Lesson 9 - R

Sampling Distributions

Objectives• Define a sampling distribution• Contrast bias and variability• Describe the sampling distribution of a sample

proportion (shape, center, and spread)• Use a Normal approximation to solve probability

problems involving the sampling distribution of a sample proportion

• Describe the sampling distribution of a sample mean• State the central limit theorem• Solve probability problems involving the sampling

distribution of a sample mean

Vocabulary• None New

Effect of Sample Size on σ

• Sampling distribution of sample means have the same mean as population, but have tighter spreads (less variance) than the population

• Sampling distribution of sample proportions have the same proportion as the population, but also have less variance than the population

• As n increases, then σ decreases

DistributionStandard Deviation

Example

Population σ σ = 10

Sampling of x-bar

σ/√nn = 9

σx-bar = σ/√n = 10/√9 = 10/3 = 3.333

Central Limit Theorem

Regardless of the shape of the population, the sampling distribution of x-bar becomes approximately normal as the sample size n increases.

Caution: – only applies to shape and not to the mean or

standard deviation

Conditions: (requirements for x-bar)– SRS– Rule of Thumb: n ≥ 30 (big enough to apply)

P-hat Requirements

Sampling Distribution of p-hat

• SRS

• For a simple random sample of size n such that n ≤ 0.1N (sample size is ≤ 10% of the population size) [keeps binomial vs going to hypergeometric]

• The shape of the sampling distribution of p-hat is approximately normal provided n(1 – p) ≥ 10 and n(1-p) ≥ 10 [allows use of normal approximations of binomial]

X vs X-bar

• When problem talks about a single instance or event, then we are dealing with X and not with X-bar.– If normally distributed use σ, μ, and X– If nothing is said about the distribution of

population, then you cannot assume normality (problem is not workable)

• When problem talks about a sample of size n, then we are dealing with X-bar and not with X– Use σX-bar = σ/√n and X-bar

X vs P-hat

• When problem talks about values that are not in percentages or in numbers greater than 1, then they are talking in terms of X– Convert x to percentage (divide by n) before using– Work in percentages (p-hat)

• When problem talks about percentages or values less than one, then they are (more than likely) talking in terms of p-hat.– Use values directly

• Read problem carefully

Chapter 9 Summary• The sample mean and the sample proportion can be

considered as random variables in sampling distribution• The sampling distribution of the sample mean is

approximately normal with– A mean equal to the population mean, μx-bar = μ

– A standard deviation equal to

• The sampling distribution of the sample proportion is approximately normal with– A mean equal to the population proportion , μp-hat = p

– A standard deviation equal to

• Central Limit Theorem: for large n (>30), the sampling distribution of x̄� is approximately normal for any population (with a finite σ)

n/)p(pp̂ 1

nx /

Summary and Homework

• Summary

• Homework– pg 607 – 609; 9.47, 49-53, 56, 58

Problem 1

Based on a simple random sample of size 100, a researcher calculated the standard deviation associated with a sample proportion to be 0.08. If she increases the sample size to 400, what will be the new standard deviation associated with the sample proportion?______________0.04

0.08 ~ n = 100 n = 400 is an increase of 4Inverse Square root relationship0.08/ √4 = 0.08/ 2 = 0.04

Problem 2

We know that p-hat is a/an _______________ statistic because the mean of the sampling distribution of p-hat is equal to the true population proportion p.

unbiased

Problem 3aAccording to the manufacturer’s specifications, the mean time required for a particular anesthetic drug to produce unconsciousness is 7.5 minutes with a standard deviation of 1.8 minutes. A random sample of 36 patients is to be selected and the average time for the drug to work will be computed for the sample. Find the probability that 

(a) the mean time for the sample will be less than 7.0 mins

x-bar ~ N(7.5, 1.8/√36) because 36 is sufficiently large to invoke CLT (Central Limit Thrm)

P( x-bar < 7.0) = 0.0478

normcdf(-E99, 7.0, 7.5, 1.8/√36)

Problem 3bAccording to the manufacturer’s specifications, the mean time required for a particular anesthetic drug to produce unconsciousness is 7.5 minutes with a standard deviation of 1.8 minutes. A random sample of 36 patients is to be selected and the average time for the drug to work will be computed for the sample. Find the probability that 

(b) a randomly selected patient requires less than 7.0

If X ~ N(7.5, 1.8) then it can be computed, otherwise not. No statement of normality.

