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IS 310 Business Statistics CSU Long Beach

IS 310 Business Statistics CSU Long Beach. Interval Estimation. In chapter 7, we studied how to estimate a population parameter with a sample statistic. We used a point estimate as a single value.

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IS 310 Business Statistics CSU Long Beach

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  1. IS 310 Business Statistics CSU Long Beach

  2. Interval Estimation In chapter 7, we studied how to estimate a population parameter with a sample statistic. We used a point estimate as a single value. To increase the level of confidence in estimation, we use a range of values (rather than a single value) as the estimate of a population parameter. Let’s take a real-life example. Suppose, someone asks you how long it takes to go from CSULB to LAX. What would be a more reliable estimate – 30 minutes or between 25 and 40 minutes? If you use 30 minutes as estimate, you are using a point estimate. On the other hand, if you use between 25 and 40 minutes as estimate, you are using an interval estimation.

  3. Interval Estimation Interval estimation uses a range of values. The width of the range indicates the level of confidence. The narrower the range, lower the confidence. Wider the range, higher the confidence. Once a confidence level is specified, the interval estimate can be calculated using the formula, 8.1, in the book. If one wants 95% confidence, the interval estimate is known as 95% Confidence Interval. For a 90% confidence, the interval is known as 90% Confidence Interval.

  4. Interval Estimation • In this chapter, we will cover the following: • Interval Estimate for a Population Mean (known σ) • Interval Estimate for a Population Mean (unknown σ) • Determining Size of a Sample • Interval Estimate for a Population Proportion

  5. Margin of Error and the Interval Estimate The general form of an interval estimate of a population mean is

  6. Margin of Error The Margin of Error implies both Confidence level and Amount of error

  7. where: is the sample mean 1 - is the confidence coefficient z/2 is the z value providing an area of /2 in the upper tail of the standard normal probability distribution s is the population standard deviation n is the sample size Interval Estimate of a Population Mean:s Known • Interval Estimate of m

  8. Interval Estimate for Population Mean (Known σ) • Example: • Lloyd’s Department Store wants to determine a 95% confidence interval on the average amount spent by all of its customers (population mean). • Lloyd took a sample of 100 customers and found that the average amount spent is $82 (this is sample mean or x‾ ). Lloyd assumes that the population standard deviation ( σ ) is $20.

  9. Interval Estimation for Population Mean(Known σ ) • Using Formula 8.1, the 95% confidence interval is: • _ • x ± z (σ/√ n)  = 0.05 • 0.025 • 82 ± 1.96 (20/√100) • 82 ± 3.92 • 78.08 to 85.92 • We are 95% sure that the average amount spent by all Lloyd customers is between $78.08 and $85.92.

  10. Sample Problem Problem # 5 (10-Page 306; 11-Page 315) _ Given: x = 24.80 n = 49 σ = 5 confidence level = 95%  = 0.05 or 5% • Margin of Error = z x (σ / √n) /2 = 1.96 (5/√49) = 1.4 • 95% Confidence Interval is: _ x ± Margin of Error 24.8 ± 1.4 23.4, 26.2 It means that we are 95% confident that the average amount spent by all customers for dinner at the Atlanta restaurant is between $23.40 and $26.20

  11. Interval Estimation of a Population Mean:s Unknown • If an estimate of the population standard deviation s cannot be developed prior to sampling, we use the sample standard deviation s to estimate s . • This is the s unknown case. • In this case, the interval estimate for m is based on the t distribution. • (We’ll assume for now that the population is normally distributed.)

  12. t Distribution The t distribution is a family of similar probability distributions. A specific t distribution depends on a parameter known as the degrees of freedom. Degrees of freedom refer to the sample size minus 1.

  13. t Distribution A t distribution with more degrees of freedom has less dispersion. As the number of degrees of freedom increases, the difference between the t distribution and the standard normal probability distribution becomes smaller and smaller.

  14. t Distribution t distribution (20 degrees of freedom) Standard normal distribution t distribution (10 degrees of freedom) z, t 0

  15. t Distribution Standard normal z values

  16. Interval Estimation of a Population Mean:s Unknown • Interval Estimate where: 1 - = the confidence coefficient t/2 = the t value providing an area of /2 in the upper tail of a t distribution with n - 1 degrees of freedom s = the sample standard deviation

  17. Interval Estimate for Population Mean (Unknown σ) • Example: • We want to estimate the mean credit card debt of all customers in the U.S. with a 95% confidence. • A sample of 70 households provided the credit card balances shown in Table 8.3. The sample standard deviation (s) is $4007. • Using formula 8.2, • _ • x ± t (s/√n)  = 0.05 • 0.025 • 9312 ± 1.995 (4007/√70)

  18. Interval Estimate (Continued) • 9312 ± 955 • 8357 to 10,267 • We are 95% confident that the average credit card balance of all customers in the U.S. are between $8,357 and $10,267.

