Stats exam 3
standard error
Standard deviation of the distribution of sample means is also called STANDARD ERROR standard error = the average amount that you can expect a sample (of size n) to deviate from the population mean
What do you need to know when using a z-test:
The distribution of sample means is normal The population standard deviation σ is known
Hypothesis testing procedures assumes what?
assumes we are trying to reject the null, not trying to prove the alternative hypothesis... because it is easier to show something is not true, than prove it is
Alpha level is used for what?
based on the probabilities of making a certain criterion. How big of a difference is big enough to SAFELY conclude that there is an effect of the experiment In psychology 'a' is usually set at 0.05 (this indicates a 5% chance of a Type 1 error...not too horrible, so we can accept that risk)
why is udbar always 0?
because that is what we are predicting tin the null
Standard error is influenced by
by the variability of the population and the sample size Variability: the bigger the variability of the population, the more variability you will have in the sample means
For z-test when you are rejecting the null you should always ask
can we be confident that we didn't make a type 1 error? Is our sample z probability less than our standard of .05?
Matched subject study
collect data from 2 groups, then you form pairs based on - Naturally occurs: twins, spouses - Researcher matches/ creates pairs: on traits they want to control (IQ, introversion etc.)
Type 2 error (b)
conclude e that there is not an effect when there actually is You are being conservative
What are the t-test varieties?
o 1- sample t-test o Related sample t-test o Independent sample t-test
What is the formula for independent t-test: standard error
sXbar A - Xbar B = square root Sp2 (pool variance)/ nA Sp2 (pool variance)/ nA
What does fail to reject the null mean?
to reject = no support for our theory, can not Not a big enough difference that shows that it works, rather the differences could be due to chance.
estimation with sample does what
using sample statistics to estimate the population parameters
What is the key for one-sample t-test
when we do not know the population standard deviation... so we can not use a z-test, we have to use a t-test
Properties of the Distribution of Sample Means: shape
The same of the distribution of sample means tends to be a normal distribution. When n is large (30>) the distribution of sample means is almost perfectly normal
What is the formula for one-sample t-test:
Xbar (sample mean) - comparison distribution mean (population)/ Sxbar (standard error - how much difference we expect on average)
Mean of the distribution of sampling means
μM) = the mean of population
The standard deviation of sample means (or standard error) formula
σ x= σ/sqroot n
What are the assumptions of independent t-test: Leven's test of equality of variance, tests this assumption steps
(1) Sig-value tests equal group variance, and we compare the sig value to the alpha level of .05. Null: variance for group a = variance for group b (2) If levene's test is NOT significant (meaning the sig value is greater than .05). FAIL TO REJECT! Then this means the variance are equal, so you would stay in the upper row to find t-test p-value If levene's test IS significant (.05 is greater than sig) then we would reject the null and be penalized, therefore, going to the second row and loosing degrees of freedom.
How to compute t-statistic for independent
(1) find pool variance (2) find standard error (3) find t-test
What is the formula for independent t-test
(Xbar A - Xbar B) - (population mean A - population mean B 0) / (sXbar A -Xbar B) --> difference between A and B (expected by chance, random error)
Related sample t-test (or paired sample t-test in SPSS): define and what is it looking at?
- Related sample t-test = compare 2 sample means from related with an unknown population standard deviation - Considers the differences between scores from a related pair of subjects. Because the two scores for each pair are related, the differences are based on differences between each individual or matched pair
What are the assumptions of independent t-test: Equality of variance:
- This is a new assumption relevant only for independent t-test. - Because we have tow ppol the 2 samples' variance, we need to make sure they are at least relatively similar or equal. - The test assumes that they are and we will be penalized if they are not actually equal
Z-score vs distribution of sample mean
- Z-scores: looked at a single score, but instead in z sample we are looking at the samples scores that were randomly selected from the population --> Sample = sample consist of scores that were randomly selected from population - Sampling distribution: is the huge set of possible samples forms a simple, orderly and predictable pattern ---> The distribution of sample means = the collection of sample means for all the possible random samples of a particular size (n) that can be obtained from a population
One-tail vs two tail: z-test
1 tailed test: because you have one direction you will have 1 'cuttoff point' with 1 rejection region --> if specifies increase, then .05 would be on the positive side and everything lower would reject the null and everything higher than .05 would fail to reject the null 2 tailed tests: is whether it has an effect or not, so there are 1 'cuttoff' points to form 2 rejection regions --> has two rejection regions and anything between the 2 rejection regions would fail to reject
What are the 4 possible conclusions that can be made
2 ways of making mistake - Type 1 Null is correct (conclude that there is an effect), but you reject the null - Type 2: Null is wrong, fail to reject the null 2 chances of being correct - Null is correct, and you fail to reject the null (cannot support the alternative) - Null is wrong, so you reject the null (support alternative)
One-sample T-test
= tests whether our sample mean differs significantly from a known population mean with an UNKNOWN STANDARD DEVIATION OR VARIANCE
Type 1 (significance level) concludes what? How do we minimize this?
