ppt hypothesis testing

Hypothesis Testing

Significant DifferencesHypothesis testing is designed to detect significant differences: differences that did not occur by random chance.

In the one sample case: we compare a random sample (from a large group) to a population.

We compare a sample statistic to a population parameter to see if there is a significant difference.

Our Problem:The education department at a university has been accused of grade inflation so education majors have much higher GPAs than students in general.

GPAs of all education majors should be compared with the GPAs of all students.There are 1000s of education majors, far too many to interview. How can this be investigated without interviewing all education majors?

What we know:The average GPA for all students is 2.70. This value is a parameter.

To the right is the statistical information for a random sample of education majors: = 2.70

=3.00s = 0.70n = 117

Questions to ask:Is there a difference between the parameter (2.70) and the statistic (3.00)?

Could the observed difference have been caused by random chance?

Is the difference real (significant)?

Two Possibilities:The sample mean (3.00) is the same as the pop. mean (2.70). The difference is trivial and caused by random chance.

The difference is real (significant).Education majors are different from all students.

The Null and Alternative Hypotheses:Null Hypothesis (H0)The difference is caused by random chance.The H0 always states there is no significant difference. In this case, we mean that there is no significant difference between the population mean and the sample mean.Alternative hypothesis (H1)The difference is real.(H1) always contradicts the H0.

One (and only one) of these explanations must be true. Which one?

Test the ExplanationsWe always test the Null Hypothesis. Assuming that the H0 is true:What is the probability of getting the sample mean (3.00) if the H0 is true and all education majors really have a mean of 2.70? In other words, the difference between the means is due to random chance.If the probability associated with this difference is less than 0.05, reject the null hypothesis.

Test the HypothesesUse the .05 value as a guideline to identify differences that would be rare or extremely unlikely if H0 is true. This alpha value delineates the region of rejection.

Use the Z score formula for single samples

If the probability is less than .05, the calculated or observed Z score will be beyond 1.96 (the critical Z score).

Two-tailed Hypothesis Test:

When = .05, then .025 of the area is distributed on either side of the curve in area (C )The .95 in the middle section represents no significant difference between the population and the sample mean.The cut-off between the middle section and +/- .025 is represented by a Z-value of +/- 1.96.

Chart1

0.0000669151

0.0001762978

0.0004363413

0.001014524

0.0022159242

0.0045467813

0.0087641502

0.0158698259

0.0269954833

0.0431386594

0.0647587978

0.0913245427

0.1209853623

0.1505687161

0.1760326634

0.1933340584

0.1994711402

0.1933340584

0.1760326634

0.1505687161

0.1209853623

0.0913245427

0.0647587978

0.0431386594

0.0269954833

0.0158698259

0.0087641502

0.0045467813

0.0022159242

0.001014524

0.0004363413

0.0001762978

0.0000669151

68.26%

95.44%

99.72%

=NORMDIST(B2,$F$1,$H$1,FALSE).

