At UF, there are always a few days between the end of classes and the beginning of final exams. These days are meant as a study period, but some students would prefer to take the exams as soon as possible, to have a longer vacation – in fact, some students even leave Gainesville during those days as a mini-vacation. To see if the student body supports abolishing “dead week”, the Student Government decides to conduct a survey. They conduct a phone survey (with local numbers selected at random from the student directory) calling people during “dead week”. Will this sample be representative of all UF students?
No, since not all students have local phone numbers.
The cost of tuition is a very important topic. The Alligator surveyed 500 randomly selected students and asked them if they supported a 5% tuition increase. What kind of study is this?
Survey
A high positive correlation is found between college students’ age and their GPA. However, if one student aged 44 with a high GPA is omitted from the study, the correlation all but disappears. This is an example of:
influential outlier
Suppose that you recorded the number of television sets per person and the average life expectancy for the world’s nations. There is a high positive correlation: nations with many TV sets have higher life expectancies.
The response variable is: the nation’s life expectancy, in years
The explanatory variable is: the number of TV sets per person
The experimental units are: the nations
The lurking variable that acts as a confounding factor in this study is: the nations’ wealth
The explanatory variable is: the number of TV sets per person
The experimental units are: the nations
The lurking variable that acts as a confounding factor in this study is: the nations’ wealth
A survey asks participants if they support or oppose the building of new football stadium. If there are 80 respondents (selecting using a simple random sampling method) to a survey, what is the approximate margin of error?
0.1118
The Alligator, UF’s Independent Newspaper, included a poll question on Friday, September 14th, 2023, “Do you have health insurance with the new health care plan?” Any student could then go and answer the survey question.Seventy two students answered this question. Will this sample be representative of all UF students?
No, since this is an example of voluntary response sample.
A study found a strong positive correlation between ice cream consumption and the number of violent crimes committed each month.What kind of study is this?
Observational Study.
P(A’)
= 1 – P(A)
P(A or B)
= P(A) + P(B) – P(A and B)
P(B|A)
= P(A and B)/P(A)
P(A and B)
P(A) * P(B|A)
Events are ____ if they have no outcomes in common.
disjoint
Events are ___ if the occurrence of one does not impact the probability of occurrence of the other.
independent
If events Y and Z are disjoint, P(Y and Z) =
Disjoint events are ____ independent.
never
The collection of all possible outcomes of an experiment is called the ____.
sample space
For a random process, each attempt is called a ____ which generates an outcome.
trial
Suppose that in Los Angeles, 66% of homes have a garage, 23% have a pool and 12% have both. Use a Venn Diagram or contingency table to help answer the questions that follow.
SPACER (use to answer questions below) From the information above, we can identify the following probabilities: Probability that a house has a garage = P(G) = 0.66 Probability that a house has a pool = P(P) = 0.23 Probability that a house has a pool AND a garage = P(G and P) = 0.12
What is the probability that a home chosen at random has a pool?
P(pool) = 0.23
What is the probability that a home chosen at random has a garage?
P(garage) = 0.66
What is the probability that a home chosen at random has neither a pool nor a garage?
P(p’ and g’) = 0.23 For this question we will use the general addition rule and the definition of a compliment P(P or G) = P(P) + P(G) – P(P and G) P(P or G) = 0.23 + 0.66 – 0.12 = .77 <- this is the probability that the house has either a garage OR a pool. The opposite of “having a garage or a pool” is “not having a garage and not having a pool”. Thus to solve this we take the compliment of P(P or G). As is stated in the notes, P(P or G)’ (the compliment of having P or G) = 1 – P(P or G) P(P or G)’ = 1 – 0.77 = 0.23
What is the probability that a home chosen at random has a garage but not a pool?
P(G and P’) = 0.54 (get this value directly from the table) The probability that a home has a garage but not a pool is equal to “the probability that a home has a garage” minus “the probability that a home has a garage and a pool” = P(G) – P(G and P) = 0.66-0.12 = 0.54
Suppose that you have applied to two graduate schools and believe that you have a 0.6 probability of being accepted by school C, a 0.7 probability of being accepted by school D, and a 0.5 probability of being accepted by both.
SPACER (use to answer questions below) *instead of using a contingency table, the above problem can be solved with probability rules* Probability of being accepted by school C = P(C) = 0.6 Probability of being accepted by school D = P(D) = 0.7 Probability of being accepted by both = P(C and D) = 0.5
Determine the probability of being accepted by at least one of these two schools.
0.8 This can be solved using the union of C and D P(C or D). P(C or D) = P(C) + P(D) – P(C and D) = 0.6 + 0.7 – 0.5 = 0.8 P(accepted by at least 1) = 1 – P(accepted by neither C or D) = 1 – 0.2 = 0.8
Determine the probability of being rejected by both schools.
0.2 The probability of being rejected by both schools is the compliment (opposite) of being accepted by at least one of these schools. P(rejected by C and D) = 1 – (probability of being accepted by at least 1 school) = 1 – P(C or D) = 1 – 0.8 = 0.2 This can be taken directly from the table; find the intersection of “not accepted by C” and “not accepted by D) = 0.2
Determine the probability of acceptance by one school but not both schools.
0.3 The probability of being accepted by 1 school but not both is equivalent to the probability of C or D but not both. This is written as: P(C or D) – P(C and D) = 0.8 – 0.5 = 0.3 OR We will take the sum of two probabilities here. Find the box where “accepted by c” intersects “not accepted by d”. This is 0.1. Now find the box where “not accepted by c” intersects “accepted by d”. This is 0.2. 0.1 + 0.2 = 0.3
Events (accepted by school C) and (accepted by school D) are independent. True False
False
Events (accepted by school C) and (accepted by school D) are disjoint True False
False
Determine the conditional probability of acceptance by D given acceptance by C.
0.83 This conditional probability can be calculated as: P(D|C) = P(D and C) / P(C) (read “probability of D given C”) P(D|C) = 0.5/0.6 = 0.83
