Chapter Nine McGraw-Hill/Irwin © 2006 The McGraw-Hill Companies, Inc., All Rights Reserved. Estimation and Confidence Intervals.

Chapter

Nine

McGraw-Hill/Irwin

© 2006 The McGraw-Hill Companies, Inc., All Rights Reserved.

Estimation and Confidence Estimation and Confidence IntervalsIntervals

We cannot be sure that Point estimate is the mean. But we can calculate an interval around this estimate and assert with a certain confidence that the true population mean will lie inside it.

A Confidence Interval is a range of values within which the population parameter (eg. μ ) is expected to occur at a specified level of confidence generally expressed as a percent.

A Point estimate is a single value (statistic) used to estimate a population value (parameter).

Eg. μx is a point estimate of μ

Level of confidence

Confidence Interval

Let us recall from Chapter 8 that …

σ/√n

xμ

σ / √n

3.(σ / √n)3.(σ / √n)

•The best estimator of μ is X

•The SD of X distribution is σ/√n

Any X you calculate based on a sample will have to be within 3.(σ/√n) of μ (based on the Empirical rule)

We also know from Chapter 8, Z = (X – μ) / (σ/√n)

From Chapter 8, Sampling Error = X – μ

X + Z . (σ / √n)- Z . (σ / √n)

How much width around X ?How much width around X ?

If σ is not known and n >30, the SD of the sample s is used.

CI for the population mean μ is:n

szX

Combining the two,Sampling Error, X – μ = Z . (σ / √n)

So, if we add & subtract the above Sampling Error factor to X, we can estimate the range (called, CI ) within which μ must lie.

Problem (page 250)

The AM Association wants info on the mean income of managers working in the retail industry. A random sample of 256 managers had a mean of $45420 with a standard deviation of $2050. What is the interval in which the population mean would lie with a 95% confidence level.

n

sX 96.1

Since Z for 95% is 1.96*, the formula for CI can rewritten as:

= 45420 ± 1.96 (2050 / √256) = 45420 ± 251

So, the CI is $45169 - $45671

*See next slide

Because, area under the curve between Z = +1.96 and – 1.96, is 95% (see Appendix D)

Why use Z=1.96 for CI at 95% ?

Question: What would be the value of Z for CI at 99%? Z = 2.58 !

Notice that the CI widens when confidence level is increased from 95% to 99%

What does the CI at a 95% level of confidence mean ?It means that 95% of the sample intervals will contain the population mean μ

Try experimenting With Visual Statistics software

How do we increase our confidence?How do we increase our confidence?

1. Widen the interval (Z )

Let us say, based on past exams, I claim with 75% confidence that in the coming test, the class average (μ ) will be between 70-80 points.

If I want to raise my confidence to 95%, I can do two things:

1) widen the CI from 70-80 to 60-902) increase n to reduce dispersion of

the distribution

μX

2. Increase the sample size (n )

Larger n squishes the area (and therefore, the probabilities) into a thinner peak; so, the level of confidence will be a high percentage even with a smaller interval.

SD = σ/√n

n

stX

Use t-distribution when:•n < 30 (eg. You are crash-testing expensive autos!)

•only s is known (ie. σ is unknown)•underlying population is approximately normal

t-Distribution

In general, if you see n<30 in the exam problem, you must think t-distribution!

The Story of t-Distribution

Once upon a time, there was a statistician called Gosset …

When you don’t know σ, you have to use s instead. But the problem is, when n is small (n<30), s has a wide dispersion and is not a good estimator of σ

Gosset created a new distribution called ‘t’ that spreads the area under the curve wider when s is small but automatically converges to normal when n increases beyond 30!

Compare with Chart 9-2 in text (page 255)

Note:n=5

Z=1.96

t=2.776

Visual Statistics Demo

Using Continuous Distribution module

Look at it this way: Since n is small, we are not sure s would be a good estimate of σ; so, we play it safe by increasing CI for the same confidence level.

Observe how the ± 1.96 (95%) in Z in stretched outward to ± 2.776 in t to keep

the area under the curve same at 0.95, when sample size is only 5.

