1

Level of Significance

• α is a predetermined value by convention usually 0.05

• α = 0.05 corresponds to the 95% confidence level

• We are accepting the risk that out of 100 samples, we would reject a true null hypothesis five times

2

Sampling Distribution Of Means

• A sampling distribution of means is the relative frequency distribution of the means of all possible samples of size n that could be selected from the population.

3

One Sample Test

• Compares mean of a sample to known population mean– Z-test– T-test

This lecture focuses onone sample t-test

4

The One Sample t – Test

Testing statistical hypothesis about µ when σ is not known OR sample size is

small

5

An Example Problem

• Suppose that Dr. Tate learns from a national survey that the average undergraduate student in the United States spends 6.75 hours each week on the Internet – composing and reading e-mail, exploring the Web and constructing home pages. Dr. Tate is interested in knowing how Internet use among students at George Mason University compares with this national average.

• Dr. Tate randomly selects a sample of only 10 students. Each student is asked to report the number of hours he or she spends on the Internet in a typical week during the academic year.

Populaon mean

Small sample

Population variance is unknown & estimated from sample

6

Steps in Test of Hypothesis

1. Determine the appropriate test 2. Establish the level of significance:α3. Determine whether to use a one tail or two

tail test4. Calculate the test statistic5. Determine the degree of freedom6. Compare computed test statistic against a

tabled value

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1. Determine the appropriate test

If sample size is more than 30 use z-test If sample size is less than 30 use t-test

Sample size of 10

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2. Establish Level of Significance

• α is a predetermined value

• The convention• α = .05

• α = .01

• α = .001

• In this example, assume α = 0.05

9

3. Determine Whether to Use a One or Two Tailed Test

• H0 :µ = 6.75

• Ha :µ ≠ 6.75 A two tailedtest because it

can be either larger

or smaller

10

4. Calculating Test Statistics

StudentNumber of Hours(X)

A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61

Σ=148.90

2)( XX )( XX 2)( XX

90.9X

StudentNumber of Hours(X)

A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61

Σ=148.90

2)( XX )( XX 2)( XX

90.9X

Sample mean

11

4. Calculating Test Statistics

StudentNumber of Hours(X)

A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61

Σ=148.90

2)( XX )( XX 2)( XX

90.9X

StudentNumber of Hours(X)

A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61

Σ=148.90

2)( XX )( XX 2)( XX

90.9X

Deviationfrom sample

mean

12

StudentNumber of Hours(X)

A 6 -3.9 15.21B 9 -0.9 0.81C 12 2.1 4.41D 3 -6.9 47.61E 11 1.1 1.21F 10 0.1 0.01G 18 8.1 56.61H 9 -0.9 0.81I 13 3.1 9.61J 8 -1.9 3.61

Σ=148.90

)( XX

4. Calculating Test Statistics

Squared deviation

from samplemean

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4. Calculating Test Statistics

Standard deviation

of observations

14

4. Calculating Test Statistics

Calculated

t value

15

4. Calculating Test Statistics

Standard deviationof sample means

16

4. Calculating Test Statistics

Calculated t

17

5. Determine Degrees of Freedom

• Degrees of freedom, df, is value indicating the number of independent pieces of information a sample can provide for purposes of statistical inference.

• Df = Sample size – Number of parameters estimated

• Df is n-1 for one sample test of mean because the population variance is estimated from the sample

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Degrees of Freedom• Suppose you have a sample of three observations:

2 -1 1

2 -1 1

5 +2 4

-------- ------ ------

Σ=0 6

-------- ----------------2)( XX )( XX X

3X

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Degrees of Freedom Continued

• For your sample scores, you have only two independent pieces of information, or degrees of freedom, on which to base your estimates of S and xS

20

6. Compare the Computed Test Statistic Against a Tabled Value

• α = .05

• Df = n-1 = 9

• Therefore, reject H0

21

Decision Rule for t-Scores

If |tc| > |tα| Reject H0

22

Decision Rule for P-values

If p value < α Reject H0

Pvalue is one minusprobability of observing

the t-value calculated from our sample

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Example of Decision Rules

• In terms of t score:

|tc = 2.449| > |tα= 2.262| Reject H0

• In terms of p-value:

If p value = .037 < α = .05 Reject H0

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Constructing a Confidence Interval for µ

Sam

ple m

ean

Standard deviation of sample means

Critical t value

25

Constructing a Confidence Interval for µ for the Example

• Sample mean is 9.90

• Critical t value is 2.262

• Standard deviation of sample means is 1.29

• 9.90 + 2.262 * 1.29

• The estimated interval goes from 6.98 to 12.84