We are given a data set with the results of an experiment about whether water with added calcium affects the yields of yellow squash plants.

	Yield (kilograms)
Control Group	$0.9$	$0.9$	$1.4$	$0.6$	$1.0$	$1.1$	$0.7$	$0.6$	$1.2$	$1.3$
Treatment Group	$1.0$	$1.2$	$1.2$	$1.3$	$1.0$	$1.8$	$1.7$	$1.2$	$1.0$	$1.9$

We will repeat the tasks from Exploration $1$ and $2$ using this data set. Let's start with Exploration $1,$ in which we will be resampling the data.

Resampling Data

First, we were asked to find the means of the control and treatment groups and their difference.

Difference of the Means

Let's look at the given data of the control group.

	Yield (kilograms)
Control Group	$0.9$	$0.9$	$1.4$	$0.6$	$1.0$	$1.1$	$0.7$	$0.6$	$1.2$	$1.3$

We can start with adding all the yields of the control group from table.

Sum = 0.9 + 0.9 + 1.4 + 0.6 + 1.0 + 1.1 + 0.7 + 0.6 + 1.2 + 1.3 = 9.7

Next, to find the mean of the control group we need to divide the sum by the number of data records, which is

10 .

\overline{x}_{control} = \frac{9 . 7}{1 0} = 0.97

We found the mean yield of the control group,

\overline{x}_{control} = 0.97

kilograms. Now, to find the mean of the treatment group

\overline{x}_{treatment},

we can repeat this process using the treatment group's yield data.

	Yield (kilograms)
Treatment Group	$1.0$	$1.2$	$1.2$	$1.3$	$1.0$	$1.8$	$1.7$	$1.2$	$1.0$	$1.9$

Let's add all the yields from the treatment's group table.

Sum = 1.0 + 1.2 + 1.2 + 1.3 + 1.0 + 1.8 + 1.7 + 1.2 + 1.0 + 1.9 = 13.3

To calculate the mean of the treatment group, we will divide this sum by the number of records,

10 .

\overline{x}_{control} = \frac{1 3 . 3}{1 0} = 1.33

The mean of the treatment group is

1.33

kilograms. Now we can calculate the difference in the means.

\overline{x}_{treatment} - \overline{x}_{control} = 1.33 - 0.97 = 0.36

We found that the difference of the means is equal to

0.36 .

Since the difference is based on the result of the experiment, we will call it the experimental difference. Now, let's move to the second part of Exploration

1,

which is to perform the resampling experiment.

Resampling

To perform the resampling experiment, we have to choose $10$ yields at random from the control and treatment groups. Let's write each yield measure on a piece of paper and place them in a bag.

Now, we can select $10$ pieces of paper at random from the bag and call them the control group. The $10$ pieces left in the bag will be the treatment group. Let's look at an example of a random selection.

We created a new control group and a new treatment group.

	Control Group	Treatment Group
Yields	$0.9,$ $0.9,$ $1.0,$ $1.1,$ $0.6,$ $1.3,$ $1.0,$ $1.0,$ $1.8,$ $1.2$	$1.4,$ $0.6,$ $0.7,$ $1.2,$ $1.2,$ $1.2,$ $1.7,$ $1.3,$ $1.0,$ $1.9$

Let's calculate their means!

	Control Group	Treatment Group
Yields	$0.9,$ $0.9,$ $1.0,$ $1.1,$ $0.6,$ $1.3,$ $1.0,$ $1.0,$ $1.8,$ $1.2$	$1.4,$ $0.6,$ $0.7,$ $1.2,$ $1.2,$ $1.2,$ $1.7,$ $1.3,$ $1.0,$ $1.9$
Sum	$10.8$	$12.2$
Mean	$\overline{x}_{control} = \frac{1 0 . 8}{1 0} = 1.08$	$\overline{x}_{treatment} = \frac{1 2 . 2}{1 0} = 1.22$

Next, we can calculate the difference of the means.

\overline{x}_{treatment} - \overline{x}_{control} = 1.22 - 1.08 = 0.14

We found the first difference,

0.14 .

This process must be repeated four more times. Let's look at the example results of resampling. Since the process is random, your results might differ.

