Math / Data & Statistics

QuestionIn a large clinical trial, 398,307 children were randomly assigned to two groups. The treatment group consisted of 197,816 children given a vaccine for a certain disease, and 37 of those children developed the disease. The other 200,491 children were given a placebo, and 146 of those children developed the disease. Consider the vaccine treatment group to be the first sample. Identify the values of $n_{1}, \hat{p}_{1}, \hat{q}_{1}, n_{2}, \hat{p}_{2}, \hat{q}_{2}, \bar{p}$ , and $\bar{q}$ . $\mathrm{n}_{1}=\square$

Studdy Solution

STEP 1

What is this asking? We need to calculate some important values from a clinical trial where one group got a vaccine and the other a placebo.
These values will help us understand how effective the vaccine was! Watch out! Don't mix up the groups!
Keep the vaccine group and the placebo group separate in your mind.
Also, remember that $\hat{p}$ represents the observed proportion of successes (developing the disease), and $\hat{q}$ is the observed proportion of failures (not developing the disease).

STEP 2

1. Define the variables for group 1 (vaccine).
2. Calculate $\hat{q}_1$ .
3. Define the variables for group 2 (placebo).
4. Calculate $\hat{q}_2$ .
5. Calculate $\bar{p}$ .
6. Calculate $\bar{q}$ .

STEP 3

The first group, $n_1$ , is the vaccine group.
We're told there are $n_1 = \mathbf{197,816}$ children in this group.

STEP 4

Out of these, $\mathbf{37}$ children developed the disease.
This gives us $\hat{p}_1$ , the proportion of children in the vaccine group who developed the disease.
So, $\hat{p}_1 = \frac{37}{197,816} \approx \mathbf{0.000187}$ .

STEP 5

$\hat{q}_1$ represents the proportion of children in the vaccine group who didn't develop the disease.
Since everyone either developed the disease or didn't, $\hat{p}_1$ and $\hat{q}_1$ add to one.
So, $\hat{q}_1 = 1 - \hat{p}_1 = 1 - 0.000187 \approx \mathbf{0.999813}$ .

STEP 6

The second group, $n_2$ , is the placebo group.
There are $n_2 = \mathbf{200,491}$ children in this group.

STEP 7

In this group, $\mathbf{146}$ children developed the disease.
So, $\hat{p}_2 = \frac{146}{200,491} \approx \mathbf{0.000728}$ .

STEP 8

Just like before, $\hat{q}_2$ is the proportion of the placebo group who didn't develop the disease.
So, $\hat{q}_2 = 1 - \hat{p}_2 = 1 - 0.000728 \approx \mathbf{0.999272}$ .

STEP 9

$\bar{p}$ is the overall proportion of children (from both groups) who developed the disease.
The total number of children who developed the disease is $37 + 146 = \mathbf{183}$ .
The total number of children in the study is $197,816 + 200,491 = \mathbf{398,307}$ .
Therefore, $\bar{p} = \frac{183}{398,307} \approx \mathbf{0.000459}$ .

STEP 10

$\bar{q}$ is the overall proportion of children who didn't develop the disease.
So, $\bar{q} = 1 - \bar{p} = 1 - 0.000459 \approx \mathbf{0.999541}$ .

STEP 11

$n_1 = \mathbf{197,816}$ , $\hat{p}_1 \approx \mathbf{0.000187}$ , $\hat{q}_1 \approx \mathbf{0.999813}$ $n_2 = \mathbf{200,491}$ , $\hat{p}_2 \approx \mathbf{0.000728}$ , $\hat{q}_2 \approx \mathbf{0.999272}$ $\bar{p} \approx \mathbf{0.000459}$ , $\bar{q} \approx \mathbf{0.999541}$

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Studdy Solution

STEP 1

STEP 2

1. Define the variables for group 1 (vaccine).
2. Calculate $\hat{q}_1$ .
3. Define the variables for group 2 (placebo).
4. Calculate $\hat{q}_2$ .
5. Calculate $\bar{p}$ .
6. Calculate $\bar{q}$ .

STEP 3

The first group, $n_1$ , is the vaccine group.
We're told there are $n_1 = \mathbf{197,816}$ children in this group.

STEP 4

Out of these, $\mathbf{37}$ children developed the disease.
This gives us $\hat{p}_1$ , the proportion of children in the vaccine group who developed the disease.
So, $\hat{p}_1 = \frac{37}{197,816} \approx \mathbf{0.000187}$ .

STEP 5

$\hat{q}_1$ represents the proportion of children in the vaccine group who didn't develop the disease.
Since everyone either developed the disease or didn't, $\hat{p}_1$ and $\hat{q}_1$ add to one.
So, $\hat{q}_1 = 1 - \hat{p}_1 = 1 - 0.000187 \approx \mathbf{0.999813}$ .