Conclude that there is an effect, but there really is not - False positive - How to minimize this: Try to minimize this by picking a low level for alpha (.05 AND .01 are most common) 5% or 1% chance of getting type 1 - WANT TO BE AVOIDED MOST IN SOCIAL SCIENCES - minimize chances of making type 1 error
Two tailed test
EX: our theory is that the treatment has an effect on memory - No direction specified - Use: if you have conflicting results in other research NULL: µ treatment = µ no treatment ALTERNATIVE: µ treatment ≠ no treatment
What are the assumptions of independent t-test
Equality of variance: Leven's test of equality of variance, tests this assumption
What does the one-sample z-test ask:
Find a sample mean and ask what is the probability that my sample mean is the same as the comparison (population) mean - Some differences is expected, given sampling error
In step 4 for related sample t-test how do you solve for numerator?
Given information on pre-test and post and we have to find difference between the pre-test and post then we add up those differences and divide by the number of different scores then get d-bar
Because we are not given information on the population, what do we have to do for independent test
Groups may have different sample sizes and we still don't have information about the population variance or standard deviation, so we must estimate it from the samples Because we have two samples, we have to find pooled variance (s2p)
Hypothesis for independent sample t-test:
Hypothesis will be different because the situation is different. You are now making hypotheses about two different populations, not just comparing treatment group to a known population comparison group/control group You want to compare two groups, so you are hypotheses is about population A (men) and population B (women) and how they are DIFFERENT FROM ONE ANOTHER
Standard deviation of the distribution of sample means
IS SAME AS STANDARD ERRROR = is the average amount you would expect a sample to deviate from the population mean just from random chance
How to reject null for 1 sample t-test
If our sample statistic (sample t) is beyond cutoffs which is set by the critical t-value... then we would reject the null If too extreme or above FAIL TO REJECT: if test statistic (sample t) is in middle of distribution
Independent sample t-test:
Independent sample t-test = experiment that uses a separate independent sample for each treatment condition (or each population) 2 independent samples, one score per subject. We will compare the group means for two independent groups
t-test negative positive
Like the z table these are all positive t's but you can make them negative if using lower tail how do you know if you are using lower tail (alternative) • If 2-tailed then have two rejection zones and would be + -
We want to determine if management training impacts stress levels. We measure stress for 4 pairs of leaders - each pair consists of 2 identical twins who both happen to be leaders in the same organization. We randomly assign 1 twin from each pair to receive training and the other twin from each pair does not. We measure their stress levels once. What is this type of test, what would the null and alternative be?
MATCHED - Look at the two means that are systemically choosen to be in such groups based on their match or twin (no training vs training measures of stress) and examine the averages across pairs 2 tailed test: • Null: stress of trained group is equal to the stress of the matched no train group (uD = 0) • Alternative: Stress of the trained group is not equal to stress of the matched not training group (uD not equal to 0)
What is sample distribution
Need to find our 'comparison distribution of sample means' - this is a distribution of means based on repeatedly creating a sample of size 'n" from our population - it will look like a normal distribution and basically show us which means are most common/most likely and which sample means are unlikely or less likely - Helps us decide on whether to reject or fail to reject the nul
null vs alternative hypothesis
Null = predicts that the independent variable (treatment) has no effect on the dependent variable for the population (or differences between treatment and groups) --> This is the one we test and hope to reject Alternative = predicts that the independent variable (treatment) WILL HAVE AN effect on the dependent variable for the population --> If we REJECT the null, we are left with support for this
Independent sample t-test: one tailed test hypotheses to test question of whether fatigue increases errors on mental alertness tasks.
Null: group 2 will be greater than or equal to the number of errors compared to group 1 Alternative: group 1, in the fatigue group, will have an increase in errors on mental alertness compared to group 2
Independent sample t-test: two tailed test hypotheses for our interest in how tall men and women are
Null: men and women are the same height (A = B) no difference Alternative: men and women are different mean heights (A is not = to B) Difference
Once we check level's test to determine _______________ ... what do we do next?
