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Psychology

PSYC 1010

Benjamin Giguere

Winter

Description

Chapter 6
Probability Tests and Z-Scores
1) avg. employee in Canada spends u=22 min commuting to work each day.
Assume distribution of commute times is normal with a variance of o2=49
minutes. For each of the following questions, make sure you:
i) sketch the distribution
ii) state the formula you used
You will want to use the z-score with an x in it rather than z. find the
probability using the z-table.
a. what proportion of Canadian employees spend less than 17 minutes a
day commuting?
X=17, u=22
First step: get sigma
Square root of the variance
Sqrt of sigma2=sqrt49=7.00
You don’t need to label mean for full marks but it helps
Upside down parabola with 22 at the bottom of line in middle
Change raw score into z score
Z=x-u/sigma= (z score of .71) were looking for proportion of the tail
so we go to the right column and find out it’s 23.89%
We take that proportion and probability=0.2389
Proportions show all decimals, when looking something up in the
table write all values.
Shes not worried about the probability of x being less than 17
You would lose a half mark if you leave as probability and don’t turn it
into proportion.
You just take the z value you get and look it up in the table. b. what is the probability of randomly selecting an employee who
spends less than 33 minutes commuting each day?
U=22, sigma=7
Now that you have sigma it is now a 3 mark question.
Same sketch, mean still 22
Looking for area below 33, left hand body of the distribution.
Here we know that the left hand side is 50% as well as the right
Now we can calculate the z-score for 33
33-22/7=1.57
proportion from table is 44.18
we add that to the 50% on the left to get overall proportion then turn
it into a decimal!
P=/5
Area between 22 and 33 is .4418
Add together to get .9418
1 mark for probability, half for z-score and half for formula.
2) u=50, t=5
what z-score is at the 60 percentile? Drawing a sketch is helpful and look at
our table as well.
Remember that the 60% is the percentage below (60% of scores therefore
fall below)
Above is 40%
Left of mean is 50%
Right of mean is 10%
What z-score puts 10% in the mean (same that puts 40% in the tail) The closest thing in the z-table is
0.25 for the 60 percentile
we cant look up 60% in the body because there’s no table for that. This is
why we do the z score for 40.13% in tail because that’s the closest.
Change question now: what mean is at the 60% for a sample size of .25
You want to use a formula that turns a z-score into a sample mean
M=um+zisgmaM
sigmaM=sigmaoversqrtM=5/sqrt25=1.00
50+(0.25)(1)
=50.25
Sample mean at the 60% percentile
Z=0.25
Figure out what the question is asking and then figure out if you’re dealing
with sample means or scores.
M,x can be turned into z-scores=probability and percentages
3) researcher measures perceived stress using a standardized measure. Scores
on the measure of perceived stress are normally distributed with u=100 or
0=15
For each of the following questions make sure you sketch distribution, shade
area of interest, then state any formulas used.
a. if the researcher takes a sample of n=4 what is the prob. That the sample
mean will be greater than 110?
Always start with sketch
Find z-score with the sample mean.
We first need the standard error
Sigmam=sigma/sqrtm=15/sqrt4=7.50
Now: z=M-um/jm=110-100/7.5=1.33 Positive z-score makes sense because were looking for upper right tail.
To find proportion in tail: .0918 prob. When you look it up
Now we move onto hypothesis testing after these probability questions.
Single sample z tests – examples and template
A researcher decides to investigate whether the short term memory of ‘night
owls’ differs from that of the general population of students.
A random sample of n=25 university students who prefer to study late at
night is obtained and each student is given a standardized memory test. The
avg score for the sample is M=40. For the general population of university
students, the distribution of memory test scores is normal with a mean of u-
35 and a standard deviation of o=10
Do the night owls have significantly different memory scores from the
general population? Use o=0.5
*on the exam, start from hypothesis and work to n (steps 2-6 in textbook)
pops. Are night owls + general pop.
-one sample and o known so it’s a single sample z-test
scale data (memory scores)
random sample and normally distributed comparison distribution
H0: the short term memory of night owls does not differ from that of the
general population
(1)H1: The short-term memory of “night owls” differs from that of the
general pop.
H0: u=35 (1)
H1: u/=35 (1)
Ou=o/sqrtn=10/sqrt25=2.00
1) for std error formula and 1) for correct answer
a=0.5, two-tailed (1)
z-crit= +/- 1.96 (1)
if z < - 1.96 or z > +1.96, reject Ho. (1)
if the z-score you calculated displays this, that’s when you reject the null z=m-um/om=40-35/2=2.50
1) for z-formula 1) for correct answer
Reject H0 and conclude that night owls have significantly better short-term memory
than the general pop. (z=2.50, p < .05).. (4)
1) reject ho (0.5) – significantly (1) direction – better (0.5) – name both
variables in conclusion (1) – APA style
less than 5% chance that we would get this result if the null was true.
Chapter 6 Textbook Notes
-The normal curve is bell-shaped and symmetrical.
-Patterns of error are similar on both sides of the curve, which is what
makes it symmetrical.
-This curve describes the distributions of many characteristics and
measures also how much they vary.
-If the population is normally distributed, the sample size emits a curve
more similar to the normal curve as the sample size increases.
-Round scores to facilitate the creation of a histogram.
-The process of standardization converts individual scores into
standard scores for which we know the percentiles.
-Standardization does this by converting individual scores from
different normal distributions into a shared normal distribution with a
known mean, standard deviation, and percentiles.
-A z score is the number of standard deviations that a particular score is
from the mean.
-A z score is part of its own distribution (a z distribution) just like height
would be (a distribution of heights).
-A z distribution always has a mean of 0 and a standard deviation of 1. -Working with easy numbers, we don’t need a formula to calculate.
-Say the mean is 70 with an SD of 10. Raw score of 50= z score of -2, etc.
-There are 2 steps in order to calculate a particular z score.
1) Determine the distance of a particular score from the population
mean.
X – u (the score is X, population mean is u)
2) Express the distance in terms of standard deviation.
Z= (X-u)
O
Let’s apply this formula to an example:
1) Subtract the mean of the population from your score (ex. 70-64.886)
2) Divide by the SD of the population = (70-64,886) = 1.25
4.086
3) Therefore, you are 1.25 SD above the mean.
-Always check to make sure your SD makes sense in correlation to the question.
Say your height is 62. You would subtract that the population mean fro your score
and divide it by the standard deviation of 4.086. As a result, you get a z score of -
0.71. You would have a SD below the mean in this case.
Say your height is 64.886. After performing your calculations, you should be getting
a z score of 0
Say someone is exactly 4.086 inches above the mean. 64,886+4.086= 68.972.
Perform the z score formula with these numbers and your SD should be 1.
Turning z scores into raw scores simply involves plugging in all of the numbers
except for x and then solving algebraically.
Ex. 1.79(SD)= (X-64.886)
4.086
The following formula helps us solve for X: X=z(o)+u
Step 1: multiply the z score by the SD of the pop. Step 2: Add the mean of the population to this product.
If your z score is negative, you would do the same process. Just don’t forget
the negative sign or it’ll fuck up all of your calculations.
The normal curve represents all 100 percentiles, with the 50 one being in
the middle.
If the percentile is to the right of the mean of the normal curve, your score is
above the 50 percentile. Vice

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