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3 years ago

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A researcher uses a repeated-measures design to compare individuals’ performance before treatment with their performance after t

reatment. If all the participants show improved performance of 8 or 9 points after treatment, what should the researcher find _______
a) a sample mean difference near zero.
b) the statistic near zero.
c) the variance of the difference scores is near zero.
d) none of the other options is correct.

1 answer:

Alenkasestr [34]3 years ago

5 0

Answer:

c. the variance of the difference scores is near zero

Step-by-step explanation:

If all the participants show improved performance of 8 or 9 points after treatment, what should the researcher find - the variance of the difference scores is near zero.

But this can be true only when the original scores had a low variance.

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Step-by-step explanation:

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Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

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Complete Question

The complete question is shown on the first uploaded image

Answer:

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Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

So

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

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