Once we check levene's test to determine whether we have equal variance or unequal variance in the 2 groups, next we go on to examine the main t-test hypothesis of equality of group means Look at sig (2-tailed) to tell if you reject or fail to reject the null
one sample t-test ___ tail hypothesis Suppose that your psychology professor, Dr. I. D. Ego, gives a 20 point true-false quiz to 9 students and wants to know if they were different from groups in the past who have tended to have an average of 9.0. Their scores from the current group were: 6, 7, 7, 8, 8, 8, 9, 9, 10. Did the current group perform differently from those in the past? We'll assume a significance level of α = 0.05.
One sample t-test 2 tailed: null: the population mean is equal to 9.0 Alternative: the population mean is not equal to 9.0
one sample t-test ___ tail hypothesis: Suppose that your physics professor, Dr. M. C. Squared, gives a 20 point true-false quiz to 9 students and wants to know if they did worse than guessing (note: if students were guessing, they'd be expected to get half right given that it's a True-False 2 option test, so 10 out of 20 correct).The students' scores were: 6, 7, 7, 8, 8, 8, 9, 9, 10. We'll assume a significance level of α = 0.05.
One sample t-test One tail Null: no difference between the treatment and population μT = 94 Alternative: treatment will be lower (reduce errors) than the control population μT < 94
One-tailed
One tail = EX: our theory is that the treatment should result in FEWER errors - Direction specified - About population so you would use µ notation Example: NULL: treatment group mean greater than (or less) or = to no treatment group mean ALTERNATIVE: mean treatment < (or >) no treatment
related t-test __ tail hypothesis: A major university would like to improve its tarnished image following a large on-campus scandal. Its marketing department develops a short television commercial and tests it on a sample of n = 7 subjects. People's attitudes about the university are measured with a short questionnaire, both before and after viewing the commercial. (Go to the lab instructions file to view the data and formulas needed)
RElated sample ONE TAIL Null: the post test scores will be same or less than the pretest Alternative: the post test scores will be greater than the pretest
related t-test __ tail hypothesis: An instructor asks her statistics class, on the first day of classes, to rate how much they like statistics, on a scale of 1 to 10 ('1'= hate it, '10'= love it). Then, at the end of the semester, the instructor asks the same students the same question. The instructor wants to know if taking the stats course had an impact on the students' feelings about statistic
Related sample 2 tailed because no direction H0: μD = 0 (there is no difference in pre-test and post-test scores) HA: μD ≠ 0 (there is a significant difference in pre-test and post-test scores)
We want to determine if management training impacts stress levels. We measure stress for 4 leaders both before and after the training. We expect a different level of stress pre- and post-training (not sure if more or less stress). What is this type of test, what would the null and alternative be?
Repeated measure - Focus on the difference, compare pair-wise differences between pre and post - Average the pairs differences: should be 0 if no effect of treatment... so we do not want uD to be 0 to reject the null 2 tailed test: • Null: no significant difference in stress levels post vs pre (uD = 0) • Alternative: stress at post training will differ significantly from pre-training level (or uD not equal 0)
EX: We have a new treatment that we hypothesize will reduce memory errors. We collect a sample of 16 patients that get the new treatment next, they take a memory test. Our treatment sample averages 55 errors. The control population averages μ control population = 60 errors. Test whether the treatment reduced errors (using α = 0.05). What test would you use, and solve using all steps
STEP 1: state hypothesis: 1 tail Null: The treatment mean = control population mean (μ treatment = μpop) Alternative: the new treatment will reduce memory errors (treatment < 60) STEP 2: set decision criteria .05 would be on the negative side of distribution STEP 3: collect data n= 16 & Xbar = 55 & POP MEAN = 60 (1) start with denominator START WITH DENOMINATOR: σ xbar = σ/ sqroot n (8/SQROOT 16) = (2) Find our sample z-statistic: 55 (sample) - 60 (population) / (8/SQROOT 16) = -2.5 STEP 4: take this number to the normal distribution table and find probability --> ALWAYS LOOK AT COLUMN C (because want to know extreme score) Probability for sample = .0062 VS .05 --> we reject the null and accept the alternative so the new treatment will reduce memory errors
Difference between individual-level z formula and sample-level z formula:
Sample level z formula uses the distribution sample means and examines the sample rather than an individual scores