	$\overline{x}_{treatment}$	$\overline{x}_{control}$	$\overline{x}_{treatment} - \overline{x}_{control}$
Resampling #1	$1.22$	$1.08$	$0.14$
Resampling #2	$1.1$	$1.2$	$- 0.1$
Resampling #3	$1.23$	$1.07$	$0.16$
Resampling #4	$1.18$	$1.12$	$0.06$
Resampling #5	$1.09$	$1.21$	$0.12$

Comparison of the Differences of Means

Finally, we will compare the differences in the means that we obtained by resampling with the experimental difference of the means.

Results
Difference Obtained by Resampling	Experimental Difference
$0.14$	$0.36$
$- 0.1$	$0.36$
$0.16$	$0.36$
$0.06$	$0.36$
$0.12$	$0.36$

We can see that for each difference in the means obtained by resampling, the experimental difference of the means is greater. Let's compare the results with Exploration $1$ results. If you want to know how we obtained the results from Exploration $1,$ you can check the Extra box below this section.

Exploration $1$ Results
Difference Obtained by Resampling	Experimental Difference
$0.24$	$0.18$
$- 0.06$	$0.18$
$0.02$	$0.18$
$0.14$	$0.18$
$- 0.12$	$0.18$

As we can see, in Exploration

1

the experimental difference was more similar to the resampling differences. We finished all the tasks in Exploration

1 .

Let's move to Exploration

2,

in which we will be evaluating our results.

Extra

Exploration 1 Solution

In Exploration $1$ we are analyzing the same experiment about the effect of calcium added to water on the yields of a yellow squash plant. However, the results of the experiment are different.

	Control Group	Treatment Group
Yields	$1.0,$ $1.2,$ $1.5,$ $0.9,$ $1.1,$ $1.4,$ $0.8,$ $0.9,$ $1.3,$ $1.6$	$1.1,$ $1.3,$ $1.4,$ $1.2,$ $1.0,$ $1.7,$ $1.8,$ $1.1,$ $1.1,$ $1.8$

We will follow the same Exploration $1$ steps as we did before using this data set. First, let's calculate the means of the control and treatment groups.

	Control Group	Treatment Group
Yields	$1.0,$ $1.2,$ $1.5,$ $0.9,$ $1.1,$ $1.4,$ $0.8,$ $0.9,$ $1.3,$ $1.6$	$1.1,$ $1.3,$ $1.4,$ $1.2,$ $1.0,$ $1.7,$ $1.8,$ $1.1,$ $1.1,$ $1.8$
Sum	$11.7$	$13.5$
Mean	$\overline{x}_{control} = \frac{1 0 . 8}{1 0} = 1.17$	$\overline{x}_{treatment} = \frac{1 2 . 2}{1 0} = 1.35$

Now we can calculate the difference of the means.

\overline{x}_{treatment} - \overline{x}_{control} = 1.35 - 1.17 = 0.18

Next, we will perform the resampling experiment five times and display the results in a table. The detailed explanation of how to perform resampling can be found in the previous part.

	$\overline{x}_{treatment}$	$\overline{x}_{control}$	$\overline{x}_{treatment} - \overline{x}_{control}$
Resampling #1	$1.38$	$1.14$	$0.24$
Resampling #2	$1.23$	$1.29$	$- 0.06$
Resampling #3	$1.27$	$1.25$	$0.02$
Resampling #4	$1.33$	$1.19$	$0.14$
Resampling #5	$1.2$	$1.32$	$- 0.12$

Finally, we can compare the resampling differences with the experimental difference.

Results
Difference Obtained by Resampling	Experimental Difference
$0.24$	$0.18$
$- 0.06$	$0.18$
$0.02$	$0.18$
$0.14$	$0.18$
$- 0.12$	$0.18$

As we can see, in some cases the experimental difference is similar to the differences obtained by resampling.

Evaluating Results

In this part, we are asked to find strong evidence to reject or accept the following hypothesis.

Water dissolved in calcium has no effect on the yields of yellow squash plants.

To do that, we will first create a histogram.

Histogram

If you have the resampling differences of means from your class, use them to create the histogram. In case you do not have them, we will make a histogram based on data from $100$ resampling differences generated by a computer.

