# Project #3128 - Psychology Statistics

INDIVIDUAL ASSIGNMENT: WEEK THREE

The focus of this assignment is on the Five-Step Hypothesis Testing Process.  In hypothesis testing we are always making comparisons.  We’ll be looking at two different kinds of comparisons.  Both of these comparisons use z tests.  In real life, this type of research is very rare but it’s a good starting place for us to learn some basic concepts.  This assignment has 3 parts to it.

In Part One, we’re using the Five-Step Hypothesis Testing Process to compare an individual to a population.  How would we know if an individual is like the general population on some specific variable or not?  In the scenarios we’re using, we know that population mean and standard deviation so we’re just comparing this one person’s score to the population’s mean as the Comparison Distribution.

In Part Two, we’re using the Five-Step Hypothesis Testing Process to compare a sample of people to a population.  Is this sample of people different on some variable from the general population or not?  Since we’re comparing a sample of people to a population, we have to use that sample’s mean and we use the Distribution of Means as a Comparison Distribution.

In Part Three, we’re looking at a concept called Effect Size.

For Parts One and Two, I’m going to first give you an example and walk you through the process and then I’m going to give you one to do on your own.  You will need the Major Formulas Handout that I told you to print out the first day of class.  You will need to reference our textbook.

Part One

Lottery Winner Example

(from text)

A study is done in which a randomly selected person is given \$10 million. This person's happiness, measured 6 months later, is 80. It is known in advance that happiness in the general population is normally distributed with m = 70 and s = 10. Could this result have occurred by chance?

Œ   Restate question as a research hypothesis and a null hypothesis about the populations.

Research hypothesis:     There will be a statistically significant difference between the happiness ratings on a self-report scale of this lottery winner six months after winning the lottery when compared to the happiness ratings of the general population. (μ1 > μ2)

Null hypothesis:        There will not be a statistically significant difference between the happiness ratings on a self-report scale of this lottery winner six months after winning the lottery when compared to the happiness ratings of the general population. (μ1μ2)

You’re always going to write the research and null hypothesis in terms that a) operationally define your variable of study (how you’re going to measure it) and b) indicate that the research hypothesis will find a difference and the null hypothesis will not find a difference.

?   Determine the characteristics of the comparison distribution.

Known normal distribution, with m = 70 and s = 10.

When this is known, it will be given to you in the question.  In later weeks, we will be working with scenarios where this is unknown but we’ll deal with that later.  For right now, you can get this information from the question.  If you’re not sure what those symbols mean, there is a Glossary of Symbols that will be helpful in the textbook.  For your assignments, it’s OK to either use the symbols or write it out in narrative.

Ž   Determine the cutoff sample score (critical value) on the comparison distribution at which the null hypothesis should be rejected.

For a .05 probability (top 5% of comparison distribution), Z needed is +1.64.

There is a table on p. 121 of our textbook.  It will have the “Conventional Levels of Significance” which you can use for this assignment.  In the future, this step will require you to use the tables at the end of our textbook.  The question will usually tell you which significance level to use and whether it’s a one-tail or two-tailed test.  In this case, we’re using .05 as our significance level and it’s a one-tailed test.

?   Determine your sample's score on the comparison distribution.

The person's score is 80 on the happiness test. (X = 80)

Z = (XM) / SD  à  Z = (80 – 70) / 10 = 10 / 10 = 1. (Z = +1.00)

This formula is on the Major Formulas Handout.  You’re just taking that one person’s individual score and turning it into a z score so you can compare it to your cutoff score.

?   Decide whether to reject the null hypothesis.

Score at Step 4 (Z = +1.00) is not more extreme than score at Step 3 (Z = +1.64). Therefore, fail to reject the null hypothesis.

Here you will either reject the null hypothesis or fail to reject the null hypothesis by comparing Step 3 and Step 4.  If Step 4 is larger than Step 3, you can reject the null hypothesis.  If Step 3 is not larger than Step 4, you fail to reject the null hypothesis.

OK.  Now your turn.  I’ll give you the scenario.  You work through all five steps.   Show your work.

Polluted Region Example

(fictional data)

Does pollution affect amount of sleep? A person living in a polluted region was randomly selected; his sleep the night before was 10.2 hours. In the general population, amount of sleep is normally distributed with μ = 8 and σ = 1.   Use a two-tailed test at the .05 significance level.

Part Two

Personality Attractiveness Example

(from text)

A social psychologist is interested in whether being told a person has positive personality qualities increases ratings of the physical attractiveness of that person. The psychologist asks 64 randomly selected students to rate the attractiveness of a particular person in a photograph. Prior to rating the attractiveness of the person, each student is told that the person has positive personality qualities (kindness, warmth, a sense of humor, and intelligence). On a scale of 0 (the lowest possible attractiveness) to 400 (the highest possible attractiveness), the mean attractiveness rating given by the 64 students is 220. From previous research, the psychologist knows that the attractiveness ratings of the person in the photograph (when no mention is made of the person’s positive personality qualities) have a mean of 200 and a standard deviation of 48.  Use a one-tailed test at the .05 level of significance.

Œ   Restate question as a research hypothesis and a null hypothesis about the populations.

Research hypothesis:     There will be a statistically significant difference between the attractiveness ratings of students who are told the person has positive personality qualities when compared to the general population who are told nothing about the person’s personality qualities.  1 > μ2)

Null hypothesis:        There will not be a statistically significant difference between the attractiveness ratings of students who are told the person has positive personality qualities when compared to the general population who are told nothing about the person’s personality qualities. (μ1 ≤ μ2)

See explanation above.  It’s the same.

?   Determine the characteristics of the comparison distribution.

Sample: N = 64, M = 220

Population 2: µ = 200, σ= 48, shape = unknown

μM = μ                   à μM = 200

σ2M = σ2 / N          à σ2M = 482 / 64 = 2304 / 64 = 36

σM = √σ2M                         à σM = √36 = 6

This step is different from above.  The question is providing you with the information in

blue; however, you now have to take that information and calculate your Distribution of

Means.

Ž   Determine the cutoff sample score (or Z score) on the comparison distribution at which the null hypothesis should be rejected.

For a .05 probability (top 5% of comparison distribution), Z needed is +1.64.

See explanation above.

?   Determine your sample's score on the comparison distribution.

Sample: N = 64, M = 220

Z = (MμM) / σM  à  Z = (220 – 200) / 6 = 20 / 6 = 3.33. (Z = +3.33)

You’ll need your numbers from Step 2 to calculate this z score.

?   Decide whether to reject the null hypothesis.

Score at Step 4 (Z = +3.33) is more extreme than score at Step 3 (Z = +1.64). Therefore, reject null hypothesis.

See explanation above.

OK.  Now your turn.  I’ll give you the scenario.  You work through all five steps.   Show your work.

Stress Reduction Example

(fictional data)

A health psychologist wants to test the effectiveness of a new stress-reduction method. In the general population, stress level is normally distributed with m = 40 and s = 10. A randomly selected group of 20 individuals participated in this experiment with a m = 25.  Use a .01 level of significance and a one-tailed test.

Part Three

Explain in narrative form the concept of Effect Size and how to calculate it.

What is the effect size for the personality-attractiveness example discussed previously?

 Subject Science Due By (Pacific Time) 03/11/2013 02:00 pm
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