Normal Curve

Chart2

0.0000669151

0.0001762978

0.0004363413

0.001014524

0.0022159242

0.0045467813

0.0087641502

0.0158698259

0.0269954833

0.0431386594

0.0647587978

0.0913245427

0.1209853623

0.1505687161

0.1760326634

0.1933340584

0.1994711402

0.1933340584

0.1760326634

0.1505687161

0.1209853623

0.0913245427

0.0647587978

0.0431386594

0.0269954833

0.0158698259

0.0087641502

0.0045467813

0.0022159242

0.001014524

0.0004363413

0.0001762978

0.0000669151

95.25%

95.25% of the area lies to the left of (below) a Z score of 1.67


Scores

Frequency

Chart3

0.0000669151

0.0001762978

0.0004363413

0.001014524

0.0022159242

0.0045467813

0.0087641502

0.0158698259

0.0269954833

0.0431386594

0.0647587978

0.0913245427

0.1209853623

0.1505687161

0.1760326634

0.1933340584

0.1994711402

0.1933340584

0.1760326634

0.1505687161

0.1209853623

0.0913245427

0.0647587978

0.0431386594

0.0269954833

0.0158698259

0.0087641502

0.0045467813

0.0022159242

0.001014524

0.0004363413

0.0001762978

0.0000669151

Z= -1.96

c

Z = +1.96

c


Sheet1

=NORMDIST(B2,$F$1,$H$1,FALSE).mean10stdev2

-420.0000669151

-3.752.50.0001762978

-3.530.0004363413

-3.253.50.001014524

-340.0022159242

-2.754.50.0045467813

-2.550.0087641502

-2.255.50.0158698259

-260.0269954833

-1.756.50.0431386594

-1.570.0647587978

-1.257.50.0913245427

-180.1209853623

-0.758.50.1505687161

-0.590.1760326634

-0.259.50.1933340584

0100.1994711402

0.2510.50.1933340584

0.5110.1760326634

0.7511.50.1505687161

1120.1209853623

1.2512.50.0913245427

1.5130.0647587978

1.7513.50.0431386594

2140.0269954833

2.2514.50.0158698259

2.5150.0087641502

2.7515.50.0045467813

3160.0022159242

3.2516.50.001014524

3.5170.0004363413

3.7517.50.0001762978

4180.0000669151

Sheet2

Sheet3

Testing Hypotheses:Using The Five Step ModelMake Assumptions and meet test requirements.State the null hypothesis.Select the sampling distribution and establish the critical region.Compute the test statistic.Make a decision and interpret results.

Step 1: Make Assumptions and Meet Test RequirementsRandom samplingHypothesis testing assumes samples were selected using random sampling. In this case, the sample of 117 cases was randomly selected from all education majors.

Level of Measurement is Interval-RatioGPA is I-R so the mean is an appropriate statistic.

Sampling Distribution is normal in shapeThis is a large sample (n100).

Step 2 State the Null HypothesisH0: = 2.7 (in other words, H0: = )You can also state Ho: No difference between the sample mean and the population parameter(In other words, the sample mean of 3.0 really the same as the population mean of 2.7 the difference is not real but is due to chance.)The sample of 117 comes from a population that has a GPA of 2.7. The difference between 2.7 and 3.0 is trivial and caused by random chance.

Step 2 (cont.) State the Alternate HypothesisH1: 2.7 (or, H0: )Or H1: There is a difference between the sample mean and the population parameterThe sample of 117 comes a population that does not have a GPA of 2.7. In reality, it comes from a different population.The difference between 2.7 and 3.0 reflects an actual difference between education majors and other students.Note that we are testing whether the population the sample comes from is from a different population or is the same as the general student population.

Step 3 Select Sampling Distribution and Establish the Critical RegionSampling Distribution= Z

Alpha () = .05

is the indicator of rare events.

Any difference with a probability less than is rare and will cause us to reject the H0.

Step 3 (cont.) Select Sampling Distribution and Establish the Critical RegionCritical Region begins at Z= 1.96

This is the critical Z score associated with = .05, two-tailed test.

If the obtained Z score falls in the Critical Region, or the region of rejection, then we would reject the H0.

Step 4: Use Formula to Compute the Test Statistic (Z for large samples ( 100)

When the Population is not known,use the following formula:

Test the HypothesesWe can substitute the sample standard deviation S for s (pop. s.d.) and correct for bias by substituting N-1 in the denominator.Substituting the values into the formula, we calculate a Z score of 4.62.

Step 5 Make a Decision and Interpret ResultsThe obtained Z score fell in the Critical Region, so we reject the H0. If the H0 were true, a sample outcome of 3.00 would be unlikely. Therefore, the H0 is false and must be rejected.

Education majors have a GPA that is significantly different from the general student body (Z = 4.62, = .05).*

*Note: Always report significant statistics.

Looking at the curve:(Area C = Critical Region when =.05)

