Applying a Level of Significance In this task, you’ll study the data from two experiments. For each data set, you will determine whether the difference of means between the treatment group and the control group is statistically significant and interpret the result in the context of the situation.

Question 1
Researchers want to find a way to increase the growth rate of corn so it can be harvested more often. To test the effectiveness of a new type of spray fertilizer, plants grown from 200 corn seeds were divided equally into two groups: a treatment group and a control group. The treatment group was frequently sprayed with the new type of fertilizer. The growth in centimeters of each corn plant was noted before and after the treatment.

The results showed that the mean change in the growth rate by the treatment group is 8 points more than that of the control group. To test whether the results could be explained by random chance, researchers created a table that summarizes the results of 1,000 re-randomizations of the data with differences of means rounded to the nearest 2 points.

Consider the significance level to be set at 5%, so results less than 5% can be considered statistically significant.

The calculation of the probability and interpretation of the difference in mean and the relation to level of significance between and the treatment and control groups are as follows;

Question 1;   Part A

The probability that the treatment group mean is greater than the control group mean by 8 points or more is 3.6%

The completed statement is as follows;

The significance level is set at 5%, and the probability of the result is 3.6%, which is lower than significance level. The result is statistically significant

Part B

The result is statistically significant, which implies that spraying the corn plant with the new type of fertilizer does increase the growth rate

Question 2; Part A

The probability of the treatment group mean is lower than the control group mean by 15 points or more is 7.8%

The completed statement is as follows;

The significance level is set at 5%, and the probability of the result is 7.8% which is higher than the significance level. The result is not statistically significant

Part B;

The result is not statistically significant which implies that wearing a watch does not make people manage their time better

The reason for the above solutions is as follows;

Question 1;

The known parameters;

The significance level of the test, α = 5%

The frequencies of the difference in mean between the treatment and control groups given in the table

The number of randomizations, ∑f = 1,000

The required parameter

The probability of the treatment group mean is greater than the control group mean by 8 points or more

The significance of the result

The comparison between the given level of significance and the probability

Method

We find the probability for the required difference in mean from the data given in the table

The probability that the treatment group mean is greater than the control group mean by 8 points or more is given by the following formula;

From the table, we have;

∑f(8 or more) = 26 + 8 + 2 = 36

∑f = 1 + 10 + 28 + 58 +...+ 114 + 57 + 28 + 8 + 2 = 1,000

Therefore;

The probability, P(8 or more) 3.6%

Given that the significant level, α = 5% is higher than the probability of 3.6%, the result is statistically significant

Part B

The true statement is that the result is statistically significant, which implies that spraying the corn plant with the new type of fertilizer does increase the growth rate

Question 2;

The known parameters are;

The significance level of the test, α = 5%

The frequencies of the difference in mean between the treatment and control groups given in the table

The number of randomizations, ∑f = 1,000

The required parameter

The probability of the treatment group mean is lower than the control group mean by 15 points or more

The significance of the result

The comparison between the given level of significance and the probability

Method

Using the method from Question 1, we have;

Part A

From the table, we have;

∑f(15 or more) = 50 + 18 + 10 = 78

∑f = 10 + 18 + 50 + 127 + 195 + 226 + 180 + 117 + 48 + 17 + 12 = 1,000

Therefore;

The probability, P(15 or more) = 7.8%

Given that the significant level, α = 5% is lower than the probability of 7.8%, the result is not statistically significant

Therefore, we have;

The significance level is set at 5%, and the probability of the result is 7.8% which is higher than the significance level. The result is not statistically significant

Part B

The true statement is that the result is not statistically significant which implies that wearing a watch does not make people manage their time better

Learn more about hypothesis testing here;

answer: c i think

step-by-step explanation:

the 1st digit is in the 100s n its 1

step-by-step explanation:

3,589÷18 = 199.389

so the first digit of the quotient is in the 100s n its 1