1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
tigry1 [53]
3 years ago
10

Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o

ften. This has the obvious downside of more frequent colds and other illnesses, but it also serves to challenge the immune system of children at a critical stage in their development. A study by Gilham et al. (2005) tested whether social activity outside the house in young children affected their probabilty of later developing the white blood cell disease acute lymphoblastic leukemia (ALL), the most common cancer of children. They compared 1272 children with acute lymphoblastic leukemia to 6238 children without acute lymphoblastic leukemia. Of the acute lymphoblastic leukemia kids, 1020 had significant social activity outside the home (including day care) when younger. Of the kids without acute lymphoblastic leukemia, 5343 had significant social activity outside the home. The rest of the children in both groups lacked regular contact with children who were not in their immediate surroundings.
a.) Is this an experimental or observational study?

b.) What are the proportions of children with significant social activity in children with and without acute lymphoblastic leukemia?

c.) What is the odds ratio for acute lymphoblastic leukemia, comparing the groups with and without significant social activity?

d.) What is the 95% confidence interval for this odds ratio?

e.) Does this confidence interval indicate that the amount of social activity is associated with acute lymphoblastic leukemia? If so, did the children with more social activity have a higher or lower occurence of acute lymphoblastic leukemia?

f.) The researchers interpreted their study results in terms of the differing immune system exposure of the children, but gave several alternative explanations for the pattern. Can you think of any possible cofounding variables?
Mathematics
1 answer:
Ostrovityanka [42]3 years ago
6 0

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is \hat p_1 = 1020/1272 = 85/106 = 0.8\overline {0188679245283} ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is \hat p_2 = 5343/6238 ≈ 0.8565

c) We have the following two way table;

\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}

The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

95 \% \ CI \approx  0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

You might be interested in
If the area of a circle is 58 square feet, find the circumference
Blizzard [7]

Answer:

C≈27 feet

Step-by-step explanation:

A=πr²

A/π=r²

√A/π =r or –√A/π=r (refused)

then r =√(A/π) =√(58/π) =4.297829403 feet

C=2πr=2π(4.297829403)=27.00405856≈27 feet

7 0
3 years ago
Read 2 more answers
Log(5) + log(3) rewrite the following in the form log(c).
Dennis_Churaev [7]
Log (5) + log (3) = log (15)
3 0
3 years ago
Read 2 more answers
Apply the rule (-x, y - 2) to the following point, what is the result?
BaLLatris [955]

Answer: Second Choice. (3, 3)

Step-by-step explanation:

<u>Given point</u>

(x, y) = (-3, 5)

<u>Given the rule for application</u>

(-x, y - 2)

<u>Determine the new x-value</u>

x = -3

-x = - (-3) = 3

<u>Determine the new y-value</u>

y = 5

y - 2 = (5) - 2 = 3

Therefore, the result is \Large\boxed{(3,~3)}

Hope this helps!! :)

Please let me know if you have any questions

5 0
2 years ago
Read 2 more answers
GIVING BRAINILEST-
Nutka1998 [239]

Answer:

To construct a dilation, we need the center of dilation, which is the point from which we make the image smaller or larger. We also need a scale factor, which is the number that determines how large or small we are making the new image. If the scale factor is greater than 1, we have an enlargement, because the new image will be larger than the original image. If the scale factor is less than 1, we have a reduction because the new image will be smaller than the original image.

Enlargement

Step-1

Let's dilate triangle PQR by a scale factor of 2 with a center of dilation at point C.

Step-2

First, we will use a straightedge to connect the center of dilation C to each vertex. We are going to extend those lines across the whole page.

Step-3

Next, we'll place the point of the compass on the center of dilation C and the pencil on vertex Q and measure that distance.

Step-4

Now, without changing the size of the compass, we will move the point of the compass to vertex Q, and make a mark on the line that is extended through Q.

Step-5

Notice that since we measured the distance from C to Q and then used that same distance to mark the line, we have now created a new point that is twice the distance from C that Q was.

We now repeat the process for each of the other vertices of the triangle.

Step-6

These marks represent the vertices of our dilated image. We often use the same letters with an apostrophe to show that it is the corresponding vertex.

Step-7

Finally, we use the straightedge to connect all of the new vertices.

Now,See how triangle P

′

Q

′

R

′

is twice the size of the original triangle.

Reduction

Step-1

Given triangle PQR, now let's dilate the image with a center of dilation at C and with a scale factor of

2

1

Step-2

Using the straightedge, we will connect all of the vertices to the center of dilation C. In this case, since our scale factor is less than 1, we are reducing the size of the triangle.

Step-3

The green lines are the new lines that we have drawn.

Since we have to reduce the size of the triangle by half, we want to cut the distance in half from each vertex to the center of dilation. In order to do this, we will set our compass to a distance that is more than halfway between vertex P and point C but less than the total distance.

Step-4

Keeping the compass at that same distance, start with the point of the compass on vertex P, then draw a curve that intersects the line between P and C. Next, keeping the compass at the same distance, place the point of the compass on C and draw a curve that intersects the line between P and C and intersects the previously drawn curve.

Step-5

Now, we will take the straightedge and connect the points where the curves overlap.

4 0
3 years ago
Keiko received an $80 gift card for a coffee store. She used it in buying some coffee that cost $8.04 per pound. After buying th
Kryger [21]

Answer:

6 pounds of coffee.

Step-by-step explanation:

First, we determine how much money she spent. 80-31.76 = 48.24

Next, we divide the money spent by the price per unit. 48.24/8.04=6

Therefore, she bought 6 pounds of coffee.

3 0
3 years ago
Read 2 more answers
Other questions:
  • Bryson buys a bag of 64 plastic miniature dinosaurs.could he distribute them equally into six storage containers and not have an
    6·1 answer
  • Perform the following operation. Reduce if possible<br> X/9+y + 9+y/x
    7·1 answer
  • Of 40 people surveyed, 40% named grape juice as their favorite juice.
    5·2 answers
  • Given that y=5x + 12 determine the value of y when x=4
    8·2 answers
  • what is the area of the rectangle wxyz with vertices w(0,1),x(3,4),y(-1,8),z(-4,5) to the nearest unit
    7·1 answer
  • Can someone please help me:)
    5·1 answer
  • Please Help!
    9·1 answer
  • If you translate the cubic parent function, F(x) = x3, down 2 units, what is the
    6·1 answer
  • As a fraction (all of the bellow) <br> 6.7<br> -0.92<br> -0.1<br> -5.05
    9·2 answers
  • What is the sum of 7x/ x^2 - 4 and 2 / x + 2?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!