Hypothesis Testing - Basic Statistics for Sociology - Lecture Slides, Slides of Statistics for Psychologists

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Chapter 8

Hypothesis Testing:

One Sample Cases

Outline:

• The logic of hypothesis testing
• The Five-Step Model
• Hypothesis testing for single sample means (z test and t test)
• Testing sample proportions
• One- vs. Two- tailed tests

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:

μ = 2.

= 3.

s = 0. n = 117

X

Two Possibilities:

1. The sample mean (3.00) is the same as the pop. mean (2.70).

• The difference is trivial and caused by random chance.

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

The Null and Alternative Hypotheses:

1. Null Hypothesis (H 0 ) - The difference is caused by random chance. - The H 0 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. 2. Alternative hypothesis (H 1 ) - “The difference is real”. - (H 1 ) always contradicts the H (^) 0.

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

Test the Hypotheses

• Use the .05 value as a guideline to identify differences that would be rare or extremely unlikely if H 0 is true. This “alpha” value delineates the “region of rejection.”
• Use the Z score formula for single samples and Appendix A to determine the probability of getting the observed difference.
• 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 α = .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.

Z= -1. c

Z = +1. c

Step 1: Make Assumptions and Meet

Test Requirements

• Random sampling
  • Hypothesis 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-Ratio
  • GPA is I-R so the mean is an appropriate statistic.
• Sampling Distribution is normal in shape
  • This is a “large” sample (n≥100).

Step 2 State the Null Hypothesis

• H 0 : μ = 2.7 (in other words, H 0 : = μ)
  • You can also state H (^) o : 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 3 Select Sampling Distribution and Establish the Critical Region

• Sampling Distribution= Z
  • Alpha (α) =.
  • α is the indicator of “rare” events.
  • Any difference with a probability less than α is rare and will cause us to reject the H 0.

Step 3 (cont.) Select Sampling Distribution and Establish the Critical Region

• Critical Region begins at Z= ± 1.
  • This is the critical Z score associated with α = .05, two-tailed test.
  • If the obtained Z score falls in the Critical Region, or “the region of rejection,” then we would reject the H 0.

When the Population σ is not known,

use the following formula:

s n

Z

Test the Hypotheses

• We can substitute the sample standard deviation S for σ (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.

4. 62

117 1

. 7

3. 0 2. 7 =

−

− Z =