Practice! (problem on page 256)

A tire manufacturer wishes to investigate the tread life of its tires. A sample of 10 tires driven 50000 miles revealed a sample mean of 0.32 inch of tread remaining with a standard deviation of 0.09 inch. Construct a 95% CI for the population mean.

n

stX

= 0.32 ± 2.262 ( 0.09 / √10) = 0.32 ± 0.064

= 0.256 to 0.384

What is the formula to be used?

What is the value of t for df=9* and CI=95% (page 498) = 2.262

What is the 95% CI?

*df = (n -1)

Degrees of Freedom

You are in a room with 10 chairs and you are sitting in one of them. The other chairs are empty. How many other chairs can you move to? Ans: 9

So in general, df = n-1

CI for a population proportion

•So far we studied variables that use a ratio scale. There we can calculate the means. Eg. Manager’s $ income & Tire wear

•What if we have to work with a nominal scale variable where values are categorized into one of two groups?

Eg. CSUN career center reports that 75% of its graduates get a job related to their major.

You cannot calculate the mean of Yes & No’s. But, you can calculate a proportion of students who said Yes.

Getting the job in your major can be termed as ‘success’; if the student got a job in a different field, then it is a ‘failure’.

So, Binomial distribution formulas we studied in Chapter 6 can be used to describe sampling distribution of a proportion RV!

Mean successes in a Binomial distribution is nπ [Ch 6; Page 167]

SD for Binomial is √nπ(1-π) [Page 167]

Binomial Distribution (See Page 170)

No. of heads (successes) in 10 trials of throwing a coin

Mean (expected number of heads) = 5 [notice the peak at X=5 ]

If X-axis is redrawn as X/10 (ie proportion of successes), the curve will squish by 10 times; and so will its SD.

X/n 0 .1 .2 .3 ... ... 1.0

Estimating population proportion

Here, we focus on the proportion of successes; so, we divide the number of successes, x, by the total number of trials, n.

Xn π

√p(1-p)/n

Note: p=x/n

pπ

π has to be within 3σ’s(Empirical rule)

σp = √p(1-p)/n

CI for the population proportion π

CI = p ± Z . √p(1-p)/n

(Note the pattern: CI = Sample Mean ± (Confidence level) * (SD of Sample Distrbn)

0497.35. 500

)65)(.35(.33.235.

A sample of 500 executives who own their own home revealed 175 planned to sell their homes and retire to Arizona. Develop a 98% confidence interval for the proportion of executives that plan to sell and move to Arizona.

A word of caution

Binomial approximation works well when the following two conditions are satisfied:

n.p ≥ 5 & n.(1-p) ≥ 5.

Here is why: (see page 170)

Calculating the sample size

3 factors affect the sample size:

•The level of confidence desired

•The margin of error the researcher will tolerate.

•The variability in the population being studied.

2

E

szn

where

n is the size of the sample

E is the allowable error

z is the z- value corresponding to the selected level of confidence

(for 99%, from Appendix, Z=2.58)

s the sample deviation of the pilot survey

The formula for estimated sample size is:

Z = X – μ / ( s/√n )

X - μ = Z. ( s/√n )

E = Z. ( s/√n )

E2 = Z2. s2 / n

n = Z2.s2 /E2

n = Z.s E

2

P(r)oof !

[Ch 8; Page 235]

1075

)20)(58.2(2

n

A utility company would like to estimate the mean monthly electricity charge for a single family house within $5 using a 99% level of confidence. The standard deviation is estimated to be $20.00. How large a sample is required?

n p pZ

E

( )1

2The formula for determining the

sample size in the case of a proportion is

p is the estimated proportion, based on past experience or a pilot survey

z is the z value associated with the degree of confidence selected

E is the maximum allowable error the researcher will tolerate

where

Study the example worked out in Page 267

[You can derive this by rearranging Formula 9-6 in page 262]

Finite population Correction

If the population is finite (ie, a known number), multiply the SD by the following term.

N, population sizen, sample size

nN

N

1

When n is small, the value of the factor is close to 1.

As n gets larger, the value of the correction factor, gets smaller; the logic is that if the sample is a substantial percentage of the population, the estimate of SD is more precise (Table 9-1,p.264)

Rule of thumb: Ignore correction factor if n/N < 0.05