Chart1

0.0000669151

0.0001762978

0.0004363413

0.001014524

0.0022159242

0.0045467813

0.0087641502

0.0158698259

0.0269954833

0.0431386594

0.0647587978

0.0913245427

0.1209853623

0.1505687161

0.1760326634

0.1933340584

0.1994711402

0.1933340584

0.1760326634

0.1505687161

0.1209853623

0.0913245427

0.0647587978

0.0431386594

0.0269954833

0.0158698259

0.0087641502

0.0045467813

0.0022159242

0.001014524

0.0004363413

0.0001762978

0.0000669151

68.26%

95.44%

99.72%


Normal Curve

Chart2

0.0000669151

0.0001762978

0.0004363413

0.001014524

0.0022159242

0.0045467813

0.0087641502

0.0158698259

0.0269954833

0.0431386594

0.0647587978

0.0913245427

0.1209853623

0.1505687161

0.1760326634

0.1933340584

0.1994711402

0.1933340584

0.1760326634

0.1505687161

0.1209853623

0.0913245427

0.0647587978

0.0431386594

0.0269954833

0.0158698259

0.0087641502

0.0045467813

0.0022159242

0.001014524

0.0004363413

0.0001762978

0.0000669151

95.25%



Scores

Frequency

Chart3

0.0000669151

0.0001762978

0.0004363413

0.001014524

0.0022159242

0.0045467813

0.0087641502

0.0158698259

0.0269954833

0.0431386594

0.0647587978

0.0913245427

0.1209853623

0.1505687161

0.1760326634

0.1933340584

0.1994711402

0.1933340584

0.1760326634

0.1505687161

0.1209853623

0.0913245427

0.0647587978

0.0431386594

0.0269954833

0.0158698259

0.0087641502

0.0045467813

0.0022159242

0.001014524

0.0004363413

0.0001762978

0.0000669151

Z= -1.96

c

Z = +1.96

c z= +4.62 I


Sheet1


-420.0000669151

-3.752.50.0001762978

-3.530.0004363413

-3.253.50.001014524

-340.0022159242

-2.754.50.0045467813

-2.550.0087641502

-2.255.50.0158698259

-260.0269954833

-1.756.50.0431386594

-1.570.0647587978

-1.257.50.0913245427

-180.1209853623

-0.758.50.1505687161

-0.590.1760326634

-0.259.50.1933340584

0100.1994711402

0.2510.50.1933340584

0.5110.1760326634

0.7511.50.1505687161

1120.1209853623

1.2512.50.0913245427

1.5130.0647587978

1.7513.50.0431386594

2140.0269954833

2.2514.50.0158698259

2.5150.0087641502

2.7515.50.0045467813

3160.0022159242

3.2516.50.001014524

3.5170.0004363413

3.7517.50.0001762978

4180.0000669151

Sheet2

Sheet3

Summary:

The GPA of education majors is significantly different from the GPA of the general student body.

In hypothesis testing, we try to identify statistically significant differences that did not occur by random chance.

In this example, the difference between the parameter 2.70 and the statistic 3.00 was large and unlikely (p < .05) to have occurred by random chance.

Summary (cont.)We rejected the H0 and concluded that the difference was significant.

It is very likely that Education majors have GPAs higher than the general student body

Rule of Thumb:If the test statistic is in the Critical Region ( =.05, beyond 1.96):Reject the H0. The difference is significant.

If the test statistic is not in the Critical Region (at =.05, between +1.96 and -1.96):Fail to reject the H0. The difference is not significant.

Using the Students t Distribution for Small Samples (One Sample T-Test)When the sample size is small (approximately < 30) then the Students t distribution should be usedThe test statistic is known as t.The curve of the t distribution is flatter than that of the Z distribution but as the sample size increases, the t-curve starts to resemble the Z-curve

Degrees of FreedomThe curve of the t distribution varies with sample size (the smaller the size, the flatter the curve)In using the t-table, we use degrees of freedom based on the sample size.For a one-sample test, df = n 1.When looking at the table, find the t-value for the appropriate df = n-1. This will be the cutoff point for your critical region.

Formula for one sample t-test:(Note that it is identical to z-test, but uses a different distribution)

Example A random sample of 26 sociology graduates scored 458 on the GRE advanced sociology test with a standard deviation of 20. Is this significantly different from the population average ( = 440)?

Solution (using five step model)Step 1: Make Assumptions and Meet Test Requirements:

1. Random sample2. Level of measurement is interval-ratio3. The sample is small (

Solution (cont.)Step 2: State the null and alternate hypotheses.

H0: = 440 (or H0: = )

H1: 440 (H1: = )

Solution (cont.)

Step 3: Select Sampling Distribution and Establish the Critical Region

Small sample, I-R level, so use t distribution.Alpha () = .05Degrees of Freedom = n-1 = 26-1 = 25Critical t = 2.060

