Point Estimator

• A point estimator cannot be expected to provide the exact value of the population parameter. WHY NOT?– A point estimate comes from a single sample, which may

or may not be representative of the population and is unlikely to give us the same value as the parameter.

• We can instead use interval estimates which are less precise, but have a better chance of being correct. These interval estimates are referred to as confidence intervals.

Confidence Interval

• The purpose of an interval estimate, or confidence interval, is to provide information about how close the point estimate, provided by the sample, is to the value of the population parameter.

• Confidence intervals can be computed by adding and subtracting a margin of error to a point estimate.

Interval Estimate:Point Estimate +/- Margin of Error

• A confidence interval has an upper and lower value.

Margin of Error

• A margin of error is the difference between the point estimate and the upper or lower end of the confidence interval.

• The interpretation is that if I were to find an interval estimate a large number of times, then (confidence level)% of the intervals would cover the true (but unknown) parameter being estimated.

Margin of Error

• In other words, if we took every possible sample and calculated a confidence interval of the mean for each of them, (90%)* of the confidence intervals would contain the true population mean.

• So, we can say we are (90%)* confident that the true population mean falls within the confidence interval.

*or whatever the confidence level is

Estimating the Mean

• The first interval estimate we’ll talk about is for the mean μ.

• Let’s see if we can figure out what the general interval estimate will look like for mean:

Population Standard deviation known Case

• When the population standard deviation is known, we can use the following formula:

• Let’s take a second and identify the parts of this formula.x=point estimatezα/2 =margin of error

α= 1-confidence level

nzx

2/

Confidence Interval Example!• Discount Sounds has 260 retail outlets throughout

the United States. The firm is evaluating a potential location for a new outlet, based in part, on the mean annual income of the individuals in the marketing area of the new location.

• A sample of size n = 36 was taken; the sample mean income is $41,100. The population standard deviation is estimated to be $4,500, and the confidence level to be used in the interval estimate is .95. Determine the confidence interval.

• Remember the formula: n

zx

2/

Solution

• x = 41100; σ=4500; n=36• α=1-.95=.05; zα/2=1.96• =41100+1.96

• Interval = (39630 < μ < 42570)

• This means that we are 95% sure that the actual population mean falls in between 39630 and 42570.

nzx

2/

σ Unknown Case:• If an estimate of the population standard

deviation σ cannot be developed prior to sampling, we use the sample standard deviation s to estimate σ. This case is much more common in practice. We’ll use this formula in these cases:

n

stx 2/

tα/2

• In the σ unknown case, the interval estimate for μ is based on the t distribution. The t distribution, like the z distribution, is another normal distribution.

• The t-distribution is used frequently in statistical

inference. It is similar to the normal z distribution in shape, but has fatter tails.

• The t-distribution is characterized by its degrees of freedom. When finding a confidence interval on the mean of a normal distribution, the degrees of freedom are n-1.

Curve: t Distribution vs. z Distribution

• We use the t distribution because the population standard deviation is unknown. t handles the uncertainty of estimating σ.

When To Use t

σ known? Sample Size t or z

Yes n<30 z

Yes n>30 z

No n<30 t

No n>30 z

Example 2

• Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 25 weekly reports showed a sample mean of 19.5 customer contacts per week. The sample standard deviation was 5.2. Provide 90% confidence interval for the population mean number of weekly customer contacts for the sales personnel.

Solution

• xbar=19.5, s=5.2• n=25, df=25-1=24• CL=.9, α=.1, α/2=.05

• Look at t table t=1.71

• Confidence interval= 19.5 +/- • =19.5+1.78 and 19.5–1.78.• Confidence interval=(17.72, 21.28)

n

stx 2/

Estimating the Population Proportion p (Section 8.4)

• When estimating the population proportion p, we will use the following formula:

pbar + zα/2

• Remember that we use proportions when we have binomial data that we’d like to make continuous. ()

Example

• A sample of 30 certified public accountants in one geographical region was asked if the Financial Accounting Standards Board (FASB) is too slow in issuing new accounting rules. 20 of those surveyed thought that FASB was too slow. Find the 90% confidence interval estimate for the proportion of CPAs in this region who feel that FASB is too slow in issuing new accounts.

pbar + zα/2

Solution

• p = (20/30)=.667• n=30• α=1-.90=.1, α /2=.05• Zα/2=1.645

• Confidence interval=.667+(1.645)(.086)=.667+.142(.525<p<.808)

We’re 90% sure that the true population proportion of CPAs that think the FASB is too slow in issuing new rules is between .525 and .808.

p + zα/2

Trading off precision and confidence:

• What happens when we move from a 95% confidence interval to a 90% confidence interval? Is the interval wider or narrower?

Trading off precision and confidence:

• To be able to be more confident, we need to have a wider range.

• If we want a narrower range, we will not be able to be as confident.

• So, if we move from a 95% interval to a 90% confidence level, the interval will get narrower. To be more confident, we need a wider interval to cover more possibilities.

Trading off precision and confidence:

• The more precise we want to be, the less confident we can be about our answer.

• The more confident we want to be, the less precise our estimate (wide interval).

Higher Confidence Wider Interval

Lower Confidence Narrower Interval

Section 8.3: Determining the Sample Size

• Sometimes, a desired margin of error is selected prior to sampling.

• When this is the case, we will need to determine the sample size necessary to satisfy the selected margin of error requirements.

• Looking at the equation, how might we accomplish this (E is the desired margin of error)?

nzE

2/

Finding Sample Size Equation

• If we rearrange the formula, we can come up with the following:

n==

With E=desired (selected) margin of error--remember this value is chosen beforehand—

Sample Size Requirementsσ Unknown • When we do not know the population standard deviation, we

can use the same formula with one of the following procedures:

1. Use the estimate of the population standard deviation computed form data of previous studies as the planning value for σ.

2. Use a pilot study to select a preliminary sample. The sample standard deviation from the preliminary sample can be used as the planning value for σ.3. Use judgment or a best guess for the value of σ.

Example: σ Unknown

• Assume that we are interested in estimating the mean height of the head of beer we are producing. Our desired margin of error is .02 inches and our confidence level is .99. A preliminary sample resulted in a standard deviation of .04. What sample size do we need to take?

• What other methods could we have used to determine a value for σ?

n=

Solution

• E=.02• Planning value for σ=.04• CL=.99 α=1-CL=1-.99=.01 α/2=.005• z α/2=2.576• n= OR n=• n=26.5• ROUND UP• n=27

n=

Sample Size Requirements for Estimating a Proportion

• If a desired margin of error is selected beforehand and we are working with proportions, the formula to determine sample size is

• Again to start this we need an initial estimate of . We can use a preliminary sample, can use our best guess, or if nothing else is possible, use p=0.5, which will give the highest value of n for any value of p.

2

2/2

E

)p(1p )(zn

Example: Proportion

• Suppose in the FASB example above we desired the width of the interval to be .1. What sample size would we need to take? Remember that we are working with a 90% confidence level.

• How can we solve this?– We need to do a pilot study, use our best guess, or use .5 for

pbar.

• Let’s say we conduct a preliminary (pilot) study of 30 people and 20 thought the FASB was too slow.

2

2/2

E

)p(1p )(zn

Solution

• E=half of width=.05• p = = = .667

• CL=.90 α=.1 α/2=.05• z α/2=1.645

• n==240.536

• n=241

2

2/2

E

)p(1p )(zn