Review
first question is a test of independence,
when given that the student pop is 60% female; 7/100 were male and were wearing attire (is this probability of A and B or A given B? this is A and B); when 1 in 10 female student were wearing attire (Pr[CSU I F], in other words, A given B)
if 60% are female then 40% are male; if 10% of females wear CSu then 90% dont; if 7% of men dont then the propability of 0.07/.4 = .175 and 1 - .175 = 0.825 is the prob that men dont
remember that Pr M and CSU = PrM x P[CSU I M] → /07 = .4 x P[CSU I M] for findng .175
now multiply the probability of male or female divided by the probabilities to find the given outcoms
then add the propability of female and attire with male and attire and we find that .06 doesnt equal .078, therefore not independent
or use a table
check out slides, i like this wayyyy better
Binomial Probability: n trials; p probability of successes
X = 20; n = 20; p = 0.5
20! / 18!(2!) = 190
then continue with the other formula
Hypothesis testing
recognize the definition given on the slide, its asking for the p value
understand how these questions are written
Errors
If null is true and we reject → Type I
If null is false and we reject → Correct
If null is true and we fail to reject → Correct
If null is false and we fail to reject → Type II
remember that making our critical value too small will increase the Type II error
no what effect error rates
the differneces in means can determine the overlap of the distributions (Type II)
as difference in means goes up Type II goes down, as smaple size increases, Type II error decreases as well, leading to more reliable results in our hypothesis testing. As variance decreases Type II also decreases
Conditional Probabilites and Pr[A and B] vs Pr[A I B]
Interquartile range
the boxes in a box and whisker plot tells us the concentration in variation of the 1st and 3rd quartiles, the solid line is the mean
Question about pea pods
mutuall exclusive if the prob of A and B occuring together is 0, so to be short and yellow is not mutuall exclusive
are yellow and short independent?
Pr[yellow] = .25
Pr[short] = .25
.25 x .25 × 1600= 100
so yes independent bc observed matches expected
identify types of data
blood type - categorical and nominal
age - numerical (discrete [nearest integer value])
Pain Level (1-5 scale) - categorical and ordinal (theres a specific order)
Temperature - Numerical (continuous [nearest decimal place])
weight - numerical (continuous)
number or partners - numerical discrete
Energy level (low, medium high) - categorical ordinal