Solution (cont.)Step 4: Use Formula to Compute the Test Statistic

Looking at the curve for the t distribution Alpha () = .05

Chart1

0.0000669151

0.0001762978

0.0004363413

0.001014524

0.0022159242

0.0045467813

0.0087641502

0.0158698259

0.0269954833

0.0431386594

0.0647587978

0.0913245427

0.1209853623

0.1505687161

0.1760326634

0.1933340584

0.1994711402

0.1933340584

0.1760326634

0.1505687161

0.1209853623

0.0913245427

0.0647587978

0.0431386594

0.0269954833

0.0158698259

0.0087641502

0.0045467813

0.0022159242

0.001014524

0.0004363413

0.0001762978

0.0000669151

68.26%

95.44%

99.72%


Normal Curve

Chart2

0.0000669151

0.0001762978

0.0004363413

0.001014524

0.0022159242

0.0045467813

0.0087641502

0.0158698259

0.0269954833

0.0431386594

0.0647587978

0.0913245427

0.1209853623

0.1505687161

0.1760326634

0.1933340584

0.1994711402

0.1933340584

0.1760326634

0.1505687161

0.1209853623

0.0913245427

0.0647587978

0.0431386594

0.0269954833

0.0158698259

0.0087641502

0.0045467813

0.0022159242

0.001014524

0.0004363413

0.0001762978

0.0000669151

95.25%



Scores

Frequency

Chart3

0.0000669151

0.0001762978

0.0004363413

0.001014524

0.0022159242

0.0045467813

0.0087641502

0.0158698259

0.0269954833

0.0431386594

0.0647587978

0.0913245427

0.1209853623

0.1505687161

0.1760326634

0.1933340584

0.1994711402

0.1933340584

0.1760326634

0.1505687161

0.1209853623

0.0913245427

0.0647587978

0.0431386594

0.0269954833

0.0158698259

0.0087641502

0.0045467813

0.0022159242

0.001014524

0.0004363413

0.0001762978

0.0000669151

t= -2.060

c

t = +2.060

c t= +4.50 I


Sheet1


-420.0000669151

-3.752.50.0001762978

-3.530.0004363413

-3.253.50.001014524

-340.0022159242

-2.754.50.0045467813

-2.550.0087641502

-2.255.50.0158698259

-260.0269954833

-1.756.50.0431386594

-1.570.0647587978

-1.257.50.0913245427

-180.1209853623

-0.758.50.1505687161

-0.590.1760326634

-0.259.50.1933340584

0100.1994711402

0.2510.50.1933340584

0.5110.1760326634

0.7511.50.1505687161

1120.1209853623

1.2512.50.0913245427

1.5130.0647587978

1.7513.50.0431386594

2140.0269954833

2.2514.50.0158698259

2.5150.0087641502

2.7515.50.0045467813

3160.0022159242

3.2516.50.001014524

3.5170.0004363413

3.7517.50.0001762978

4180.0000669151

Sheet2

Sheet3

Step 5 Make a Decision and Interpret ResultsThe obtained t score fell in the Critical Region, so we reject the H0 (t (obtained) > t (critical)If the H0 were true, a sample outcome of 458 would be unlikely. Therefore, the H0 is false and must be rejected.

Sociology graduates have a GRE score that is significantly different from the general student body (t = 4.5, df = 25, = .05).

Testing Sample Proportions:When your variable is at the nominal (or ordinal) level the one sample z-test for proportions should be used.If the data are in % format, convert to a proportion first.The method is the same as the one sample Z-test for means (see above)

Formula for Proportions:

Note: Ps is the sample proportionPu is the population proportion

ExampleIn a recent statewide (or provincial) election, 55% of voters rejected lotteries. A random sample of 150 urban (or rural communities) precincts showed that 49% of voters rejected lotteries. Is the difference significant?

Use the formula for proportions and 5 step method to solve

Solution:Step 1: Random sampleL.O.M. is nominalThe sample is largeStep 2:H0: Pu = .55 (convert % to proportion)(Note you can also say H0: Ps = Pu )H1: Pu .55 (H1: Ps Pu Step 3:The sample is large, use Z distributionAlpha () = .05Critical Z = 1.96

Solution (cont.)Step 4

Step 5Z (obtained) < Z (critical)Fail to reject Ho. There is no significant difference between the state population and the precincts.

Main Considerations in Hypothesis Testing:Sample sizeUse Z for large samples, t for small (

Two-tailed vs. One-tailed TestsIn a two-tailed test, the direction of the difference is not predicted.A two-tailed test splits the critical region equally on both sides of the curve.In a one-tailed test, the researcher predicts the direction (i.e. greater or less than) of the difference.All of the critical region is placed on the side of the curve in the direction of the prediction.

The Curve for Two- vs. One-tailed Tests at = .05:

Two-tailed test:is there a significant difference?

One-tailed tests:is the sample meangreater than or Pu?

is the sample meanless than or Pu?

Type I and Type II ErrorsType I, or alpha error: Rejecting a true null hypothesisType II, or beta error:Failing to reject a false null hypothesis

ppt hypothesis testing

Documents