🗊Презентация Confidence interval estimation

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Confidence Interval Estimation
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Confidence Interval Estimation

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Types of Estimates Point Estimate A single number used to estimate an unknown population parameter Interval Estimate A range of values used to estimate a population parameter Characteristics Better idea of reliability of estimate Decision making is facilitated
Описание слайда:
Types of Estimates Point Estimate A single number used to estimate an unknown population parameter Interval Estimate A range of values used to estimate a population parameter Characteristics Better idea of reliability of estimate Decision making is facilitated

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Point Estimates
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Point Estimates

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Point and Interval Estimates A point estimate is a single number, a confidence interval provides additional information about variability
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Point and Interval Estimates A point estimate is a single number, a confidence interval provides additional information about variability

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Confidence Interval Estimate An interval gives a range of values: Takes into consideration the variation in sample statistics from sample to sample Based on observation from 1 sample Gives information about closeness to unknown population parameters Stated in terms of level of confidence Can never be 100% confident
Описание слайда:
Confidence Interval Estimate An interval gives a range of values: Takes into consideration the variation in sample statistics from sample to sample Based on observation from 1 sample Gives information about closeness to unknown population parameters Stated in terms of level of confidence Can never be 100% confident

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Confidence Level, (1-) Suppose confidence level γ = 95% Also written γ =(1 - ) = .95 Where  is the risk of being wrong A relative frequency interpretation: In the long run, 95% of all the confidence intervals that can be constructed will contain the unknown parameter A specific interval either will contain or will not contain the true parameter No probability involved in a specific interval
Описание слайда:
Confidence Level, (1-) Suppose confidence level γ = 95% Also written γ =(1 - ) = .95 Where  is the risk of being wrong A relative frequency interpretation: In the long run, 95% of all the confidence intervals that can be constructed will contain the unknown parameter A specific interval either will contain or will not contain the true parameter No probability involved in a specific interval

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Estimation Process
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Estimation Process

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General Formula The general formula for all confidence intervals is:
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General Formula The general formula for all confidence intervals is:

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Confidence Intervals
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Confidence Intervals

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Confidence Interval for μ (σ Known) Assumptions Population standard deviation σ is known Population is normally distributed If population is not normal, use large sample Confidence interval estimate for μ
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Confidence Interval for μ (σ Known) Assumptions Population standard deviation σ is known Population is normally distributed If population is not normal, use large sample Confidence interval estimate for μ

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Finding the Critical Value
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Finding the Critical Value

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Finding the Critical Value Consider a 95% confidence interval:
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Finding the Critical Value Consider a 95% confidence interval:

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Margin of Error Margin of Error (e): the amount added and subtracted to the point estimate to form the confidence interval
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Margin of Error Margin of Error (e): the amount added and subtracted to the point estimate to form the confidence interval

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Factors Affecting Margin of Error Data variation, σ : e as σ Sample size, n : e as n Level of confidence, 1 -  : e if γ =1 - 
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Factors Affecting Margin of Error Data variation, σ : e as σ Sample size, n : e as n Level of confidence, 1 -  : e if γ =1 - 

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Example Example A sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is .35 ohms. Determine a 95% confidence interval for the true mean resistance of the population.
Описание слайда:
Example Example A sample of 11 circuits from a large normal population has a mean resistance of 2.20 ohms. We know from past testing that the population standard deviation is .35 ohms. Determine a 95% confidence interval for the true mean resistance of the population.

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Solution –
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Solution –

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Interpretation We are γ=95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms
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Interpretation We are γ=95% confident that the true mean resistance is between 1.9932 and 2.4068 ohms

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Confidence Interval for μ (σ Unknown) If the population standard deviation σ is unknown, we can substitute the sample standard deviation, s as an estimate In these case the t-distribution is used instead of the normal distribution
Описание слайда:
Confidence Interval for μ (σ Unknown) If the population standard deviation σ is unknown, we can substitute the sample standard deviation, s as an estimate In these case the t-distribution is used instead of the normal distribution

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Student’s t Distribution
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Student’s t Distribution

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Confidence Interval for μ (σ Unknown) Assumptions Population standard deviation is unknown Population is not highly skewed Population is normally distributed or the sample size is large (>30) Use Student’s t Distribution
Описание слайда:
Confidence Interval for μ (σ Unknown) Assumptions Population standard deviation is unknown Population is not highly skewed Population is normally distributed or the sample size is large (>30) Use Student’s t Distribution

