 and 47 mph through the construction zone.
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Example: Time that college students spend studying per week have a distribution that is Skewed to the Right with a mean of 8.4 hours and a standard deviation of 2.7 hours. Find the probability that the mean studying time for a random sample of 45 students is between 8 and 9 hours of studying.
Skewed Right distribution but CLT allows us to use Normal since n ≥30
μ = 8.4
σ = 2.7
n = 45 P(8 < xbar < 9) = standardize
P((8­8.4)/(2.7/√45) < Z < (9­8.4)/2.7/√45)) = P(­.99 < Z < 1.49)
z = .1611 and .9319 so
P = .9319 ­ .1611 = .7708
therefore a 77.08 % chance that the students would study between 8 and 9 hours.
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Example: College Study Time ­ Less than 8 hours
μ = 8.4
σ = 2.7
n = 45
P(xbar < 8) = P (Z < (8­8.4)/(2.7/√45)) = P(Z < ­.99)
z = .1611 so 16.11% chance of studying less than 8 hours.
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Example: Assume the weights of all packages of a brand of cookies are Normally distributed with μ = 32 ounces and σ = .3 ounces. Find the probability that the mean weight of a random sample of 20 packages will be between 31.8 ounces and 31.9 ounces from this brand.
Normal μ = 32
σ = .3
n = 20 P(31.8 < xbar < 31.9)
P((31.8­32)/(.3/√20) < Z < (31.9­32)/(.3/√20))
P( ­2.98 < Z < ­1.49) go to chart z = .0681 ­ .0014 = .0667
6.67% that the cookies will weigh between 31.8 and 31.9 ounces.
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Example: Consumers in the US owed an average of $7868 on their credit cards in 2004. Suppose the shape of the prob. distribution is unknown but its mean is $7868 and σ = $2160. What is the probability that the mean of the current cc debt from a random sample of 81 US consumer is lower than population mean by $320 or more.
μ = 7868
σ = 2160
n = 81 Dist type unknown
but CLT allows us to use Normal
P(xbar < (μ­320)) = P(xbar < 7548) = P(Z < (7548­7868)/(2160/√81) = P(Z < ­1.33)
chart = .09181
So 9.181% chance that the debt is lower than pop. mean by $320 or more
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Example for Sample Proportions: A survey of all medium and large corporations showed that 64% of them offer a retirement plan. Suppose 50 corporations are randomly sampled. What is the probability tat the sample proportion of the corporations that offer retirement plans will be between .54 and .61?
p = .64
n = 50
need to check criteria for approx. normal
np >5 and n(1­p) >5
so np = (.64)(50) = 32 >5 and n(1­p) = 50(.36) = 18 > 5
We can apply CLT and standardize
P(.54 < p­hat < .61) app. Normal = Z = (p ­ hat ­ p)/√(p(1­p)/n)
P((.54 ­ .34)/√((.64)(.36)/50) < Z < (.61­.64)/√((.64)(.36)/50)
P( ­1.47 < Z < ­.44) .3300 ­ .0708 = .2592
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Ex: Retirement What is the probability that the sample proportion is greater tahn .71
P(p­hat > .71) appr. P(Z > (.71 ­ .64)/√(.64)(.35)/50)) = P(Z > 1.03)
1 ­ ch(1.03) = 1 ­ .8485 = .1515
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Ex: A mayor in city running for re­election. He claims that he is favored by 53% of voters in city. Assume his claim is true. What is the probability that in a random sample of 400 registered voters that less than 49% will favor the mayor?
p = .53
n = 400
np = 212>5
need to check np>5 and n(1­p) > 5 so we can use CLT
n(1­p) = 188 > 5 approx. Normal
P(p­hat < .49) = approx P(Z < (.49­.53)/√(.53)(.47)/400)
P(Z < ­1.60)
ch(­1.60) = .0548
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Ex: 38% of adult Americans say they are very satisfied with their lives. Suppose this is true. In a random sample of 1000 adults, what is the probability that between .40 and .42 adults will respond that they are satisfied with their lives?
p = .38
n = 1000 check np and n(1­p)
P(.4 < p­hat < .42) approx. P ((.4­.38)/√(.38)(.62)/1000) < Z< (.42 ­ .38)/√(.38)(.62)/1000
P(1.30 < Z < 2.61) = .9955 ­ .9032 = .0923
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Ex: It is known taht 85% of all orders that a warehouse receives from customers are delivered on time. In a random sample of 100 orders, what is the probability that the proportion of orders delivered on time is less than .87?
p = .85 n = 100 check np and n(1­p)
P (p­hat < .87) approx. P(Z > (.87­.85)/√(.85)(.15)/100)
P(Z < .56) = chart (.56) = .7123
71.23% 24
Homework:
p439 and p454 43­46, 49­55odd
p454 57­63odd, 65­68
Chapter 7 Review
Chapter 7 Practice Test Strive for 5 ­ MC and Frappy
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