the larger your sample size (n)... the more _______
Size: the larger your sample size (n), the more accurately the sample represents the population
Common themes for one-sample t-test (what is unknown/ known)
Standard deviation is unknown so we can't use z-test (because σ must be known) - It is common that you don't have access to population Will estimate σ by using the sample SD (S) - The sample should be relatively close to population
Dr. M develops a new treatment for patients with a memory disorder. He isn't sure what impact, if any, it will have. To test it he randomly assigns 8 patients to one of two samples (4 in each group). He then gives one sample the new treatment but not the other. Following the treatment period, he gives both groups a memory test. Use α = 0.05. SSa = 250 SSb = 155 Xbar a = 50 Xbar b = 44.5 Define what type of test you would use Use all steps to solve and make a conclusion
Step 1: 2 tailed Null: memory of the treatment group and the non-treatment group is equal Alternative: memory of the treatment group and the non-treatment group Is not equal Step 2: decision criteria a = .05 Step 3: data collection Step 4: compute statistic test: (1) find pool variance: 250 + 155 / (4-1) + (4-1) = 67.4 (2) find standard error: 67.4/ 4 + 67.4/ 4 = sqroot (33.7) = 5.81 (3) Find t-test : (50-44.5) - (0) /5.81 = 0.95 Step 5: decision making (1) find t-distribution: Df = 4+4 -2 = 6 (2) 2 tailed (df = 6) Tcritical = +- 2.447 T-observer = .95 (3) conclusion: Fail to reject the null, there is no significant difference between the 2 sample. Not a big enough difference that shows that it works, rather the differences could be due to chance.
Steps one sample z-test:
Step 1: state null and alternative Step 2: set decision criteria for when to reject the null Alpha level = .05 helps find distribution of sample mean Step 3: collect data methods Step 4: compute your sample test statistic to help make decision about the null Step 5: make a decision about the null --> use normal distribution table Null: the memory treatment sample mean is the same as the control population mean no effect Alternative: the treatment reduces memory errors
Steps of hypothesis testing
Step 1: state the hypothesis and select a criterion for the decision - Hypothesis testing procedures assumes we are trying to reject the null, not trying to prove the alternative hypothesis... because it is easier to show something is not true, than prove it is - Null & alternative Criteria: alpha level helps us decide whether the difference that we see is due to a 'real' difference. Refrain from getting type 1 error Step 2: collect a data Step 3: compute a test statistic Step 4: compare the test statistic to a distribution to make an inference about the parameter and hence draw a conclusion about the sample
What steps do you take forrelated t-test:
Step 1: state your hypothesis Step 2: Set you decision criteria Step 3: collect your data Step 4: compute you test statistic (1) Compute you estimate standard error (sDbar) in denominator (2) Compute your t-statistic (3) compute your degrees of freedom Step 5: make a decision about your null hypothesis
What differs from z-test to one-sample t-test?
Step 4 Estimate your standard error (different from z-test because you are given that) Compute your T-statistic Compute degrees of freedom (1 penalty for estimating something) N-1
General statistic tests do what: (step 4) compute your test statistic, what is the general formula
Test statistic (z or t) = observed differences (between a sample and a population or between different samples ___________________________________________ difference expected by chance (based on error)
SPSS: data view compare means (one sample t-test)
Test values: set the null, so you need to ENTER THE POPULATION MEAN Mean = sample mean Std. error mean = estimated standard error of the sample mean o T= observed t o Df = degrees of freedom o Sig = p-value WHAT YOU ARE COMPARING TO! How to reject of fail to reject null with SPSS You must compare the 2 tailed p-value with you're a-level If p-value is equal to or smaller than you a-level (its in the rejection region), then you should reject the null Round for degrees of freedom
ABC Corp. gives a group of employees training on a new software program. They hope that the training will reduce errors. They compare a sample of 16 employees who took the training and have a mean # errors = 87 and std dev (s) = 9 with the untrained population mean of 94 (μ). Determine whether the sample made significantly fewer errors on average than the population. WHAT TEST WOULD YOU USE? AND SOLVE
USE A 1 SAMPLE T-TEST: The population is not known and have the (s) sample standard deviation N= 16, (s) = 9, sample mean = 87, population mean = 94 T= sample mean - population/ sxbar Sxbar = s (sample standard deviation) / square root of n Step 1: state null and alternative: (1 tail test) - Null: Training samples make same number errors as control population of all employees OR Mean treatment = 94 - Alternative: Sample make fewer errors than control population of employees OR Treatment < 94 Step 2: Criterion for decision - 1 tailed or 2 tailed? ANSWER: 1 tailed (a = .05) Step 3: sample statistic - (a = .05) Step 4: Test statistic 1st: Solve for sx 9/ sqroot 16 = 2.25 2nd: solve for numerator 87-94/ 2.25 = -3.11 2nd: Find degree of freedom 16-1 = 15 3rd: go to T-distribution table Because 1 tailed and looking at the fewer errors would be a negative -1.753 Step 5: compare observed to critical value and make a decision about your null T observed = -3.11 T critical = -1.753 You would reject the null, so there was a significant difference. The trained group did have significantly fewer errors
Sample-level z formula
Xbar = sample mean μ xbar = mean of the distribution of sample means (population) σ xbar = standard error FORMULA: z xbar = Xbar - μ xbar/ σ xbar
Z-test, what scores reject the null and fail to reject the null
Z statistic should be very large/extreme, which mean there is a very small chance that it equal to the comparison mean Reject the null = if the probability is in the rejection region, so smaller than .05 THEN reject the null Fail to reject the null = if the z probability is outside of the rejection region, so larger than .05
z-test test ____ tail hypothesis: Census Bureau data show that the mean household income in the area served by a shopping mall is $52,500 per year. A market research firm questions shoppers at the mall. The researchers suspect the mean household income of mall shoppers is higher than that of the general population.
Z-test 1 taile H0: the mean household income is less than or equal to that of the general population HA: researchers suspect that mean household income of mall shoppers is higher than that of the general population
z- test ____ tail hypothesis: The diameter of a spindle in a small motor is supposed to be 5mm. If the spindle is either too small or too large, the motor will not work properly. The manufacturer measures the diameter in a sample of motors to determine whether the mean diameter has moved away from the target.
Z-test 2 tail H0: Spindle are equal to 5 HA: Spindle is not equal to 5 mm
When we have a ONE-tailed test, what do we have to consider when finding difference between pre and post
consider your order of variables in forming D, this impacts the direction of 'D'. ONLY CONSIDER IF USING ONE-TAIL TEST (Pre - post) = use this order if LOWER scores mean improvement (fewer errors, less stress) (Post - Pre) = use this order if HIGHER scores mean 'improvement' (more items correct)
Null for z-test
sample treatment = population
In related sample t-test what is the formula for standard error? what are the steps to get standard error?
sd = sd/ sqroot nd nd = number of different scores First have to find numerator (Dbar) (1) Find sum of squares of D: • (D - Dbar (mean of differences) >2 • (2-5.5)>2 + (6- 5.5) >2 + (5-5.5) >2 + (9-5.5) = 25 • ssD = 25 (2) Find SD (the numerator for standard error formula) Sd = Square root of SSD / nd -1 (3) Find standard error: Sdbar = Sd/ square root (Nd)
EX: TESTING EFFECTIVENESS OF A NEW MEMORY TREATMENT FOR PATIENTS WITH MEMORY PROBLEMS No treatment = 64 errors Treatment = 60 errors What are we trying to look for?
so there is a 4 error difference, and we are determining if the difference is due to effect of treatment OR IS IT JUST A SAMPLING ERROR
What does rejecting the null hypothesis mean?
support for our theory (alternative)
Sxbar (1 sample t-test)
sx (s (standard deviation for sample/ sqroot n)
What is the formula for related sample t-test:
t = Dbar - u dbar (ALWAYS = 0) / Sdbar
Hypothesis testing (sampling and inferential statistics does what)
test claims about populations based on data from the samples Sampling - to make data collection manageable Inferential statistics - used to generalize back to the population from the sample
Standard error tells us
the average amount we'd expect a sample size n to deviate from our population mean. (an estimate of the error you'd expect by chance just from taking a sample) Tells us we will see some differences in the distribution and sample FORMULA: σ (how much erorr you can expect)/ sqroot n (gives standard for comparison)
Properties of the Distribution of Sample Means: Mean
the average of all the sample means will equal the mean of the population
distribution of sample means
the collection of sample means for all the possible random samples of a particular size (n) that can be obtained from a population sample consist of scores that were randomly selected from population
Comparison distribution for z test assumes what?
the null hypothesis is true
Properties of the Distribution of Sample Means: variability (standard deviation) Size?