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Confidence Interval Estimate: where t is the critical value of the t-distribution with n-1 degrees of freedom and an area of α/2 in each tail)
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Confidence Interval Estimate: where t is the critical value of the t-distribution with n-1 degrees of freedom and an area of α/2 in each tail)

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Define t from equation  – Confidence Coefficient. t - obtain with using Excel function TINV. t = TINV(1- γ; n-1) =T.INV.2T (1- γ; n-1)
Описание слайда:
Define t from equation  – Confidence Coefficient. t - obtain with using Excel function TINV. t = TINV(1- γ; n-1) =T.INV.2T (1- γ; n-1)

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Example A random sample of n = 25 has X = 50 and S = 8. Form 95% confidence interval for μ degrees of freedom = n – 1 = 24,  =0,95.
Описание слайда:
Example A random sample of n = 25 has X = 50 and S = 8. Form 95% confidence interval for μ degrees of freedom = n – 1 = 24,  =0,95.

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To get a t - value use the TINV function. To get a t - value use the TINV function. The value of alpha =(1-confidence) and n-1 degrees of freedom are the inputs needed. For 95% confidence use alpha =0.05 and for a sample size of 25 use 24 df t= TINV(0,05; 24)=2,0639 t= T.INV.2T(0,05;24)= 2,0639
Описание слайда:
To get a t - value use the TINV function. To get a t - value use the TINV function. The value of alpha =(1-confidence) and n-1 degrees of freedom are the inputs needed. For 95% confidence use alpha =0.05 and for a sample size of 25 use 24 df t= TINV(0,05; 24)=2,0639 t= T.INV.2T(0,05;24)= 2,0639

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Confidence interval estimation, слайд №25
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Confidence Interval on the Variance and Standard Deviation of a Normal Distribution
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Confidence Interval on the Variance and Standard Deviation of a Normal Distribution

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Confidence Interval on the Variance and Standard Deviation of a Normal Distribution
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Confidence Interval on the Variance and Standard Deviation of a Normal Distribution

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Confidence interval estimation, слайд №28
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Confidence Interval on the Variance and Standard Deviation of a Normal Distribution
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Confidence Interval on the Variance and Standard Deviation of a Normal Distribution

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Confidence interval estimation, слайд №30
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We can use 2 –Table for solving next equation Or EXCEL function CHIINV (q; n-1), =CHISQ.INV.RT(q;n-1).
Описание слайда:
We can use 2 –Table for solving next equation Or EXCEL function CHIINV (q; n-1), =CHISQ.INV.RT(q;n-1).

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EXAMPLE According to the 20 measurements found standard deviation S = 0,12. Find precision measurements with reliability 0.98.
Описание слайда:
EXAMPLE According to the 20 measurements found standard deviation S = 0,12. Find precision measurements with reliability 0.98.

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With using CHIINV (q; n-1) we obtain 12 і 22 . With using CHIINV (q; n-1) we obtain 12 і 22 . For degrees of freedom n - 1=19 and probability α2=(1-0,98)/2=0,01 define 22 =36,2, after that for n - 1=19 and probability α1=(1+0,98)/2=0,99 define 12 =7,63. 22 = CHIINV(0,01; 19)=36,2 ; =CHISQ.INV.RT(0,01;19). 12 = CHIINV(0,99;19)=7,63. =CHISQ.INV.RT(0,01;19).
Описание слайда:
With using CHIINV (q; n-1) we obtain 12 і 22 . With using CHIINV (q; n-1) we obtain 12 і 22 . For degrees of freedom n - 1=19 and probability α2=(1-0,98)/2=0,01 define 22 =36,2, after that for n - 1=19 and probability α1=(1+0,98)/2=0,99 define 12 =7,63. 22 = CHIINV(0,01; 19)=36,2 ; =CHISQ.INV.RT(0,01;19). 12 = CHIINV(0,99;19)=7,63. =CHISQ.INV.RT(0,01;19).

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calculate Confidence Interval for σ is (0,09;0,19).
Описание слайда:
calculate Confidence Interval for σ is (0,09;0,19).

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