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LekaFEV [45]
2 years ago
6

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 probability of later developing the white blood cell disease acute lymphoblastic leukemia (ALL), the most common cancer of children. They compared 1272 children with ALL to 6238 children without ALL. Of the ALL kids, 1020 had significant social activity outside the home (including day care) when younger. Of the kids without ALL, 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 families.
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:
USPshnik [31]2 years ago
8 0

Answer:

a) This is an Observational Study because in this kind of study investigators observe subjects and measure variables of interest without assigning treatments to the subjects. Here, the Gilham et al. (2005) studied two different groups where no treatment or intervention was done. These groups were independent of each other.

b) proportions of children with significant social activity in children with acute lymphoblastic leukemia = 1020/1272 = 0.80

proportions of children with significant social activity in children without acute lymphoblastic leukemia = 5343/6238 = 0.86

c) Odds ratio can be calculated using the following formula:

OR= \frac{a/b}{c/d}

where: a - Number in exposed group with positive outcome(here this means number of children with significant social activity associated with acute lymphoblastic leukemia)

b- Number of children without social activity having with acute lymphoblastic leukemia

c- Number of children with social activity having without acute lymphoblastic leukemia

d- Number of children without social activity having without acute lymphoblastic leukemia

OR= \frac{1020/252}{5343/895}

OR= 0.6780

d) The 95% confidence interval of this Odds Ratio is 0.5807 to 0.7917.

e) Since the odds ratio lies in this confidence interval indicate that the amount of social activity is associated with acute lymphoblastic leukemia. The children with more social activity have a higher occurrence of acute lymphoblastic leukemia.

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charmaine and tesha each have a number of cards, in the ratio 2:3. tesha and holly have a number of cards in the ratio 2:1. if t
Sav [38]
C:t=2:3
t:h=2:1

multiply first equation by 2 and 2nd by 3
c:t=4:6
t:h=6:3
so
c:t:h=4:6:3

if t is 4 more than c
t is 6 units and c is 4
6-4=2
so 4=2 units
2=1 unit

hollly is 3 units
3*2=6
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3 0
2 years ago
1).He gastado 5/8 de mi dinero. Si en lugar de gastar los 5/8 hubiera gastado los 2/5 de mi dinero, tendria ahora 72 soles mas d
TiliK225 [7]

Answer:

1. The amount that was not spent is: 1080 soles

2. The answer is 5.

Step-by-step explanation:

Question 1. There are two situations:

If you spent 5/8 of the money then notice, that u did not spend 3/8.

If you spend 2/5 of the money, then you would have 72 more.

So think, this equation:

3/8x + 72 = 2/5x

2/5 of the money is the value for what you did not spend plus 72, the amount you would have, where x is the answer for the total of money.

72 = 2/5x - 3/8x

72 = 1/40x

x = 2880 soles

As you did not spend 3/8 of 2880, the answer is 1080 soles

Notice that if u spent 2/5 of 2880, you would have 1152 soles, so, the 1080 + 72, as the problem said.

Question 2.

We think letters for this excersise.

N = My born year

2020 (this year) - N = E (my age, now)

So N + 10 = when I get 10 years old

N + 20 = when I get 20 years old

N + 30 = when I get 30 years old

N + 40 = when I get 40 years old

The problem says:

(N+10) + (N+20) - (N+E) = 2020

((N+30)+E) - (N-40) = X  

So if 2020 - N = E, then notice that 2020 - E = N. Let's replace this in the equation form:

((2020-E)+10) + ((2020-E)+20) - ((2020-E)+E) = 2020

No minus in the first two terms, so we can break the ( )

2020-E+10 + 2020-E+20 - ((2020-E)+E) = 2020

We apply the distributive property for the second term

2020-E+10 + 2020-E+20 - 2020+E-E = 2020

We can cancel two E, and the 2020, so the new form will be:

2020 - 2E + 30 = 2020

We can also cancel the 2020, so if we reorder the equation, we have:

-2E = -30

E = 15 That's my age, so my born's year is 2020 - 15 = 2005 (N)

Look:

(2005 + 10) + (2005+20) - (2005+15) = 2020

So now, the last part

((2005+30)+15) - (2005+40) = 5

3 0
2 years ago
The sum of 23.8 and a number is 28.17, as shown below. What number should go in the box to complete the addition problem?
gregori [183]
Since there is no picture to show the box, all I can say is that 28.17- 23.8 = 4.37 and 23.8 + 4.37 = 28.17
5 0
2 years ago
Read 2 more answers
A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3
butalik [34]

Answer:

The answer is below

Step-by-step explanation:

A cell tower is located 3 miles east and 4 miles north of the center of a small town. The cell tower has a coverage radius of 3 miles.

a) Start by drawing a diagram of this situation. Your diagram might include a coordinate plane, the cell tower, and a circle representing the cell tower's coverage boundary.

a) A house is located 5 miles north of the center of the town and is to the east of the cell tower. If the house lies on the boundary of the cell tower's coverage, how far east of the center of the town is the house?

Answer:

Let the center of the town represent the origin (0,0). Also 1 unit = 1 mile. Since the cell tower is located 3 miles east and 4 miles north of the center of a small town, it is represented by A(3, 4)

The cell tower has a coverage of radius 3 miles. This can be represented by a circle with equation:

(x - a)² + (y - b)² = r². where (a,b) is the center of the circle and r is the radius. Hence:

(x - 3)² + (y - 4)² = 3²

(x - 3)² + (y - 4)² = 9

The diagram is drawn using geogebra.

b) The house is 5 miles north. It can be represented by y = 5 line.

To find the distance east of the house we have to substitute y = 5 and solve for x, hence:

(x - 3)² + (y - 4)² = 9

(x - 3)² + (5 - 4)² = 9

(x - 3)² + 1 = 9

(x - 3)² = 8

(x - 3) = √8

x - 3 = ±2.83

x = 3 ± 2.83

x = 5.83 or 1.83

Since it is to the east to the cell tower, hence x = 5.83.

Therefore the house is located 5.83 miles to the east of the cell tower

8 0
2 years ago
Of 162 students honored at an academic awards banquet, 48 won awards for mathematics and 78 won awards for English. There are 14
ella [17]

Answer: 56/81

Step-by-step explanation:

in the attachment

3 0
2 years ago
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