Understanding Inferential Statistics—Estimation

Types of Statistics The choice of a type of analysis is based on: Research questions. The type of data collected. Audience who will receive the results. 

Descriptive & Inferential Statistics Organize Summarize Data presentation Descriptive Statistics Generalize findings from samples to populations Hypothesis testing Assess relationships among variables Inferential Statistics

Statistical Methods Statistical Methods Descriptive Statistics Inferential Estimation Hypothesis testing

Estimation & Hypothesis testing Sample statistic (`X, Ps ) Inference Process Population Estimation & Hypothesis testing Sample statistic (`X, Ps ) Sample

Population Parameters Point Estimating &Population Parameters Population Parameters µ = Population mean σ = Population standard deviation σ2 = Population variance π = Population proportion N = The size of the population you can generalize to Sample Statistics (Point Estimates) = Mean point Estimate S = Standard deviation point estimate S2 = Variance point estimate P = Proportion point estimate n = The size of a sample taken from a population Population Parameter is Unknown Sample Statistics   Population Parameters are usually represented by Greek letters Point estimates are usually represented by Roman letters

Characteristic measures Point Estimating &Population Parameters Characteristic measures Point estimates (Sample) Parameters (Population) Mean µ Standard deviation S σ Variance S2 σ2 Proportion P π Population Parameters are usually represented by Greek letters Point estimates are usually represented by Roman letters

Statistical Methods Descriptive Statistics Inferential Estimation Point Estimation Interval Hypothesis testing Point estimation involves the use of sample data to calculate a single value (known as a statistic) which is to serve as a "best guess" for an unknown population parameter. Interval estimation is the use of sample data to calculate an interval of possible (or probable) values of an unknown population parameter.

Interval Estimation of Population Mean Interval Estimation of Population Mean (µ) with Known Variance (σ Known)   *For α = 0.05 (95% CI), we get Zα/2 = Z0.025 = 1.96. Example 1: The College Board reports that the scores on the 2010 SAT mathematics test were normally distributed. A sample of 25 scores had a mean of 510. Assume the population standard deviation is 100. Construct a 95% confidence interval for the population mean score on the 2010 SAT math test.

Solution: n = 25, = 510, σ = 100 Interpretation: We are 95% confidant that the population mean SAT score on the 2010 mathematics SAT test lies between 470.8 and 549.2

Interval Estimation of Population Mean (µ) with Unknown Variance (σ Unknown) Example 2: Estimate with 95% confidence interval the mean cholesterol level for freshman nursing students using a sample of 30 students who have an average cholesterol of 180mg/dl and a standard deviation of 34mg/dl.   Recall, Note: Since σ ( Standard deviation of the population) is unknown, we will use s (standard deviation of the sample) in place of σ. When s is used instead of σ, an error is introduced because s is only an estimate of σ. We will substitute the Z value with a another value called the student’s t or just t to account for this additional error.

Thus: If σ is known: If σ is unknown:

= 180mg/dl, σ = unknown s = 34mg/dl n = 30 Solution: = 180mg/dl, σ = unknown s = 34mg/dl n = 30 d.f * = n-1 = 30-1=29 * Degrees of freedom (d.f) is the number of values that are free to vary when computing a statistic Interpretation: we are 95% confidant that the freshman nursing students population mean cholesterol level is between 167.31 and 192.69

Effect of Increase in Sample size in Estimating Population Parameters Example 3 a: Estimate with 95% confidence interval the mean cholesterol level for freshman nursing students using a sample of 30 students who have an average cholesterol of 180mg/dl. Assume the population standard deviation to be 33 mg/dl. Solution: n = 30, = 180mg/dl, σ = 33mg/dl Interpretation: we are 95% confident that the freshman nursing students population mean cholesterol level is between 168.19 and 191.81

Example 3 b: Estimate with 95% confidence interval the mean cholesterol level for freshman nursing students using a sample of 60 students who have an average cholesterol of 180mg/dl. Assume the population standard deviation to be 33 mg/dl. n = 60 = 180mg/dl σ = 33mg/dl Interpretation: we are 95% confident that the freshman nursing students population mean cholesterol level is between 171.65 and 188.35

Effect of Increasing Sample Size in Estimating Population Parameters Using a sample size of 30 the 95% confidence interval is 168.19 and 191.81 Using a sample size of 60 the 95% confidence interval is 171.65 and 188.35 Since the confidence interval using a larger sample size is more narrow then it is more precise in estimating the population mean than using a small sample size.

Sample Size for Estimation For 95% CI, Z = 1.96 Get from literature σ could also be estimated by Range/4 if the distribution is normal OR = error we are willing to accept (difference between point estimate and parameter)

Example 4: For freshman nursing students: estimate, with 95% confidence the minimum sample size needed to estimate their mean cholesterol to within 10 mg/dl. A best estimate of σ is 33 mg/dl Interpretation: The minimum sample size needed to estimate their mean cholesterol to within 10 mg/dl is 42 subjects.

Interval Estimation of Population Proportion π Example 5: In a sample of n = 400 households, 80 households had participated in the recent elections. Estimate, with 95% confidence, the proportion of all households that will participate in the next election.

Solution:  

Sample Size for Estimation Example 6: If 50 out of 100 LLU students in a recent survey preferred alcohol free beverages, and you want to estimate the proportion, π, of LLU students who favor alcohol- free beverages, within ±3 percentage points 95% of the time, you would need a sample of ?? at least:

Solution: Interpretation: Therefore, we need at least 1068 subjects who favor alcohol free beverages to within 3 percentage points 95% of the time.