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LekaFEV [45]
1 year 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]1 year 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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Let X denote the number of bars of service on your cell phone whenever you are at an intersection with the following probabiliti
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Answer:

Step-by-step explanation:

Hello!

Given the variable

X: number of bars of service on your cell phone.

The posible values of this variable are {0, 1, 2, 3, 4, 5}

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P(X): 0.15; 0.25; 0.25; 0.15; 0.1; 0.1

F(X): 0.15; 0.4; 0.65; 0.8; 0.9; 1

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b.

The mean of the sample is:

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c.

P(X<2)

You can read this expression as the probability of having less than 2 bars of service. This means you can have either one or zero service bars, you can rewrite it as:

P(X<2)= P(X≤1)= F(1)= 0.4

d.

P(X≤3.5)

This expression means "the probability of having at most (or less or equal to) 3.5 service bars", this variable doesn't have the value 3.5 in its definition range so you have to look for the accumulated probability until the lesser whole number. This expression includes the probabilities of X=0, X=1, X=2, and X=3, you can express it as the accumulated probability until 3, F(3):

P(X≤3.5)= P(X≤3)= F(3)= 0.80

I hope it helps!

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2 years ago
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Answer:

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Answer:

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