Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 1 Understandable Statistics S eventh Edition By Brase and Brase Prepared by: Lynn Smith Gloucester County College Chapter Eight Part 1 (Sections 8.1 to 8.3) Estimation

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 2 Point Estimate an estimate of a population parameter given by a single number

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 3 Examples of Point Estimates

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 4 Examples of Point Estimates is used as a point estimate for .

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 5 Examples of Point Estimates is used as a point estimate for . s is used as a point estimate for .

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 6 Error of Estimate the magnitude of the difference between the point estimate and the true parameter value

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 7 The error of estimate using as a point estimate for  is

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 8 Confidence Level A confidence level, c, is a measure of the degree of assurance we have in our results. The value of c may be any number between zero and one. Typical values for c include 0.90, 0.95, and 0.99.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 9 Critical Value for a Confidence Level, c the value z c such that the area under the standard normal curve falling between – z c and z c is equal to c.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 10 Critical Value for a Confidence Level, c P(– z c < z < z c ) = c – z c 0 z c This area = c.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 11 Find z 0.90 such that 90% of the area under the normal curve lies between z and z P(-z 0.90 < z < z 0.90 ) = 0.90 – z.90 0 z.90.90

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 12 Find z 0.90 such that 90% of the area under the normal curve lies between z and z P(0< z < z 0.90 ) = 0.90/2 = – z.90 0 z

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 13 Find z 0.90 such that 90% of the area under the normal curve lies between z and z P( z < z 0.90 ) = =.9500 – z.90 0 z

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 14 Find z 0.90 such that 90% of the area under the normal curve lies between z and z According to Table 5a in Appendix II, lies exactly halfway between two area values in the table (.9495 and.9505). Averaging the z values associated with these areas gives z 0.90 =

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 15 Common Levels of Confidence and Their Corresponding Critical Values

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 16 Confidence Interval for the Mean of Large Samples (n  30)

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 17 Confidence Interval for the Mean of Large Samples (n  30)

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 18 Confidence Interval for the Mean of Large Samples (n  30)

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 19 Confidence Interval for the Mean of Large Samples (n  30)

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 20 Confidence Interval for the Mean of Large Samples (n  30)

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 21 Confidence Interval for the Mean of Large Samples (n  30)

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 22 Create a 95% confidence interval for the mean driving time between Philadelphia and Boston. Assume that the mean driving time of 64 trips was 6.4 hours with a standard deviation of 0.9 hours.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 23 = 6.4 hours s = 0.9 hours c = 95%, so z c = 1.96 n = 64

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 24 = 6.4 hours s = 0.9 hours Approximate  as s = 0.9 hours. 95% Confidence interval will be from

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 25 = 6.4 hours s = 0.9 hours c = 95%, so z c = 1.96 n = 64

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved % Confidence Interval: 6.4 –.2205 <  < <  < We are 95% sure that the true time is between 6.18 and 6.62 hours.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 27 What if it is impossible or impractical to use a large sample? Apply the Student’s t distribution.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 28 Student’s t Variable

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 29 The shape of the t distribution depends only only the sample size, n, if the basic variable x has a normal distribution. When using the t distribution, we will assume that the x distribution is normal.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 30 Table 6 in Appendix II gives values of the variable t corresponding to the number of degrees of freedom (d.f.)

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 31 Degrees of Freedom d.f. = n – 1 where n = sample size

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 32 The t Distribution has a Shape Similar to that of the the Normal Distribution A Normal distribution A “t” distribution

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 33 Find the critical value t c for a 95% confidence interval if n = 8. d.f.=7

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 34 Confidence Interval for the Mean of Small Samples (n < 30) from Normal Populations c = confidence level (0 < c < 1) t c = critical value for confidence level c, and degrees of freedom = n - 1

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 35 The mean weight of eight fish caught in a local lake is 15.7 ounces with a standard deviation of 2.3 ounces. Construct a 90% confidence interval for the mean weight of the population of fish in the lake.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 36 Mean = 15.7 ounces Standard deviation = 2.3 ounces. n = 8, so d.f. = n – 1 = 7 For c = 0.90, Table 6 in Appendix II gives t 0.90 =

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 37 Mean = 15.7 ounces Standard deviation = 2.3 ounces. E = 1.54 The 90% confidence interval is: <  < <  < 17.24

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 38 The 90% Confidence Interval: <  < We are 90% sure that the true mean weight of the fish in the lake is between and ounces.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 39 Review of the Binomial Distribution Completely determined by the number of trials (n) and the probability of success (p) in a single trial. q = 1 – p If np and nq are both > 5, the binomial distribution can be approximated by the normal distribution.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 40 A Point Estimate for p, the Population Proportion of Successes

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 41 Point Estimate for q (Population Proportion of Failures)

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 42 For a sample of 500 airplane departures, 370 departed on time. Use this information to estimate the probability that an airplane from the entire population departs on time. We estimate that there is a 74% chance that any given flight will depart on time.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 43 Error of Estimate for “p hat” as a Point Estimate for p

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 44 A c Confidence Interval for p for Large Samples (np > 5 and nq > 5) z c = critical value for confidence level c taken from a normal distribution

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 45 For a sample of 500 airplane departures, 370 departed on time. Find a 99% confidence interval for the proportion of airplanes that depart on time. Is the use of the normal distribution justified?

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 46 For a sample of 500 airplane departures, 370 departed on time. Find a 99% confidence interval for the proportion of airplanes that depart on time. Can we use the normal distribution?

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 47 For a sample of 500 airplane departures, 370 departed on time. Find a 99% confidence interval for the proportion of airplanes that depart on time. so the use of the normal distribution is justified.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 48 Out of 500 departures, 370 departed on time. Find a 99% confidence interval.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved % confidence interval for the proportion of airplanes that depart on time: E = Confidence interval is:

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved % confidence interval for the proportion of airplanes that depart on time Confidence interval is < p < We are 99% confident that between 69% and 79% of the planes depart on time.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 51 The point estimate and the confidence interval do not depend on the size of the population. The sample size, however, does affect the accuracy of the statistical estimate.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 52 Margin of Error The margin of error is the maximal error of estimate E for a confidence interval. Usually, a 95% confidence interval is assumed.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 53 Interpretation of Poll Results The proportion responding in a certain way is

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 54 A 95% confidence interval for population proportion p is:

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 55 Interpret the following poll results: “ A recent survey of 400 households indicated that 84% of the households surveyed preferred a new breakfast cereal to their previous brand. Chances are 19 out of 20 that if all households had been surveyed, the results would differ by no more than 3.5 percentage points in either direction.”

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 56 “Chances are 19 out of 20 …” 19/20 = 0.95 A 95% confidence interval is being used.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 57 “... 84% of the households surveyed preferred …” 84% represents the percentage of households who preferred the new cereal.

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 58 “... the results would differ by no more than 3.5 percentage points in either direction.” 3.5% represents the margin of error, E. The confidence interval is: 84% - 3.5% < p < 84% + 3.5% 80.5% < p < 87.5%

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved. 59 The poll indicates ( with 95% confidence): between 80.5% and 87.5% of the population prefer the new cereal.