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Umnica [9.8K]
3 years ago
15

Repeated-measures experiments measure the same set of research participants two or more times, while matched-subjects experiment

s study participants who are matched on one or more characteristics. Which of the following are true for both a repeated-measures experiment and a matched-subjects experiment when used to compare two treatment conditions?
a. The researcher computes difference scores to compute a t statistic.
b. If the researcher has n number of participants to use in the experiment, then the degrees of freedom will be the same in a repeated-measures experiment or in a matched-subjects experiment.
c. The researcher must compute an estimated standard error for the mean difference score to compute a t statistic.
d. Participants in both types of experiments are all measured the same number of times.
Mathematics
1 answer:
levacccp [35]3 years ago
8 0

Answer:

c. The researcher must compute an estimated standard error for the mean difference score to compute a t statistic.

Step-by-step explanation:

As already defined that repeated-measures experiments measure the same set of research participants two or more times, while matched-subjects experiments study participants who are matched on one or more characteristics.

Both repeated-measures experiment and a matched-subjects experiment when used to compare two treatment conditions use the same t statistic. Therefore, researcher must compute an estimated standard error for the mean difference score to compute a t statistic.

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Step-by-step explanation:

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The value of y directly varies with the value of x. If y equals 21when x is 14, what is the value of y when x = 20?
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3 years ago
A student records the number of hours that they have studied each of the last 23 days. They compute a sample mean of 2.3 hours a
natita [175]

Answer:

the standard deviation increases

Step-by-step explanation:

Let x₁ , x₂, .   .   .  , x₂₃ be the actual data observed by the student

The sample means  = x₁  +  x₂  +  .   .   .  , x₂₃ / 23

= \frac{x_1 +x_2 +...x_2_3}{23}

= 2.3hr

⇒\sum xi =2.3 \times 23 = 52.9hrs

let x₁ , x₂, .   .   .  , x₂₃  arranged in ascending order

Then x₂₃ was 10  and has been changed to 14

i.e x₂₃ increase to 4

Sample mean  = \frac{x_1 +x_2 +...x_2_3}{23}

\frac{52.9hrs + 4}{23} \\\\= \frac{56.9}{23} \\\\= 2.47

therefore, the new sample mean is 2.47

2) For the old data set

the median is x_1_2(th) values

[\frac{n +1}{2} ]^t^h value

when we use the new data set only x₂₃ is changed to 14

i.e the rest all observation remain unchanged

Hence, sample median = [{x_1_2]^t^h value remain unchange

sample median = 2.5hrs

The Standard deviation of old data set is calculated

=\sqrt{\frac{1}{n-1} \sum (xi - \bar x_{old})^2 } \\\\=\sqrt{\frac{1}{22}\sum ( xi - 2.3)^2 }---(1)

The new sample standard sample deviation is calculated as

= \sqrt{\frac{1}{n-1} \sum (xi-2.47)^2} ---(2)

Now, when we compare (1) and (2)  the square distance between each observation xi and old mean is less than the squared distance between each observation xi and the new mean.

Since,

(xi - 2.3)²  ∑ (xi - 2.47)²

Therefore , the standard deviation increases

6 0
4 years ago
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