  19. Summary of Interval Estimation Procedures for a Population Mean Can the population standard deviation s be assumed known ? Yes No Use the sample standard deviation s to estimate s s Known Case Use Use s Unknown Case

  20. Sample Problem Problem # 16 (10-Page 315; 11-Page 324) _ Given: x = 49 n = 100 s = 8.5 Confidence Level = 90% • At 90% confidence, the Margin of Error is: t . (s/√n) 1.660 x (8.5/10) = 1.411 /2 • The 90% Confidence Interval is: _ x ± 1.411 49 ± 1.411 47.59, 50.41 It means that we are 90% confident that the average hours of flying for Continental pilots is between 47.59 and 50.41

  21. Sample Size for an Interval Estimateof a Population Mean Let E = the desired margin of error. E is the amount added to and subtracted from the point estimate to obtain an interval estimate.

  22. Sample Size for an Interval Estimateof a Population Mean • Margin of Error • Necessary Sample Size

  23. Determination of Sample Size • Example Problem (10-Page 317; 11-Page 326): • We want to know the average daily rental rate of all midsize automobiles in the U.S. We want our estimate with a margin of error of $2 and a 95% level of confidence. • What should be the size of our sample? • Given for this problem: E =2,  = 0.05, σ = 9.65 • Using Formula 8.3, • 2 2 2 • n = (z ) (σ ) / E • 0.025 • = 89.43 • The sample size needs to be at least 90.

  24. Sample Problem Problem #26 (10-Page 318; 11-Page 327) Given: µ = 2.41 σ = 0.15 Margin of Error = 0.07 Confidence Level = 95% Margin of Error = z . (σ /√n) /2 2 2 2 0.07 = 1.96 (0.15/√n) n = (1.96) (0.15) / 0.07) = 17.63 or 18 A sample size of 18 is recommended for The Cincinnati Enquirer for its study

  25. Interval Estimationof a Population Proportion The general form of an interval estimate of a population proportion is

  26. The sampling distribution of plays a key role in computing the margin of error for this interval estimate. The sampling distribution of can be approximated by a normal distribution whenever np> 5 and n(1 – p) > 5. Interval Estimationof a Population Proportion

  27. Normal Approximation of Sampling Distribution of Sampling distribution of 1 -  of all values Interval Estimationof a Population Proportion /2 /2 p

  28. where: 1 - is the confidence coefficient z/2 is the z value providing an area of /2 in the upper tail of the standard normal probability distribution is the sample proportion Interval Estimationof a Population Proportion • Interval Estimate

  29. Interval Estimate for a Population Proportion • Example (10-Page 320; 11-Page 329): • We want to estimate the proportion of all women golfers in the U.S. who are satisfied with the availability of tee times with a 95% confidence level. • We take a sample of 900 women golfers and find that 396 are satisfied with the tee times. • _ • Given: p = 396/900 = 0.44  = 0.05 • The 95% confidence interval is: • _ _ _ • p ± z √ [p (1 – p)]/n = 0.44 ± 1.96√ [0.44 (1 – 0.44)]/900 • 0.44 ± 0.0324 0.4076 to 0.4724 or 40.76% to 47.24% • We are 95% confident that the proportion of all women golfers who are satisfied with tee times is between 40.76% and 47.24%.

  30. Sample Problem #38 (10-Page 323; 11-Page 332)) • Point estimate of the proportion of companies that fell short of estimates = 29/162 = 0.179 _ • Given : p = 104/162 = 0.642 _ _ Margin of error = z √ [(p (1 – p ) / n)] 0.025 = 1.96 √ [(0.642) (1 – 0.642) / 162] = 0.0738

  31. Sample Problem (contd) • 95% confidence interval: • 0.642 ± 0.0738 • 0.5682 and 0.7158 • We are 95% confident that the proportion of companies that beat estimates of their profits is between 56.82% and 71.58%. • 2 2 • c. Required sample size, n = [(1.96) (0.642) (0.358)]/(0.05) • = 353.18 use 354

  32. End of Chapter 8

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