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

Statistical Methods

Inference Process

PopulationPopulation

SampleSampleSample Sample statistic (`statistic (`XX, , PPs s

))

Estimation & Estimation & Hypothesis Hypothesis

testingtesting

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

Point Estimating &Population Parameters

Characteristic measures

Point estimates(Sample)

Parameters(Population)

Mean µ

Standard deviation S σ

Variance S2 σ2

Proportion P π

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.

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.

Interval Estimation of Population Mean

/2x

n

*For α = 0.05 (95% CI), we get Zα/2 = Z0.025 = 1.96.

I. Interval Estimation of Population Mean (µ) with Known Variance (σ Known)

Solution:

n = 25, = 510, σ = 100

x

Interpretation: We are 95% confidant that the population mean SAT score on the

2010 mathematics SAT test lies between 470.8 and 549.2

100510 1.96

25510 39.2

(470.8,549.2)

/2xn

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.

II. Interval Estimation of Population Mean (µ) with Unknown Variance (σ Unknown)

Recall, /2xn

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 σσ.

• We will substitute the Z value with a another value called the student’s t or just t to account for this additional error.

If σ is known:

If σ is unknown:

Thus:

/2xn

/2

sx t

n

Solution:

= 180mg/dl, σ = unknown s = 34mg/dl n = 30

x

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

34180 2.045

30180 12.69

(167.31,192.69)

/2

Sx t

n

Interpretation: we are 95% confidant that the freshman nursing students population

mean cholesterol level is between 167.31 and 192.69

III.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

x

/2xn

33180 1.96

30180 11.81

(168.19,191.81)

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

x

Interpretation: we are 95% confident that the freshman nursing students population mean cholesterol level is between 171.65 and 188.35

33180 1.96

60180 8.35

(171.65,188.35)

/2x Zn

• 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.

2

x

zn

For 95% CI, Z = 1.96

Sample Size for Estimation

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)

x

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

2

x

zn

2

1.96 3342

10n

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

/2

(1 )p pp Z

n

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.

0.2(0.8)0.2 1.96

400

0.160.2 1.96

400

0.2 1.96 0.0004

0.2 1.96(0.02)

0.2 0.0392

(0.1608,0.2392)

/2

(1 )p pp Z

n

Solution:

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:

Sample Size for Estimation

2

2

( )(1 )

( )

Zn

Interpretation:

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

2

2

( )(1 )

( )

Zn

2

2

(1.96) 3.8416.5(.5) .25( ) .25(4268.44) 1,067.11

(.03) 0.0009n

Solution: