Notes on Statistics in Psychology, Study notes of Statistics

Download Notes on Statistics in Psychology and more Study notes Statistics in PDF only on Docsity! Chapter I - Frequency Distributions ☐ Statistics : branch of mathematics that focuses on the organization, analysis + interpretation of a group of numbers 1.) Descriptive statistics → summarizes a group of numbers from a research study. 2.) Inferential statistics→ draw conclusions/make inferences that go beyond the scores from a research study. ° Variables -7 condition or characteristic that can have different values . * can vary (what is being measured) ° Values → possible numbers or categories that can be assigned to a variable . ☒ Values for variables fall in a meaningful range of numbers or categories . ° Score→ a particular individual's value on a variable ex . 20 years old , 150 pounds Levels of measurement 1.) Numeric variable (quantitative) → Equal - interval variables : the numbers stand for approximately equal amounts of what is being measured. → Rank order or ordinal variables : values are ranks 2.) Categorical variables (non- quantitative) How to make a frequency table ex. eye color, gender, brand preference 1.) make a list down the page of each possible value from lowest to highest. 2.) Go one by one through scores , making a mark for each next to its value on the list . 3.) make a table showing how many times each valve on the list is used . 4.) figure the percentage of scores for each value . → take the frequency for that value and divide it by the total number of scores and multiply it by 100 . Frequency Table Histogram ° Grouped frequency tables→ frequency table using intervals based on valves * intervals must be equivalent across the range of scores ° Histogram are in numerical order * the bars do not have to be in a certain order . ° Frequency Polygons are useful for comparing sets of data Unimodal = one high area Bimodal = two high areas Chapter 3- 2 scores ° comparing one score to the mean of a distribution does indicate whether a score is above or below average. ° comparing a score to the mean & standard deviation of a distribution indicates how much above or below average a score is in relation to the spread of scores in the distribution . ° 2 score→ describe a particular score in terms of where it fits into the overall group of scores . * transforms the ordinary score to better describe the score's location in a distribution . ° 2 score→ number of standard deviations the actual score is above or below the mean. - raw score : ordinary score opposed to 2 score (which is scaled) Formula to change Raw score to 2 score 2=2 score * 2= ✗ = raw score M= mean SD = standard deviation Formula to change 2 score to a Raw score 2=2 score * ✗= G)(SD) -1M ✗ = raw score M= mean SD = standard deviation ex . Jerome's 2 score = 1.09 , what is his raw score ? m= 3.40 If a 2 score . . . SD = 1.47 ( ✗ = (2) D)+ M 1.) has a value of 0 , it is equal to ✓ ✗ = G.09)(1-47)+3.40 = 1.60+3.40 = 5 the group mean . 2.) is positive; it is above the group mean . 3.) is negative ; ° In a normal curve , scores fall near center , fewer at extremes . 4.) is equal to +1 , it is 1 SD above the mean . 5.) is equal to +2 , it is 2 SDS above the mean. 6.) is equal to -1 , it is 1 SD below the mean . 7.) is equal to -2, it is 2 Sds below the mean . ✗ = (2-0) / o) -180 Extra Practice Paper 0/4# 2213141$ z= 120,4¥ = -2 LSAT z= 52-6402 = +2 MCAT a- = -a s.it/./.# ÷h ° the shape of a normal curve is standard t well defined ; there is a known percentage of scores above or below any particular point . ° Normal curve with approximate percentages of scores between the meant 1,2 and more than 2 SDS above + below the mean . * you can figure the exact percentage of scores between any two points on the normal curve. I % in tail % mean to Z F % mean to 2 % in tail steps to figuring the percentage of scores above or below a particular Raw or 2 score 1.) If you are beginning with a raw score , first change it to a 2 score . 2.) Draw a picture of a normal curve, decide where the 2 score falls on it , and shade in the area for which you are finding the percentage. 3.) make a rough estimate of the shaded area's percentage based on the 50%-34%-14-1. percentages. 4.) Find exact percentage using normal curve table, adding 50% if necessary. 5.) Check that your exact percentage is within the range of your rough estimate from step 3 . Chapter 4 : compare 1 individual score against the population chapter 5 : compare 1 group of individuals against the population Chapter 6 : making sense of significance (decision errors , effect size , statistical power) Chapter 7 : compare sample group against hypothetical mean , compare 2 conditions within a sample (within - subjects) Chapter 8 : compare 2 groups from the sample Chapter 9 : compare 3 or more groups from the sample (between - subjects) Chapter 5 : Hypothesis testing with means of samples Hypothesis testing Process 1.) Restate the question as a research hypothesis and a null hypothesis 2.) Determine the characteristics of the comparison distribution 3.) Determine the cutoff sample score on the comparison distribution of the null hypothesis should be rejected 4.) Determine your sample's score on the comparison distribution 5.) Decide whether to reject the null hypothesis ° Hypothesis Testing with a distribution of means → procedure in which there is a single sample and the mean and population variance is known . CZ tests) ° Hypothesis testing involving means of groups of scores - a score must be compared with a distribution of scores - a mean must be compared with a distribution of means ° Distribution of means - same mean with less variance ° Have to report the mean , the spread and shape. - the mean : is the same as the mean of the population of individuals . ÑM= µ) - the spread : standard deviation of a distribution of means (ex . Standard Error of the mean) → the variance is the variance of the population divided by the number of individuals in each sample . - the shape : is it normal or not ? Gm = JEM) → the distribution of the population of individuals corresponds to a normal distribution or each sample includes 30 or more individuals Population's Distribution Particular Sample's Distribution Distribution of means mean→ µ=%- Mean → m=G# Mean → µm=µ variance→ o2=d§ Variance→ ofn =o÷ standard→ • = Toa SD -=¢(xµ-M)Z]- standard → m=fÑdeviation deviation SD = Fiz Null Hypothesis Significance Testing with a distribution of means example . Whether being told a person has positive personality qualities increases ratings of the physical attractiveness of that person • Attractiveness ratings when nothing is mentioned about positive personality qualities ④ = 200 , 0=48) • Researcher asks 64 randomly selected students to rate the attractiveness of a particular person in a photograph after being told that the person in the photograph has positive personality qualities . @ = 220) step / : Restate the question as a research hypothesis and a null hypothesis Population I → students who are told that the person has positive personality qualities Population 2→ students in general , who are told nothing about the person's personality qualities (comparison distribution) Research hypothesis → those being told about the positive personality qualities will give higher attractiveness scores for that person than those who are told nothing. Null hypothesis→ those being told about the positive personality qualities will not on average give higher attractiveness scores for that person than those who are told nothing. * the research hypothesis is stated as directional, positive one - tailed hypothesis step 2 I Determine the characteristics of the comparison distribution Population I → N=64 M= 220 Population 2→ 4=200 0=48 The shape of the distribution of means is assumed to be " m = 4=200 approximately normal because N > 30 (N=64) 02m = = "÷=Ts = 36 On = Fm =3Ñ= 6 with a predicted mean of Population 1=208 Power = 37 . 83% with a positive, one tailed hypothesis p ⇐ • 5 Step 3 Determine the cutoff sample score on the comparison distribution of the null hypothesis should be rejected ° Cutoff sample score or cutoff statistic = critical value ° Significance level = • 05 (5% , PE . 05); positive, one- tailed hypothesis testing → Use the normal curve table → 2 score cutoff starts at +1.64 . Step 4 : Determine your sample's score on the comparison distribution M= 220 From step 2 : 4M= 200 On = 6 Calculate 2 score : Z = M£µ"M_ = ⇒%°=%- = 3.33 Step 5 : Decide whether to reject the null hypothesis 2 score of sample's mean = 3.33 We set the Z score cutoff to reject the null hypothesis (+1-64) - We reject the null hypothesis - Research hypothesis is supported * the result of the 2 test is statistically significant at the PI • 05 level Confidence Intervals ° NHST is the main focus in the course . But there is another kind of statisical question related to the distribution of means that is also important in Psychology : estimating the population mean based on the scores in a sample . ° Confidence interval (G) → the range of scores that is likely to include the true population mean . ° Confidence limit→ Upper or lower value of a confidence level . Steps for figuring confidence limits : 1.) Figure the standard error of the mean (om) oñ=% om=E 2.) Figure raw scores G) for 1.96 standard errors (95% confidence interval) or 2.58 standard errors (99% confidence interval) above and below the sample mean - Upper confidence limit : ✗ = M + (2)(om) - Lower confidence limit : ✗ = M - (2)(om) 95% 99% l l l l l l IA. A 1 1 I - 3 -2) - I 0 1 ( 2 3 -3 f- 2 - I 0 1 2 13v V v - 1.96 +1.96 V +2.58 - 2.58 ex . Find the 95% confidence interval for those students who are told that the person has positive personality qualities * Population I → A- 64 M=220 * Population 2-7 4=200 0=48 1.) Figure standard error : ozµ= = 46¥ = 236¥ = 36 on =#= 536=6 Upper confidence limit : ✗ = M + (2)(om) = 220+11.76=231.76 Lower confidence limit : ✗ = M - (2)(om) = 220 - 11.76=208.24 * Based on the sample of 64 students , you can be 95% confident that an interval from 208.24 to 231.76 includes the true population . ° If the confidence interval does not include the mean of the null hypothesis distribution , then the result is statistically significant t t ✗= 208.24 ✗ = 231.76 ° Should we use confidence intervals or null hypothesis significance testing ?  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