Lesson 9 - 3 Introduction to the Practice of Statistics.

Lesson 9 - 3

Introduction to the Practice of Statistics

Objectives

• Define statistics and statistical thinking

• Understand the process of statistics

• Distinguish between qualitative and quantitative variables

• Distinguish between discrete and continuous variables

Vocabulary• Central Limit Theorem – the larger the sample size,

the closer the sampling distribution for the sample mean from any underlying distribution approaches a Normal distribution

• Standard error of the mean – standard deviation of the sampling distribution of x-bar

Sample Mean, x̄�

The behavior of x̄� in repeated sampling is much like that of the sample proportion, p-hat.•Sample mean x̄� is an unbiased estimator of the population mean μ•Spread is less than that of X. Standard deviation of x̄� is smaller than that of X by a factor of 1/√n

Sample Spread of x̄�

If the random variable X has a normal distribution with a mean of 20 and a standard deviation of 12

– If we choose samples of size n = 4, then the sample mean will have a normal distribution with a mean of 20 and a standard deviation of 6

– If we choose samples of size n = 9, then the sample mean will have a normal distribution with a mean of 20 and a standard deviation of 4

Example 1

The height of all 3-year-old females is approximately normally distributed with μ = 38.72 inches and σ = 3.17 inches. Compute the probability that a simple random sample of size n = 10 results in a sample mean greater than 40 inches.

μ = 38.72 σ = 3.17 n = 10 σx = 3.17 / 10 = 1.00244

x - μZ = ------------- σx

40 – 38.72= ----------------- 1.00244

1.28= ----------------- 1.00244

= 1.277

normalcdf(1.277,E99) = 0.1008

normalcdf(40,E99,38.72,1.002) = 0.1007 a

P(x-bar > 40)

Example 2We’ve been told that the average weight of giraffes is 2400 pounds with a standard deviation of 300 pounds. We’ve measured 50 giraffes and found that the sample mean was 2600 pounds. Is our data consistent with what we’ve been told?

μ = 2400 σ = 300 n = 50 σx = 300 / 50 = 42.4264

x - μZ = ------------- σx

2600 – 2400= ----------------- 42.4264

200= ----------------- 42.4264

= 4.714

normalcdf(4.714,E99) = 0.000015

normalcdf(2600,E99,2400,42.4264) = 0.0000001 a

P(x-bar > 2600)

Example 3Young women’s height is distributed as a N(64.5, 2.5), What is the probability that a randomly selected young woman is taller than 66.5 inches?

μ = 64.5 σ = 2.5 n = 1 σx = 2.5 / 1 = 2.5 !!

x - μZ = ------------- σx

66.5 – 64.5= ----------------- 2.5

2= --------- 2.5

= 0.80

normalcdf(0.80,E99) = 1 – 0.7881 = 0.2119

normalcdf(66.5,E99,64.5,2.5) = 0.2119 a

P(x > 66.5)

Example 4Young women’s height is distributed as a N(64.5, 2.5), What is the probability that an SRS of 10 young women is greater than 66.5 inches?

μ = 64.5 σ = 2.5 n = 1 σx = 2.5 / 10 = 0.79

x - μZ = ------------- σx

66.5 – 64.5= ----------------- 0.79

2= --------- 0.79

= 2.53

normalcdf(2.53,E99) = 1 – 0.9943 = 0.0057

normalcdf(66.5,E99,64.5,2.5/√10) = 0.0057 a

P(x > 66.5)

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

X or x-barDistribution

Population Distribution

Random Samples Drawn from Population

x x x x x x x x x x x x x x x x

Central Limit Theorem in Action

n =1

n = 2

n = 10

n = 25

Example 5The time a technician requires to perform preventive maintenance on an air conditioning unit is governed by the exponential distribution (similar to curve a from “in Action” slide). The mean time is μ = 1 hour and σ = 1 hour. Your company has a contract to maintain 70 of these units in an apartment building. In budgeting your technician’s time should you allow an average of 1.1 hours or 1.25 hours for each unit?

μ = 1 σ = 1 n = 70 σx = 1 / 70 = 0.120

x - μZ = ------------- σx

1.1 – 1= ------------ 0.12

0.1= --------- 0.12

= 0.83

normalcdf(0.83,E99) = 1 – 0.7967 = 0.2033 a

P(x > 1.1) vs P(x > 1.25)

Example 5 contThe time a technician requires to perform preventive maintenance on an air conditioning unit is governed by the exponential distribution (similar to curve a from “in Action” slide). The mean time is μ = 1 hour and σ = 1 hour. Your company has a contract to maintain 70 of these units in an apartment building. In budgeting your technician’s time should you allow an average of 1.1 hours or 1.25 hours for each unit?

μ = 1 σ = 1 n = 70 σx = 1 / 70 = 0.120

x - μZ = ------------- σx

1.25 – 1= ------------ 0.12

0.25= --------- 0.12

= 2.083

normalcdf(2.083,E99) = 1 – 0.9818 = 0.0182 a

P(x > 1.25)

Shape, Center and Spread of Population

Distribution of the Sample Means

Shape Center Spread

Normal with mean, μ and standard deviation, σ

Regardless of sample size, n, distribution of

x-bar is normalμx-bar = μ

σσx-bar = ------- n

Population is not normal with mean, μ and standard deviation, σ

As sample size, n, increases, the distribution

of x-bar becomes approximately normal

μx-bar = μ σσx-bar = ------- n

Summary of Distribution of x

Summary and Homework• Summary

– Take an SRS and use the sample proportion x̄� to estimate the unknown parameter μ

– x̄� is an unbiased estimator of μ– Increase in sample size decreases the standard

deviation of x̄� (by a factor of 1/√n)– If the population is normal, then so is x̄� – Central Limit Theorem: for large n, the sampling

distribution of x̄� is approximately normal for any population (with a finite σ)

• Homework– Day 1: pg 595-6; 9.31-4– Day 2: pg 601-4; 9.35, 36, 38, 42-44