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Understanding Inferential Statistics—Estimation

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Presentation on theme: "Understanding Inferential Statistics—Estimation"— Presentation transcript:

1 Understanding Inferential Statistics—Estimation

2 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. 

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

4 Statistical Methods Statistical Methods Descriptive Statistics
Inferential Estimation Hypothesis testing

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

6 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

7 Characteristic measures
Point Estimating &Population Parameters Characteristic measures Point estimates (Sample) Parameters (Population) Mean 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

8 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.

9 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.

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12 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 and 549.2

13 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.

14 Thus: If σ is known: If σ is unknown:

15 = 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 and

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17 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 and

18 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 = = 180mg/dl σ = 33mg/dl Interpretation: we are 95% confident that the freshman nursing students population mean cholesterol level is between and

19 Effect of Increasing Sample Size in Estimating Population Parameters
Using a sample size of 30 the 95% confidence interval is and Using a sample size of 60 the 95% confidence interval is and 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.

20 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)

21 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.

22 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.

23 Solution:

24 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:

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


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