Week8Examples.pdf - STAT 311 CI HT EXAMPLES And the...

This preview shows page 1 - 13 out of 75 pages.

STAT 311: CI & HT EXAMPLES And the Continuity Correction to the Normal Approximation for the Binomial
Examples of how to construct confidence intervals in different situations Confidence Intervals 2 Stat 311 Spring 2018 : Prof Morris
Example: Autism Stat 311 Spring 2018 : Prof Morris 3 Concerns about a link between vaccines and autism were first raised more than a decade ago by British physician Andrew Wakefield. His 1998 article, based on 12 children (!), has since been discredited and was retracted in 2010 by Lancet, the journal that published it. And Wakefield’s medical license has been revoked… But the debate continues.
January 2011 Stat 311 Spring 2018 : Prof Morris 4 Dr. Wakefield, the face of a haunted man
Autism diagnosis trends Stat 311 Spring 2018 : Prof Morris “The American Psychiatric Association did not retreat from another widely criticized proposal, to streamline the definition of autism .” (NYTimes 5/9/2012) 0 5 10 15 20 25 30 35 40 45 50 1970 1975 1980 1985 1990 1995 2000 2005 Cases per 10000 Year Reported Autism Prevalence California Autism Prevalence 1970-2006 5
1. Sample proportion, ?? and ? 1 − ? > 10 Stat 311 Spring 2018 : Prof Morris 6 Population parameter ? Sample estimate ? ?? ? ?(1 ?) ? 𝑧 ? 𝑧 90 = 1.65 𝑧 95 = 1.96 𝑧 99 = 2.58 C % CI for ?: ? ± 𝑧 ? ?? ?
1. A 95% CI for the prevalence of autism Stat 311 Spring 2018 : Prof Morris In a random sample of 10,000 children; 45 are found to have been diagnosed with autism. Point estimate: Standard error: 95% CI for the population prevalence of autism: ? ± 1.96 ?? ? = 0.0045 ± 0.0013 = 0.003, 0.006 = 0.3 − 0.6% ? = ? ? = 45 10,000 = 0.0045 ?? ? = ?(1 ?) ? = 0.0045(1 − 0.0045) 10,000 = 0.0067 7
2. Single mean, large sample Stat 311 Spring 2018 : Prof Morris 8 Population parameter 𝜇 𝑋 Sample estimate 𝒙 ? ? 𝜎 𝑋 ? ?? ? ? 𝑥 ? 𝑧 ? 𝑧 90 = 1.65 𝑧 95 = 1.96 𝑧 99 = 2.58 C % CI for 𝜇 𝑋 : ? ± 𝑧 ? ? ? (𝜎 𝑋 ?????) ? ± ? ? (𝜐) ?? ? (??ℎ???𝑖??) ? ? (100) ? 90 (100) = 1.66 ? 95 (100) = 1.98 ? 99 (100) = 2.58 Note how little difference there is in the z and t multipliers for n >100
From a recently published study (see references) 2. A 95% CI for a sample mean Stat 311 Spring 2018 : Prof Morris 9 Mercury blood level among children with ASD ? 0.49 ? 𝑥 1.08 n 332 ?? ? = 𝑠 𝑥 ? = 1.08 332 = 0.06 For the confidence interval What multiplier do we use here?
A t-table lookup Stat 311 Spring 2018 : Prof Morris 10 95 (332 1)* ( ) x t se x For a 95% CI: There is no entry for d.f.=331, so we use the value for 100 (this will give a conservative estimate)
From a recently published study (see references) 2. A 95% CI for a sample mean Stat 311 Spring 2018 : Prof Morris 11 Mercury blood level among children with ASD ? 0.49 ? 𝑥 1.08 n 332 ?? ? = 𝑠 𝑥 ? = 1.08 332 = 0.06 95% t-interval: ? ± ? ? (331) ?? ? 0.49 ± 1.98 ∗ 0.06 0.49 ± 0.12 Note: the empirical rule multiplier = 2, and the z = 1.96, would make no difference in this case
3. Paired mean difference Stat 311 Spring 2018 : Prof Morris 12 Population parameter 𝛿 Sample estimate ? ? ? 𝜎 ? ? ?? ? ? ? ? 𝑧 ? 𝑧 90 = 1.65 𝑧 95 = 1.96 𝑧 99 = 2.58 C % CI for 𝛿: ? ± 𝑧 ? ? ? 𝜎 𝑋 ????? ? ± ? ? ?? ? (??ℎ???𝑖??) Similar to the single mean CI, but the derivation is different, see the appendix

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture