Question

task 1 performing a re - randomization simulation in this task, youll perform a re - randomization simulation to determine whether the difference of the sample means is statistically significant enough to be attributed to the treatment. suppose you have 10 green bell peppers of various sizes from plants that have been part of an experimental study. the study involved treating the pepper plants with a nutrient supplement that would produce larger and heavier peppers. to test the supplement, only 5 out of the 10 peppers come from plants that were treated with the supplement. all 10 peppers were of the same variety and grown under similar conditions, other than the treatment applied to 5 of the pepper plants. your task is to examine the claim that the nutrient supplement yields larger peppers. you will base your conclusions on the weight data of the peppers. the table shows the weights of the 10 peppers, in ounces. (note: do not be concerned with which peppers received the treatment for now.) in this task, youll divide the data into two portions several times, take their means, and find the differences of the means. this process will create a set of differences of means that you can analyze to see whether the treatment was successful. part a question a random number generator assigned each pepper to one of two groups. the weights of the peppers in each group, given three randomizations, appear in the tables. for each table, calculate the mean weight for each group, ( \bar{x}_a ) and ( \bar{x}_b ), and find the difference of the mean of group a and the mean of group b (( \bar{x}_a-\bar{x}_b )). type the correct answer in each box. 1. after the first randomization, ( \bar{x}_a ) is 2. after the second randomization, ( \bar{x}_a ) is 3. after the third randomization, ( \bar{x}_a ) is

Explanation:

Step1: Calculate mean of Group A for first randomization

The weights in Group A for first randomization are \(13.4, 12.1, 11.3, 11.2\). Mean \(\bar{x}_A=\frac{13.4 + 12.1+11.3+11.2}{4}=\frac{48}{4} = 12\).

Step2: Calculate mean of Group B for first randomization

The weights in Group B for first randomization are \(9.2, 8.2, 11.8, 14.6\). Mean \(\bar{x}_B=\frac{9.2 + 8.2+11.8+14.6}{4}=\frac{43.8}{4}=10.95\).

Step3: Find difference of means for first randomization

\(\bar{x}_A-\bar{x}_B=12 - 10.95 = 1.05\).

Step4: Calculate mean of Group A for second randomization

The weights in Group A for second randomization are \(8.2, 13.4, 11.2, 9.7\). Mean \(\bar{x}_A=\frac{8.2+13.4 + 11.2+9.7}{4}=\frac{42.5}{4}=10.625\).

Step5: Calculate mean of Group B for second randomization

The weights in Group B for second randomization are \(14.6, 12.1, 11.8, 11.2\). Mean \(\bar{x}_B=\frac{14.6+12.1+11.8+11.2}{4}=\frac{49.7}{4}=12.425\).

Step6: Find difference of means for second randomization

\(\bar{x}_A-\bar{x}_B=10.625-12.425=- 1.8\).

Step7: Calculate mean of Group A for third randomization

The weights in Group A for third randomization are \(9.2, 11.3, 14.6, 11.8\). Mean \(\bar{x}_A=\frac{9.2+11.3+14.6+11.8}{4}=\frac{46.9}{4}=11.725\).

Step8: Calculate mean of Group B for third randomization

The weights in Group B for third randomization are \(13.4, 8.2, 11.2, 9.7\). Mean \(\bar{x}_B=\frac{13.4+8.2+11.2+9.7}{4}=\frac{42.5}{4}=10.625\).

Step9: Find difference of means for third randomization

\(\bar{x}_A-\bar{x}_B=11.725 - 10.625=1.1\).

Answer:

1. After the first randomization, \(\bar{x}_A\) is \(12\), \(\bar{x}_B\) is \(10.95\), and \((\bar{x}_A-\bar{x}_B)\) is \(1.05\).
2. After the second randomization, \(\bar{x}_A\) is \(10.625\), \(\bar{x}_B\) is \(12.425\), and \((\bar{x}_A - \bar{x}_B)\) is \(-1.8\).
3. After the third randomization, \(\bar{x}_A\) is \(11.725\), \(\bar{x}_B\) is \(10.625\), and \((\bar{x}_A-\bar{x}_B)\) is \(1.1\).