STEP 6

The second group, $n_2$ , is the placebo group.
There are $n_2 = \mathbf{200,491}$ children in this group.

STEP 7

In this group, $\mathbf{146}$ children developed the disease.
So, $\hat{p}_2 = \frac{146}{200,491} \approx \mathbf{0.000728}$ .

STEP 8

Just like before, $\hat{q}_2$ is the proportion of the placebo group who didn't develop the disease.
So, $\hat{q}_2 = 1 - \hat{p}_2 = 1 - 0.000728 \approx \mathbf{0.999272}$ .

STEP 9

STEP 10

$\bar{q}$ is the overall proportion of children who didn't develop the disease.
So, $\bar{q} = 1 - \bar{p} = 1 - 0.000459 \approx \mathbf{0.999541}$ .

STEP 11

$n_1 = \mathbf{197,816}$ , $\hat{p}_1 \approx \mathbf{0.000187}$ , $\hat{q}_1 \approx \mathbf{0.999813}$ $n_2 = \mathbf{200,491}$ , $\hat{p}_2 \approx \mathbf{0.000728}$ , $\hat{q}_2 \approx \mathbf{0.999272}$ $\bar{p} \approx \mathbf{0.000459}$ , $\bar{q} \approx \mathbf{0.999541}$

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Studdy Solution

STEP 1

STEP 2

1. Define the variables for group 1 (vaccine).2. Calculate q^1\hat{q}_1q^​1​.3. Define the variables for group 2 (placebo).4. Calculate q^2\hat{q}_2q^​2​.5. Calculate pˉ\bar{p}pˉ​.6. Calculate qˉ\bar{q}qˉ​.

STEP 3

The **first group**, n1n_1n1​, is the **vaccine group**.We're told there are n1=197,816n_1 = \mathbf{197,816}n1​=197,816 children in this group.

STEP 4

Out of these, 37\mathbf{37}37 children developed the disease.This gives us p^1\hat{p}_1p^​1​, the **proportion** of children in the vaccine group who developed the disease.So, p^1=37197,816≈0.000187\hat{p}_1 = \frac{37}{197,816} \approx \mathbf{0.000187}p^​1​=197,81637​≈0.000187.

STEP 5

STEP 6

The **second group**, n2n_2n2​, is the **placebo group**.There are n2=200,491n_2 = \mathbf{200,491}n2​=200,491 children in this group.

STEP 7

In this group, 146\mathbf{146}146 children developed the disease.So, p^2=146200,491≈0.000728\hat{p}_2 = \frac{146}{200,491} \approx \mathbf{0.000728}p^​2​=200,491146​≈0.000728.

STEP 8

Just like before, q^2\hat{q}_2q^​2​ is the proportion of the placebo group who *didn't* develop the disease.So, q^2=1−p^2=1−0.000728≈0.999272\hat{q}_2 = 1 - \hat{p}_2 = 1 - 0.000728 \approx \mathbf{0.999272}q^​2​=1−p^​2​=1−0.000728≈0.999272.

STEP 9

STEP 10

qˉ\bar{q}qˉ​ is the overall proportion of children who *didn't* develop the disease.So, qˉ=1−pˉ=1−0.000459≈0.999541\bar{q} = 1 - \bar{p} = 1 - 0.000459 \approx \mathbf{0.999541}qˉ​=1−pˉ​=1−0.000459≈0.999541.

STEP 11

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1. Define the variables for group 1 (vaccine).
2. Calculate $\hat{q}_1$ .
3. Define the variables for group 2 (placebo).
4. Calculate $\hat{q}_2$ .
5. Calculate $\bar{p}$ .
6. Calculate $\bar{q}$ .

The first group, $n_1$ , is the vaccine group.
We're told there are $n_1 = \mathbf{197,816}$ children in this group.

Out of these, $\mathbf{37}$ children developed the disease.
This gives us $\hat{p}_1$ , the proportion of children in the vaccine group who developed the disease.
So, $\hat{p}_1 = \frac{37}{197,816} \approx \mathbf{0.000187}$ .

The second group, $n_2$ , is the placebo group.
There are $n_2 = \mathbf{200,491}$ children in this group.

In this group, $\mathbf{146}$ children developed the disease.
So, $\hat{p}_2 = \frac{146}{200,491} \approx \mathbf{0.000728}$ .

Just like before, $\hat{q}_2$ is the proportion of the placebo group who didn't develop the disease.
So, $\hat{q}_2 = 1 - \hat{p}_2 = 1 - 0.000728 \approx \mathbf{0.999272}$ .

$\bar{q}$ is the overall proportion of children who didn't develop the disease.
So, $\bar{q} = 1 - \bar{p} = 1 - 0.000459 \approx \mathbf{0.999541}$ .