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Alborosie
3 years ago
13

A consumer wanted to compare two types of battery storage. One type is refrigerated and the other is room temperature. It is a c

ommon belief that refrigerated batteries last longer. This belief is tested by placing 10 fully charged batteries into each type of storage and then measuring the percentage of charge remaining after thirty days. A coin tip was used to determine which type of storage each battery would be placed in first. Results indicated that there was no difference in the two types of storage Complete parts (a) through in below.
What type of experimental design is this?
A) Matched pairs design
B) Completely randomized design
C) Randomized block design
D) Case-control study
Mathematics
1 answer:
bearhunter [10]3 years ago
5 0

Answer:

<u>A) Matched pairs design</u>

<u>Step-by-step explanation:</u>

In<em> </em><em>experimental design</em>, the match pairs experimental design is one that often involves only two treatment conditions; which in this case are the 'refrigerated and the room temperature battery storage types'.

Thus, these two treatment conditions form a matched pair because we are told that 10 fully charged batteries are placed into each type of storage.

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Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
Vinil7 [7]

Answer:

a)

X[bar]_A= 71.8cm

X[bar]_B= 72cm

b)

M.A.D._A= 8.16cm

M.A.D._B= 5.4cm

c) The data set for Soil A is more variable.

Step-by-step explanation:

Hello!

The data in the stem-and-leaf plots show the heights in cm of Teddy Bear sunflowers grown in two different types of soil (A and B)

To read the data shown in the plots, remember that the first digit of the number is shown in the stem and the second digit is placed in the leaves.

The two data sets, in this case, are arranged in a "back to back" stem plot, which allows you to compare both distributions. In this type of graph, there is one single stem in the middle, shared by both samples, and the leaves are placed to its left and right of it corresponds to the observations of each one of them.

Since the stem is shared by both samples, there can be observations made only in one of the samples. For example in the first row, the stem value is 5, for the "Soil A" sample there is no leaf, this means that there was no plant of 50 ≤ X < 60 but for "Soil B" there was one observation of 59 cm.

X represents the variable of interest, as said before, the height of the Teddy Bear sunflowers.

a) To calculate the average or mean of a data set you have to add all observations of the sample and divide it by the number of observations:

X[bar]= ∑X/n

For soil A

Observations:

61, 61, 62, 65, 70, 71, 75, 81, 82, 90

The total of observations is n_A= 10

∑X_A= 61 + 61 + 62 + 65 + 70 + 71 + 75 + 81 + 82 + 90= 718

X[bar]_A= ∑X_A/n_A= 218/10= 71.8cm

For Soil B

Observations:

59, 63, 69, 70, 72, 73, 76, 77, 78, 83

The total of observations is n_B= 10

∑X_B= 59 + 63 + 69 + 70 + 72 + 73 + 76 + 77 + 78 + 83= 720

X[bar]_B= ∑X_B/n_B= 720/10= 72cm

b) The mean absolute deviation is the average of the absolute deviations of the sample. It is a summary of the sample's dispersion, meaning the greater its value, the greater the sample dispersion.

To calculate the mean absolute dispersion you have to:

1) Find the mean of the sample (done in the previous item)

2) Calculate the absolute difference of each observation and the sample mean |X-X[bar]|

3) Add all absolute differences

4) Divide the summation by the number of observations (sample size,n)

For Soil A

1) X[bar]_A= 71.8cm

2) Absolute differences |X_A-X[bar]_{A}|

|61-71.8|= 10.8

|61-71.8|= 10.8

|62-71.8|= 9.8

|65-71.8|= 6.8

|70-71.8|= 1.8

|71-71.8|= 0.8

|75-71.8|= 3.2

|81-71.8|= 9.2

|82-71.8|= 10.2

|90-71.8|= 18.2

3) Summation of all absolute differences

∑|X_A-X[bar]_A|= 10.8 + 10.8 + 9.8 + 6.8 + 1.8 + 0.8 + 3.2 + 9.2 + 10.2 + 18.2= 81.6

4) M.A.D._A=∑|X_A-X[bar]_A|/n_A= 81.6/10= 8.16cm

For Soil B

1) X[bar]_B= 72cm

2) Absolute differences |X_B-X[bar]_B|

|59-72|= 13

|63-72|= 9

|69-72|= 3

|70-72|= 2

|72-72|= 0

|73-72|= 1

|76-72|= 4

|77-72|= 5

|78-72|= 6

|83-72|= 11

3) Summation of all absolute differences

∑ |X_B-X[bar]_B|= 13 + 9 + 3 + 2 + 0 + 1 + 4 + 5 + 6 + 11= 54

4) M.A.D._B=∑ |X_B-X[bar]_B|/n_B= 54/10= 5.4cm

c)

If you compare both calculated mean absolute deviations, you can see M.A.D._A > M.A.D._B. As said before, the M.A.D. summary of the sample's dispersion. The greater value obtained for "Soil A" indicates this sample has greater variability.

I hope this helps!

7 0
3 years ago
What is the value of 1 3/4 in decimal form
Luba_88 [7]
The answer would be, 1.75 because 3/4 of 1 is 75% pulse you spreading have a whole, so 1.75.
3 0
3 years ago
Read 2 more answers
Help would be nice‍♀️.
11111nata11111 [884]

Answer:

the ones with a line above the numbers are repeating decimals

5 0
3 years ago
Read 2 more answers
Answers because I need help
kramer

Answer:

f(x)

Step-by-step explanation:

the function f(x) crosses the y-axis higher than either of the other functions, therefore it has the largest y-intercept

6 0
3 years ago
Read 2 more answers
HELP PLEASE!!!!!! Find the speed of an athlete who makes 4and3/4 laps in 3mins45 seconds on a 400m field in m/s​
Rom4ik [11]

Given:

An athlete who makes 4\dfrac{3}{4} laps in 3 mins 45 seconds on a 400m field.

To find:

The speed of the athlete in m/s.

Solution:

We know that,

Distance covered in 1 lap = 400 m

Distance covered in 4\dfrac{3}{4} laps = 4\dfrac{3}{4}\times 400 m

                                               = \dfrac{19}{4}\times 400 m

                                               = 1900 m

We know that,

1 minute = 60 seconds

3 minutes = 180 seconds

3 minutes 45 second = 180 + 45 seconds

                                   = 225 second

The speed of the athlete is:

Speed=\dfrac{Distance}{Time}

Speed=\dfrac{1900}{225}

Speed\approx 8.44

Therefore, the speed of the athlete is about 8.44 m/s.

7 0
2 years ago
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