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fgiga [73]
3 years ago
11

Testing for a disease can be made more efficient by combining samples. If the samples from people are combined and the mixture t

ests​ negative, then samples are negative. On the other​ hand, one positive sample will always test​ positive, no matter how many negative samples it is mixed with. Assuming the probability of a single sample testing positive is ​, find the probability of a positive result for samples combined into one mixture. Is the probability low enough so that further testing of the individual samples is rarely​ necessary?
Mathematics
1 answer:
Snezhnost [94]3 years ago
7 0

Answer:

The probability of testing positive for one is 0.20.

The probability of testing negative for one sample is (1-0.2)=0.8.

We only save time when all five are negative, which has a probability of 0.8^5=0.32768.

This means that the expected number of tests is

combined sample tests negative = 1 with probability 0.32768

combined sample tests positive = 1+5 retests = 6 with probability 0.67232

Expected number of tests

=Σ nipi / n

=(1*0.32768+6*0.67232)/5 [divide by 5 because we tested 5 samples]

= 0.87232 < 1

So yes, there is a saving.

Note: In practice, all medical tests are not absolute, i.e. they give false-positives(α) and false-negatives (β). The ratios 1-α and 1-β are respectively measures of specificity and sensitivity.

These two parameters complicate the simplistic evaluation above.

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4 years ago
Andrew found a pan of brownies with 3/4 of the brownies left. He ate 1/3 of what was left. What portion of the whole brownies di
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Answer:

1/4

Step-by-step explanation:

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3 years ago
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lesantik [10]

Answer:

a) The sampling distribution for n=540 has a mean sample proportion of p=0.37 and a standard deviation of σs=0.0208.

b) probability = 0.99998

d) probability = 0.99874

e) You gain 0.12% in probability for an increase of 80% in sample size.

The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).

Step-by-step explanation:

a) We have a known population proportion π=0.37 and we have to describe the sampling distribution when the sample size is n=540.

The mean sample proportion is expected to be the same as the population proportion:

\bar p = \pi = 0.37

The standard deviation of the sampling will be the population standard deviation divided by the square root of the sample size:

\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{540}}=\sqrt{0.000431667}=0.0208

Then, we can say that the sampling distribution will have a p=0.37 and a standard deviation σs=0.0208.

b) We have to calculate the probability that the sample proportion will be within 0.09 of the population proportion.

We can calculate the z-value as:

z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0208}=4.3269\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0208}=-4.3269

As the distribution is symmetrical, we can calculate the probabilty that he sample proportion will be within 0.09 of the population proportion as:

P(|p-\bar p|

probability = 0.99998

d. Now the sample is smaller (n=300), so the standard deviation of the samping distribution:

\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{300}}=\sqrt{0.000777}=0.0279

We have to recalculate the z-scores:

z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0279}=3.2258\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0279}=-3.2258

And the probability is:

P(|p-\bar p|

probability = 0.99874

e. The increase in sample size is 80%

\Delta n\%=\dfrac{n_1}{n_2}-1=\dfrac{540}{300}-1=1.8-1=0.8=80\%

and the increase in probability is 0.12%

\Delta P\%=\dfrac{P_1}{P_2}-1=\dfrac{0.99998}{0.99874}-1=1.0012-1=0.0012=0.12\%

You gain 0.12% in probability for an increase of 80% in sample size.

The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).

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Answer:

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