Question 1 of 19: A wildlife biologist uses capture-recapture: catches 50 fish, tags them, releases them. Later catches 60 fish and finds 15 are tagged. Estimate the total fish population.
A wildlife biologist uses capture-recapture: catches 50 fish, tags them, releases them. Later catches 60 fish and finds 15 are tagged. Estimate the total fish population. — 200.Proportion tagged: 15/60 = 1/4. So tagged fish (50) are 1/4 of total: total = 50 ÷ (1/4) = 200.
A study samples 80 of 400 voters. 56 say they will vote yes. How many of the 400 are estimated to vote yes? — 280.56/80 = 70%. 400 × 0.70 = 280 voters.
When making an inference from a sample, what is the key assumption? — The sample is representative of the population.The key assumption is that the sample represents the population — which is most likely when the sample is random.
Three different random samples of the same population give slightly different results. This is called: — Sampling error.Sampling variability (sampling error) is the natural variation between different random samples from the same population.
An online poll where users choose to vote is an example of: — Voluntary response bias.When only those who choose to respond participate, the sample suffers from voluntary response bias — those with strong opinions are overrepresented.
A poll of 100 people at a vegetarian restaurant finds that 95% don't eat meat. Can we infer 95% of all Americans don't eat meat? — No, the sample is biased — only vegetarian restaurant customers were surveyed.The sample is biased. Customers of a vegetarian restaurant are far more likely to be vegetarian than the general population.
Which sampling method assigns a number to every member and uses a computer to randomly select from those numbers? — Simple random sampling.Simple random sampling assigns every member a number and uses a random selection process to choose the sample.
School A: mean GPA 3.2, MAD 0.1. School B: mean GPA 3.2, MAD 0.6. What does the difference in MAD tell us? — School B's students have more varied GPAs.Same mean means similar average performance, but School B's larger MAD (0.6) shows much more variation in GPAs.
Group 1 scores: 70, 75, 80, 85, 90. Group 2 scores: 50, 60, 80, 100, 110. Both have mean 80. Which comparison is most complete? — Both groups are similar in center but Group 2 has much more variability.Complete comparison includes both center (mean) and spread. Same mean but very different variability.
A data set has a large outlier. Which pair of measures would you use to compare it fairly with another data set? — Median and IQR.Median and IQR are resistant to outliers, making them better measures when extreme values exist.
What does a MAD of 0 tell you about a data set? — All data values are equal (no variability).MAD = 0 means every data value is exactly at the mean — all values are equal with no variability.
A teacher compares test scores: Class 1 median = 78, Class 2 median = 78. Class 1 IQR = 5, Class 2 IQR = 18. What can the teacher conclude? — Both classes have similar middle scores, but Class 2 scores are more spread out.Same median means similar middle performance, but Class 2's larger IQR shows more variability in scores.
If the IQR of a data set is 0, what does this tell you? — The middle 50% of the data are all the same value (Q1 = Q3).IQR = Q3 − Q1. If IQR = 0, then Q1 = Q3, meaning the middle half of the data are all the same value.
Two data sets have the same IQR. Can their spreads still differ? — Yes, the range (min to max) could still differ.IQR measures only the middle 50% spread. Two sets can have the same IQR but different overall ranges (outliers).
You flip a fair coin 3 times. What is P(all heads)? — 1/8.P(HHH) = (1/2)³ = 1/8. Independent events multiply.
A bag has 2 red and 3 blue marbles. You draw one marble and don't replace it, then draw another. What is P(red then blue)? — 3/10.P(red first) = 2/5. P(blue second | red first) = 3/4. P(red then blue) = (2/5)(3/4) = 6/20 = 3/10.
What is P(at least one head) in 3 coin flips? — 7/8.P(at least one head) = 1 − P(all tails) = 1 − (1/2)³ = 1 − 1/8 = 7/8.
P(A) = 0.4 and P(B) = 0.5, and A and B are mutually exclusive (can't both happen). What is P(A or B)? — 0.9.For mutually exclusive events: P(A or B) = P(A) + P(B) = 0.4 + 0.5 = 0.9.
There are 10 students (6 girls, 4 boys). Two are selected randomly. What is P(both girls) if selections are without replacement? — 1/3.P(first girl) = 6/10. P(second girl | first girl) = 5/9. P(both) = (6/10)(5/9) = 30/90 = 1/3.