1 / 13

Section 7-3: Confidence Intervals For the Mean

Section 7-3: Confidence Intervals For the Mean. σ is unknown AND n < 30. In Section 7.2, we used the standard normal distribution (z-table) to find confidence intervals for the mean because:. 1) σ was known and normally distributed. or. 2) σ was unknown and n ≥ 30.

carrie
Download Presentation

Section 7-3: Confidence Intervals For the Mean

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Section 7-3: Confidence Intervals For the Mean σ is unknown AND n < 30

  2. In Section 7.2, we used the standard normal distribution (z-table) to find confidence intervals for the mean because: 1) σ was known and normally distributed or 2) σ was unknown and n ≥ 30

  3. Sometimes the population standard deviation, σ, is unknown and the sample size is less than 30. The sample standard deviation, s, is used instead of σ, but we must use the t-table

  4. Characteristics of the t distribution The t distribution is SIMILAR to the standard normal distribution in the following ways: 1) It is bell-shaped 2) It is symmetrical about the mean 3) The mean, median, and mode are equal to 0 and are located at the center of the distribution 4) The curve never touches the x-axis

  5. Characteristics of the t distribution The t distribution is DIFFERS from the standard normal distribution in the following ways: 1) The variance is greater than 1 2) The t-distribution is actually a family of curves based on “degrees of freedom” (d.f.) 3) As the sample size increases, the t-distribution approaches the standard distribution

  6. Degrees of Freedom (d.f.): the number of values that are free to vary after a sample statistic has been computed and is used when a distribution consists of a family of curves Equation for Degrees of Freedom: (d.f.) = n – 1

  7. Example: • Find the critical value (tα/2) for a 99% • confidence interval when the sample size is 14. n = 14 d.f. = n – 1 = 14 – 1 = 13 C.I. = 99% t = ± 3.012

  8. Example: 2) Find the critical value (tα/2) for a 90% confidence interval when the sample size is 21. n = 21 d.f. = n – 1 = 21 – 1 = 20 C.I. = 90% t = ± 1.725

  9. Example: 3) Find the critical value (tα/2) for a 95% confidence interval when the sample size is 8. n = 8 d.f. = n – 1 = 8 – 1 = 7 C.I. = 95% t = ± 2.365

  10. Formula for a Specific Confidence Interval for the Mean when σ is unknown and n < 30 ( ) ( ) _s_ √n _s_ √n X – tα/2 < µ < X + tα/2 Note: -d.f. is used to find tα/2 -n is used in the equation You’re using t-values which are rounded to 3 decimals, so your answer should be rounded to 3 decimals

  11. Example: (ROUND TO THREE DECIMAL PLACES) • The average yearly income for 28 married • couples living in City A is $58,219. The standard • deviation of the sample is $56. Find the 95% • confidence interval of the true mean. n = 28 d.f. = 27 C.I. = 95% tα/2= 2.052 X = $58,219 s = $56 ( ) ( ) _s_ √n _s_ √n X – tα/2 < µ < X + tα/2 ( ) ( ) _ 56 √28 _ 56 √28 58,219 – 2.052 < µ < 58,219 + 2.052 58,219 – 21.716 < µ < 58,219 + 21.716 58,197.283< µ < 58,240.716

  12. Example: (ROUND TO THREE DECIMAL PLACES) • Find a 90% confidence interval for the • population mean if a sample of 20 has a sample • mean of 1462 and a sample standard deviation of 42. n = 20 d.f. = 19 C.I. = 90% tα/2= 1.729 X = 1462 s = 42 ( ) ( ) _s_ √n _s_ √n X – tα/2 < µ < X + tα/2 ( ) ( ) _ 42 √20 _ 42 √20 1462 – 1.729 < µ < 1462 + 1.729 1462 – 16.238 < µ < 1462 + 16.238 1445.762< µ < 1478.238

  13. Example: (ROUND TO THREE DECIMAL PLACES) • A sample of six adult elephants had an average • weight of 12,200 pounds, with a sample standard • deviation of 200 pounds. Find the 95% confidence • Interval of the true mean. n = 6 d.f. = 5 C.I. = 95% tα/2= 2.571 X = 12,200 s = 200 ( ) ( ) _s_ √n _s_ √n X – tα/2 < µ < X + tα/2 ( ) ( ) 200 √6 200 √6 12,200 – 2.571 < µ < 12,200 + 2.571 12,200 – 209.921 < µ < 12,200 + 209.921 11,990.079< µ < 12,409.921

More Related