$100$ Resampling Differences
$0.00$
$- 0.26$
$0.00$
$- 0.12$
$0.08$
$0.16$
$- 0.16$
$- 0.02$
$- 0.26$
$0.10$
$- 0.12$
$0.18$
$- 0.04$
$- 0.22$
$- 0.06$
$- 0.10$
$- 0.06$
$- 0.10$
$0.12$
$0.00$
$0.08$
$0.00$
$- 0.22$
$- 0.22$
$0.02$
$- 0.08$
$0.06$
$- 0.32$
$- 0.06$
$- 0.08$
$- 0.04$
$0.24$
$- 0.04$
$- 0.04$
$- 0.10$
$- 0.18$
$- 0.10$
$0.06$
$0.18$
$- 0.28$
$- 0.04$
$- 0.04$
$- 0.40$
$- 0.22$
$- 0.04$
$- 0.14$
$- 0.14$
$0.14$
$0.00$
$0.18$
$0.18$
$- 0.24$
$- 0.06$
$- 0.22$
$- 0.16$
$- 0.36$
$- 0.18$
$- 0.30$
$0.04$
$0.12$
$- 0.26$
$0.02$
$0.20$
$0.00$
$- 0.22$
$- 0.14$
$0.04$
$0.24$
$- 0.22$
$- 0.10$
$- 0.04$
$- 0.08$
$0.06$
$0.16$
$0.16$
$0.24$
$- 0.20$
$- 0.08$
$- 0.02$
$- 0.04$
$0.02$
$0.08$
$0.02$
$- 0.04$
$0.18$
$0.04$
$0.16$
$- 0.02$
$0.14$
$- 0.16$
$0.10$
$0.16$
$0.14$
$0.08$
$0.36$
$0.04$
$- 0.16$
$- 0.14$
$- 0.02$
$- 0.10$

To make drawing easier, we will make each bar of our histogram represent an interval of the length $0.1 .$ Let's count the number of differences that fall into each interval. The number of differences will represent the frequency of occurrence of each interval.

Interval	Frequency (Number of Differences)
$[- 4, - 0.3)$	$1$
$[- 0.3, - 0.2)$	$7$
$[- 0.2, - 0.1)$	$14$
$[- 0.1, - 0)$	$23$
$[0, 0.1)$	$23$
$[0.1, 0.2)$	$17$
$[0.2, 0.3)$	$8$
$[0.3, 0.4)$	$6$
$[0.4, 0.5)$	$1$

Now we are ready to draw a histogram. The bars will represent the intervals. The height of each bar should be proportional to the corresponding frequency.

We displayed the values in the histogram.

Comparing the Histogram with Experimental Difference

Next, we will draw a vertical line on the histogram that will represent the experimental difference of means. Earlier, we found that the experimental difference is equal to

0.36 .

Let's draw a vertical line at the value

0.36 .

When to Reject the Hypothesis

In this part, we are asked to find a place on the histogram where the experimental difference should lie to give us enough evidence to reject the hypothesis.

Water dissolved in calcium has no effect on the yields of yellow squash plants.

If the calcium in the water actually has no effect on the yields, the means of the control and treatment groups should be similar. Therefore, the experimental difference should be close to zero. Let's look at the histogram that we made earlier. Notice how the most common differences are in the middle, around

0 .

We will reject the hypothesis when the experimental difference is too far away from

0 .

We have to agree on how confident we want to be about our decision of rejecting the hypothesis. We will assume that we want to be

90 %

sure about our decision. Now, we have to find the middle

90 %

of the values on the histogram. Notice that the interval from

- 0.3

to

0.3

contains about

90 %

of the values.

Therefore, we will accept the hypothesis if the experimental difference is within this interval and reject it if it is not.

In conclusion, we will reject the hypothesis if the experimental difference is below

- 0.3

or above

0.3 .

This will give us

90 %

confidence about our decision.

Are We Able to Reject the Hypothesis?

Let's check where our experimental difference, which is equal to

0.36,

lies on the histogram.

As we can see, our experimental difference lies outside the interval in which we accept the hypothesis. Therefore, we have enough evidence to reject the hypothesis.

Water dissolved in calcium has no effect on the yields of yellow squash plants. $\times Rejeceted$

This means that based on our experiment, water dissolved in calcium has an effect on the yields of the plants. Let's see if this result is different from Exploration

2 .

The experimental difference in that part was

0.18 .

We can draw a vertical line at the difference

0.18 .

As we can see, the experimental difference from the data from Exploration 2 lies within the interval in which we accept the hypothesis. Therefore, in Exploration 2 we do not have enough evidence to reject the hypothesis.

Exercise