If X ~ N(7.5, 1.8) then P( X < 7.0) = 0.3906

normcdf(-E99, 7.0, 7.5, 1.8)

Problem 3cAccording to the manufacturer’s specifications, the mean time required for a particular anesthetic drug to produce unconsciousness is 7.5 minutes with a standard deviation of 1.8 minutes. A random sample of 36 patients is to be selected and the average time for the drug to work will be computed for the sample. Find the probability that 

(c) If more random samples of size 36 were selected, the middle 95% of the sample means should fall between _______________ minutes and ______________ minutes.7.0757 7.9243

middle 95% is μ 2σ

7.5 2(1.8/√72) = 7.5 0.4243

Problem 4As we have discussed in class, a one-pound (16 ounce) box of sugar generally weighs more than 1 lb. According to some state laws, producers will be fined if the mean of 5 randomly selected boxes is less than 1 lb. If the packaging equipment delivers individual weights that are N (μ, 0.4) ounces, what setting should be used for μ so the probability of being fined is 0.01? Provide a sketch to support your answer.

invnorm(0.02) = -2.054 (or from the table)

x-bar – μ 16 - μZ = ------------- -2.054 = -------- σ/√n 0.4/√5

-2.054 (0.17889) = 16 – μ-0.367431 = 16 – μμ = 16.3674 ounces

Problem 5According to the __________________________________, when a simple random sample of size n is drawn from any population with mean µ and standard deviation σ, if n is sufficiently large the sampling distribution of the sample mean is approximately normal.

Central Limit Theorem

Problem 6Place the word “true” or “false” in the blank at the end of

each of the following sentences.

(a) If the underlying population is skewed, the distribution of x-bar will be normal for n = 2. _________________

(b) If the underlying population is skewed, the distribution of x-bar will be normal for n = 100. _________________

(c) If the underlying population is normal, the distribution of x-bar will be normal for n = 2. _________________

(d) If the underlying population is normal, the distribution of x-bar will be normal for n = 100. _________________

False

True

True

True

Problem 7aWe know that 60% of the students in a large state university are male.  

(a) Determine the mean and standard deviation of the sampling distribution of the sample proportion of males (p-hat) when samples of 400 students are randomly selected from this population. 

Mean = standard deviation =

 

0.60 √p(1-p)/n√.6(.4)/400 = 0.0245

Problem 7b&cWe know that 60% of the students in a large state university are male.  

(b) Verify that the formula you used for your standard deviation computation is valid in this situation. State the condition(s) that must be satisfied and convince me that all necessary conditions are met. 

 

 

 

(c) What is the probability that a simple random sample of 400 students will contain more than 65% males?

Assume total number of male students > 4000 N > 10n

400(0.6) = 240 ≥ 10 np ≥ 10400(0.4) = 160 ≥ 10 n(1-p) ≥ 10

p ~ N(0.6, 0.0245)

P(p-hat > 0.65) = 0.0206

normalcdf(0.65, E99, 0.6, 0.0245)

Problem 8The weight of eggs produced by a certain breed of hen is N (60, 4). What is the probability that the weight of a dozen (12) randomly selected eggs falls between 700 grams and 725 grams.

X is weight of an egg12X is the weight of a dozen eggs

E(X) = 60 E(12X) = 12E(X) = 720V(X) = 4² = 16 V(12X) = 12²V(X)= 144(16) = 2304s(X) = 48

P(700 < 12X < 725) = 0.2030

normcdf(700,725,720,48)