Statistics: Parameter Estimation

• Point estimate 
• Confidence interval (CI) estimate.

For both continuous variables (e.g., population mean) and dichotomous variables (e.g., population proportion) one first computes the point estimate from a sample.

The process of finding an approximate value of some parameter—such as the mean (average)—of a population from random samples of the population.

It is the value of statistic that estimates the value of a parameter.

The sample standard deviation (s) is a point estimate of the population standard deviation (σ).

The sample mean (̄x) is a point estimate of the population mean, μ.

The sample variance (s2) is a point estimate of the population variance (σ2).

Range of values you expect your estimate to fall between if you redo your test, within certain level of confidence.

Confidence-Describes - Probability

Point Estimate  +- Margin of error 

Given : ( α = 0.05 , n ,x̄ , Population standard deviation σ )  

Tests:

1. z-test (Population standard is given & n>=30)
2. t-test (Population standard is not given & n

Population standard is given & n >= 30.

Point Estimate +- Margin of error 

x̄  +-  z  α/2 (σ/ √ n )

Upper Bound -     x̄ + z  α/2 (σ/ √ n ) = ztable(Answer)

Lower Bound -     x̄  -  z  α/2 (σ/ √ n )=ztable(Answer)

Population standard deviation is not  given & n >≠ 30.

Point Estimate +- Margin of error 

x̄ +- z  α/2 (σ/ √ n )

t = degree of freedom= n-1

Upper Bound -   x̄ + t  α/2 (s/ √ n ) = t table(Answer)

Lower Bound -   x̄  -  t  α/2 (s / √ n )=t table(Answer)

 (s / √ n ) is Standard error

The standard error is the standard deviation of a sample population. It measures the accuracy with which a sample represents a population.

Degrees of freedom, often represented by v or df, is the number of independent pieces of information used to calculate a statistic.

It’s calculated as the sample size minus the number of restrictions.

Suppose you randomly sample 10 American adults and measure their daily calcium intake. You use a one-sample t test to determine whether the mean daily intake of American adults is equal to the recommended amount of 1000 mg.The test statistic, t, has 9 degrees of freedom:

df = n − 1

df = 10 − 1

df = 9

You calculate a t value of 1.41 for the sample, which corresponds to a p value of .19. You report your results:

“The participants’ mean daily calcium intake did not differ from the recommended amount of 1000 mg, t(9) = 1.41, p = 0.19.”