Estimating the Value of a Parameter Using Confidence

Intervals

Chapter 9

Point Estimate

• A point estimate is a single value (or point) used to approximate a population parameter.

• The sample (mean) or (proportion) is the best point estimate of the population (mean) or (proportion)

Example: (mean)Pennies minted after 1982 are made from 97.5% zinc and 2.5% copper.

The following data represent the weights (in grams) of 17 randomly selected pennies minted after 1982.

2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49

2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45

Treat the data as a simple random sample. Estimate the population mean weight of pennies minted after 1982.

Answer:

x 2.46 2.47 2.45

172.464

The point estimate of is 2.464 grams.

In a sample of 1,000 people, we find that 426 are Democrats: so that's 42.6% on this Sample.

If we sample again 1,000 people, we will most likely get a different but close percentage (→ Random chance error from sampling.)

Conclusion: A Point Estimate on the Population percentage is: 42.6%

Example: (proportion)

How do we determine how good our estimate is?

Answer: Establish a Confidence Interval around that Estimate.

First: We must compute a Standard Error on our estimate.

|___________| large population

Here, p = 42.6% so q = (100-42.6) = 57.4%

= sqrt(0.426 x 0.574)

Standard Error of Sample Proportion is:

p = /sqrt(n) = sqrt(0.426 x 0.574)/sqrt(1000) = 0.016

= 1.6%We have a point estimate: 42.6% with an error of 1.6%.

Next: Establish a Confidence Interval about our estimate.

p(1) q(0) 1000

With a sample size large enough (as is the case here, n=1000), we use the Normal Approximation.

± 1 SE → 68.27% Confidence Interval

± 2 SE's → 95.45% Confidence Interval

± 3 SE's → 99.74% Confidence Interval

± 1.96 SE → 95% Confidence Interval

± 2.57 SE → 99% Confidence Interval

So, for our example: The 95% Confidence Interval is:

42.6% ± 1.96 x 1.6% = (39.46%, 45.73%)

Conclusion:We are 95% confident that the percentage of Democrats in the population will range between 39.46% and 45.73%.Or: There is a 95% probability that the actual population proportion will be within the interval (39.46%, 45.73%)

Definitions

In the previous example:

95% is the Confidence Level.

= 1-0.95 = 0.05 = 5% is the Degree of Confidence.

The Z scores corresponding to the values: 39.46%, 45.73% are called Critical Values

• Confidence Interval on a MeanConfidence Interval on a Mean:Example:The National Institute for Standards and Technology has a small weight called NB10 that is supposed to weigh 10 grams. The average of 100 weighings of NB10 on an unbiased scale was 406 micrograms below 10 grams, and the SD of the weightings was 6 micrograms. (link)

Find the 95% Confidence Interval on this estimate.

Answer:We must compute a Standard Error on our mean estimate of

x = 406g.

→ Find x then the 95% C.I. on the estimate.

Here:

= 6g and n = 100

So x = 6/sqrt(100) = 0.6

The 95% C.I is: (406 ± 1.96 x 0.6) = (404.82g, 407.17g)

Construct a 99% confidence interval about the population mean weight (in grams) of pennies minted after 1982. Assume =0.02 grams.

2.46 2.47 2.49 2.48 2.50 2.44 2.46 2.45 2.49

2.47 2.45 2.46 2.45 2.46 2.47 2.44 2.45

Another example:

• • Lower bound:

= 2.464-2.575

= 2.464-0.012 = 2.452

• Upper bound:

= 2.464+2.575

= 2.464+0.012 = 2.476We are 99% confident that the mean weight of

pennies minted after 1982 is between 2.452 and 2.476 grams.

z 2 2.575

x z 2 n

0.0217

x z 2 n

0.0217

E z 2 n

a) Find the margin of error E that corresponds to a 95% confidence level.

Example: 829 adult Minnesotans were surveyed, and 51% of them are opposed to the use of the photo-cop for issuing traffic tickets. Use these survey results to:

03403.0829

)49.0)(51.0(96.1 E

51.0ˆ p 49.051.01ˆ q 96.12/ Z 829n

Answer:

Therefore:

b) Find the 95% confidence interval for the population proportion p.

We substitute the values from Part a) to obtain:

0.51 – 0.03403 < p < 0.51 + 0.03403,

0.476 < p < 0.544

c) Based on the results, can we safely conclude that the majority of adult Minnesotans oppose use of the photo-cop?

Based on the survey results, we are 95% confident that the limits of 47.6% and 54.4% contain the true percentage of adult Minnesotans opposed to the photo-cop. The percentage of opposed adult Minnesotans is likely to be any value between 47.6% and 54.4%. However, a majority requires a percentage greater than 50%, so we cannot safely conclude that the majority is opposed (because the entire confidence interval is not greater than 50%).

Sample Size (given a Margin of Error and Confidence Level)

Suppose we want to collect sample data with the objective of estimating some population.

The question is how many sample items must be obtained?

(ie: Finding n, given a margin of error and a confidence level)

Determining the Sample Size n: For a mean

Go back to the formula for margin of error:

Now solve for n:

n z

2

E

2

E z 2 n

Determining the Sample Size n: For a proportion

2

2)ˆ1(ˆ

E

zppn

n 0.25z 2

E

2

E z 2 n

)ˆ1(ˆ pp

Back to margin of error formula:

Solving for n gives:

Solving for n gives:

A sociologist wanted to determine the percentage of residents of America that only speak English at home. What size sample should be obtained if she wishes her estimate to be within 3 percentage points with 90% confidence assuming she uses the 2000 estimate obtained from the Census 2000 Supplementary Survey of 82.4%?

Example : (Determining Sample Size)

• E=0.03

•

•

•

We round this value up to 437.

The sociologist must survey 437 randomly

selected American residents.

z 2 z0.05 1.645

ˆ p 0.824

n 0.824(1 0.824)1.645

0.03

2

436.04

Solution