Problem 1.17
In problem 1.17, the average age of pinon trees in the coast ranges of California was investigated. 500 ten-hectare plots randomly selected using a computer. The age of every pinon tree within each plot was measured and the average age of the plot was used as the unit measurement (Whitlock and Schluter, 2015, pg 20). The results of this investigation were used to estimate the average age of California pinon pines, making the population of interest in this study is all California pinon pines. The researchers took an average of the ages of trees within each plot as their unit measurement since the pinon trees contained within the 10-hectare plots cannot be described as a random sample. It is a non-random sample because the trees …show more content…
For the Coach group and Supporter group, a 3-second mean hug duration is plausible since this value falls within the 95% confidence interval. Since it falls between the two limits, there is a 95% chance of the length hug occurring in each relationship group. Coach/Athlete: Lower limit= 3.77-2(.451)=2.87 and Upper limit = 3.77+2(.451)=4.67 2.87 < µ < 4.67 Supporter/Athlete: Lower limit= 3.16-2(.319)=2.52 and Upper limit = 3.16+2(.319)=3.80 3.16 < µ < 3.80
Problem 5.31
The figure provided in problem 5.31, shows the probability density of colony diameters (in mm) in a hypothetical population of Paenibacillus bacteria (Whitlock and Schluter, 2015, pg 146). In the examples, a single colony is randomly sampled form the population. Since the probability is a bell-shape, the hypothetical population is a normal distribution.
(Figure re-created using Microsoft Word shapes and …show more content…
Since the hypothesis is looking at both extremes, the final P-value is: P = 2(0.1209) = 0.2418. This value is above the significance value of 0.05 failing to reject the null hypothesis. The woman may have ESP powers. If another individual tried to guess the ESP cards 1000 times and was correct 350 of those times, their success rate would be 0.35. This proportion is more successful than the expected success rate of 0.2, but still less than the woman’s success rate of 0.4 in part a of this question. This is because of the large sample size, 1000 vs 100 trials. The law of large numbers states that there is an improvement in precision as the sample size increases (Whitlock and Schluter, 2015, pg 185). Although the value is greater than the expected rate of success, it is starting to fall closer to the value of 0.2.
In problem 1.17, the average age of pinon trees in the coast ranges of California was investigated. 500 ten-hectare plots randomly selected using a computer. The age of every pinon tree within each plot was measured and the average age of the plot was used as the unit measurement (Whitlock and Schluter, 2015, pg 20). The results of this investigation were used to estimate the average age of California pinon pines, making the population of interest in this study is all California pinon pines. The researchers took an average of the ages of trees within each plot as their unit measurement since the pinon trees contained within the 10-hectare plots cannot be described as a random sample. It is a non-random sample because the trees …show more content…
For the Coach group and Supporter group, a 3-second mean hug duration is plausible since this value falls within the 95% confidence interval. Since it falls between the two limits, there is a 95% chance of the length hug occurring in each relationship group. Coach/Athlete: Lower limit= 3.77-2(.451)=2.87 and Upper limit = 3.77+2(.451)=4.67 2.87 < µ < 4.67 Supporter/Athlete: Lower limit= 3.16-2(.319)=2.52 and Upper limit = 3.16+2(.319)=3.80 3.16 < µ < 3.80
Problem 5.31
The figure provided in problem 5.31, shows the probability density of colony diameters (in mm) in a hypothetical population of Paenibacillus bacteria (Whitlock and Schluter, 2015, pg 146). In the examples, a single colony is randomly sampled form the population. Since the probability is a bell-shape, the hypothetical population is a normal distribution.
(Figure re-created using Microsoft Word shapes and …show more content…
Since the hypothesis is looking at both extremes, the final P-value is: P = 2(0.1209) = 0.2418. This value is above the significance value of 0.05 failing to reject the null hypothesis. The woman may have ESP powers. If another individual tried to guess the ESP cards 1000 times and was correct 350 of those times, their success rate would be 0.35. This proportion is more successful than the expected success rate of 0.2, but still less than the woman’s success rate of 0.4 in part a of this question. This is because of the large sample size, 1000 vs 100 trials. The law of large numbers states that there is an improvement in precision as the sample size increases (Whitlock and Schluter, 2015, pg 185). Although the value is greater than the expected rate of success, it is starting to fall closer to the value of 0.2.