1. John has to complete a research project on the ape population in Spain. He is trying to estimate the size of population of apes. He randomly catches 37 apes and marks them with paint. He releases apes in jungle. The following year he observes 250 apes and he found that 10 were marked with the paint that he used. Find out the best estimate for the size of the ape population?

2. The ratio of marked apes to total apes should be about the same in both the sample population and the overall population. Set up a proportion. Total number of marked Ape Estimate of Ape population = Number of marked Ape counted Total number of Ape counted

3. Plug in the numbers you know and solve for the Ape population, p. 10_ 250 37 p = 10p = 37 x 250 (find the cross products)10p = 9,250 divide both sides by 10. P = 925

4. 50_ 260 58 p = 50 p = 260 x 58 (find the cross products) 50p =15,080 divide both sides by 50. P = 301.6 1) A famous television channel put tags on 58 alligator and released them. Later, they catch 260 alligator; among those alligator, 50 were tagged. Calculate the best estimate for the alligator population?

5. 2) Management wants to open a cafeteria in the office building. So they collected data from two random samples of 100 employees regarding employees’ office lunch preference? Make at least two inferences based on the results. Sample Group Corn Dog Philly Cheese Steak Reuben Sandwich Total #156359100 #2781210100

6. Sample Group Corn Dog Philly Cheese Steak Reuben Sandwich Total #156359100 #2781210100 1. Most employees prefer Corn Dogs. 2. More people prefer Corn Dogs more than Philly Cheese Steaks and Reuben Sandwich combined. 3. Reuben sandwiches are not very popular.

7. 28_ 440 32 p = 28 p = 440 x 32 (find the cross products) 28p = 14,080 divide both sides by 28. P = 502.9 3) A computer tracking chip company wants to test their chips. They marked 32 birds by attaching computer chips to their legs. Later on, they catch 440 birds, 28 of which are marked with computer chips attached to them. They provided this data to animal welfare authorities to estimate the bird population. To the nearest whole number, what is the best estimate of the bird population?

8. Mike is a statistician. He wanted to compare the average height of some of the players on his favorite baseball and football teams. Because football is characteristically played by larger athletes, he thinks that there will be a greater variability in the heights of football players as compared to baseball players. He used the heights that were listed in the game time team programs to determine the heights of the players. Football Team – Height of Players in inches 60, 62, 65, 74, 71, 69, 64, 67 Baseball Team – Height of Players in inches 74, 73, 62, 68, 65, 70, 64, 75 1. Which team’s average player is taller? 1. The mean height of the football players is 66.5 inches as compared to the mean height of the baseball players at 68.875 inches. The average baseball player is taller in this case. 2. What is the height difference between the average players on both teams? 2. The height difference of average players is: Baseball average height – football average height = height difference 68.875 inches – 66.5 inches = 2.375 inches

9. 3. The football team’s best three players are 62, 71, and 74 inches tall. The baseball team’s best three players are 68, 70, and 74 inches tall. Which team has a greater variability of heights for their top 3 players? Explain your answer. The mean absolute deviation (MAD) tells us the degree of variability of any data set. The greater the MAD value, the more variability. We can calculate the MAD for each set by:

10. 1. Find the mean of each set: Football: 62, 71, 74 Mean = (62 + 71+ 74) / 3 = 69 Baseball: 68, 70, 74 Mean = (68 + 70 + 74) / 3 = 70.67

11. 2. Determine the deviation of each variable from the mean: Football: 62 – 69= -7 71 – 69= 2 74 – 69= 5 Baseball: 68 – 70.67 = -2.67 70 – 70.67= -.67 74 – 70.67= 3.33