the standard deviation of the distribution of sample means is called standard error of the mean. *** the bigger the variability of the population, the more variability you'll have in the sample means. Large pop standard deviation: big differences from the pop mean Small pop standard deviation: small differences from the population
Levene's null hypothesis is
variance for group a = variance for group b (1) look at sig to determine if levene is significant Less than .05 = go to lower row (significant) Greater than .05 = stay in upper row (not significant)
Repeated measures
will have two scores from one group of participants who are measure twice (pre-test or post-test- 2 different time points), (longitudinal study) The same person is measure twice, those two scores are related and can be NATRUALLY paired and can compare Each individual is measured in one treatment and measured again in a second treatment
What do z-tests tell us What do one sample t-tests tell us What do related sample t-test tell us What do independent sample t-test test us
z-test compare a treatment sample with a population for which we knew the population mean μ and the population standard deviation One-sample: examine the same situation except we don't know the population --> so we need to use an estimate (the sample standard deviation). Related sample: test for two more situations (repeated measures and matched pairs) Independent sample t-test = experiment that uses a separate independent sample for each treatment condition (or each population) - 2 independent samples, one score per subject. We will compare the group means for two independent groups
Population μ μx
• μ = pop mean • μx = mean of distribution of means
• σ = • s = • σx =
• σ = pop std deviation • s = sample std deviation • σx = std dev of distribution of means
T-test distribution: properties
(1) 1-sample t-test has to first calculate our sample t-value (or observed) - Decide if one tail or 2 THEN how many degrees of freedom (2) Then figure our a critical t-value through looking at the T distribution table (3) Mark critical value on our distribution and everything more extreme than that will become our rejection region In the body of the table is the t critical value (t crit) to use for cutoof points on distribution
We want to determine if management training impacts stress levels. We measure stress for 4 leaders both before and after the training. We expect a different level of stress pre- and post-training (not sure if more or less stress). What test is this, solve. Person: 1 2 3 4 Pre-test: 45 55 40 60 Post-test: 43 49 35 51
(1) NUMERATOR: Have to Find difference scores: (pre-test) - (post-test) Difference: 2 6 5 9 22/ 4 = 5.5 DBAR = 5.5 (2) DENOMINATOR: finding standard error Standard error formula: • sd = sd/ sqroot nd • nd = number of different scores (1) Find sum of squares: • (D - Dbar (mean of differences) >2 • (2-5.5)>2 + (6- 5.5) >2 + (5-5.5) >2 + (9-5.5) = 25 • ssD = 25 (2) Find SD (the numerator for standard error formula) • Sd = Square root of SSD / nd -1 • 25/ 4-1 = 5.25 • Square root (5.25) = 2.9 (3) Find standard error: • Sdbar = Sd/ square root (Nd) • Square root 4 = 2 • 2.9/ 2.0 = 1.45 (4) Find t observed: 5.5 - 0 /1.45 = 3.79 STEP #5: Compare t-observed with t-critical (1) df = 4-1 = 3 & 2 tailed test T-critical = +- 3.18 VS t-observed = 3.79 CONCLUSION: reject the null, there is a significant effect of stress training
How to decide to reject/fail to reject null: z-test
(1) look at comparison distribution that assumes the null hypothesis is true and find our samples z statistic --> Reject the null: the sample z-statistic has to be extremely small (2) Chosen alpha level (.05) gives us decision criteria (.05 is the standard level for our type 1 error rate)
Will 1 sample t-test s match with population standard d
(s) and population standard deviation will not match exactly
• How to find distribution of sample means: 2 4 6 8
1.) you are give the population scores: 2 + 4+ 6+8/4 = 5 2.) sample size of 2 (n=2) 3.) To create the first sample: put out one tile, record the number on the tile, then second tile (b/c sample size 2) --> Table represents all 16 possible samples that could result from this process along with the mean of each sample --> Distribution of sample means where scores are means from the sample you choose. Plott all the different sample means found in the last column ???
Properties of the Distribution of Sample Means
Mean Variability Shape
Setting up the null and alternative for 1 tail vs 2 tail
One-tailed: NULL: the treatment is equal to the population mean ALTERNATIVE: the population is greater than (or less it depends which direaction) than the population Two-tailed: NULL: mean = to population mean (comparison) ALTERNATIVE: mean is not equal to population mean (comparison
What are the two types of related sample t-test:
Repeated measures (pre-test or post-test --- different times) Matched subject study
