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M ARIO F . T RIOLA

Learn how to estimate a population mean with small samples using the Student t distribution and confidence intervals in statistics. Understand degrees of freedom and critical values.

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M ARIO F . T RIOLA

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  1. STATISTICS ELEMENTARY Section 6-3 Estimating a Population Mean: Small Samples MARIO F. TRIOLA EIGHTH EDITION

  2. If 1) n 30 2) The sample is a simple random sample. 3) The sample is from a normally distributed population. Case 1 ( is known): Largely unrealistic; Use methods from 6-2 Case 2 (is unknown): Use Student t distribution Small SamplesAssumptions

  3. If the distribution of a population is essentially normal, then the distribution of Student t Distribution x - µ t = s n

  4. If the distribution of a population is essentially normal, then the distribution of Student t Distribution x - µ t = s n • is essentially a Student t Distributionfor all      samples of size n. • is used to find critical values denoted by t/ 2

  5. Find the “t” score using the InvT command as follows: 2nd - Vars – then invT( percent, degree of freedom ) Using the TI Calculator

  6. Degrees of Freedom (df ) corresponds to the number of sample values that can vary after certain restrictions have imposed on all data values Definition

  7. Degrees of Freedom (df ) corresponds to the number of sample values that can vary after certain restrictions have imposed on all data values Definition df = n - 1 in this section

  8. Degrees of Freedom (df ) = n - 1 corresponds to the number of sample values that can vary after certain restrictions have imposed on all data values Definition Specific # Any # Any # Any # Any # Any # Any # Any # Any # Any # n = 10 df = 10 - 1 = 9 so that x = a specific number

  9. Based on an Unknown  and a Small Simple Random Sample from a Normally Distributed Population Margin of Error E for Estimate of 

  10. Based on an Unknown  and a Small Simple Random Sample from a Normally Distributed Population Margin of Error E for Estimate of  Formula 6-2 s E = t/ n 2 where t/ 2 has n - 1 degrees of freedom

  11. Confidence Interval for the Estimate of EBased on an Unknown  and a Small Simple Random Sample from a Normally Distributed Population

  12. Confidence Interval for the Estimate of EBased on an Unknown  and a Small Simple Random Sample from a Normally Distributed Population x - E < µ < x + E

  13. Confidence Interval for the Estimate of EBased on an Unknown  and a Small Simple Random Sample from a Normally Distributed Population x - E < µ < x + E s E = t/2 where n

  14. t/2 found using invT Confidence Interval for the Estimate of EBased on an Unknown  and a Small Simple Random Sample from a Normally Distributed Population x - E < µ < x + E s E = t/2 where n

  15. There is a different distribution for different sample sizes It has the same general bell shape as the normal curve but greater variability because of small samples. The t distribution has a mean of t = 0 The standard deviation of the t distribution varies with the sample size and is greater than 1 As the sample size (n) gets larger, the t distribution gets closer to the normal distribution. Properties of the Student t Distribution

  16. Student t Distributions for n = 3 and n = 12 Student t distribution with n = 12 Standard normal distribution Student t distribution with n = 3 0 Figure 6-5

  17. Using the Normal and t Distribution Figure 6-6

  18. Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.)

  19. x = 26,227 s = 15,873  = 0.05 /2 = 0.025 Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.)

  20. x = 26,227 s = 15,873  = 0.05 /2 = 0.025 t/2 = 2.201 Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.) E = t2 s =(2.201)(15,873) = 10,085.29 n 12

  21. x = 26,227 s = 15,873  = 0.05 /2 = 0.025 t/2 = 2.201 x - E < µ < x + E Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.) E = t2 s =(2.201)(15,873) = 10,085.3 n 12

  22. x = 26,227 s = 15,873  = 0.05 /2 = 0.025 t/2 = 2.201 x - E < µ < x + E Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.) E = t2 s =(2.201)(15,873) = 10,085.3 n 12 26,227 - 10,085.3 < µ < 26,227 + 10,085.3

  23. x = 26,227 s = 15,873  = 0.05 /2 = 0.025 t/2 = 2.201 x - E < µ < x + E Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.) E = t2 s =(2.201)(15,873) = 10,085.3 n 12 26,227 - 10,085.3 < µ < 26,227 + 10,085.3 $16,141.7< µ < $36,312.3

  24. x = 26,227 s = 15,873  = 0.05 /2 = 0.025 t/2 = 2.201 x - E < µ < x + E Example: A study of 12 Dodge Vipers involved in collisions resulted in repairs averaging $26,227 and a standard deviation of $15,873. Find the 95% interval estimate of , the mean repair cost for all Dodge Vipers involved in collisions. (The 12 cars’ distribution appears to be bell-shaped.) E = t2 s =(2.201)(15,873) = 10,085.3 n 12 26,227 - 10,085.3 < µ < 26,227 + 10,085.3 $16,141.7< µ < $36,312.3 We are 95% confident that this interval contains the average cost of repairing a Dodge Viper.

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