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kumpel [21]
3 years ago
9

A teacher wants to see if a new unit on fractions is helping students learn. She has five randomly selected students take a pre-

test and a post test on the material. The scores are out of 20. Suppose that you are about to compute a confidence interval for \mu_dμ d, how do you check for normality?
Mathematics
1 answer:
mylen [45]3 years ago
5 0

Answer:

Find the difference between the two scores for a number of sample distributions. Make a plot of the differences and check for outliers.

Step-by-step explanation:

Checking for Normality means basically checking if one's data distribution approximates a normal distribution.

A normal distribution is represented by a bell-shaped curve, peaking around the mean, indicating that all of the data spreads out from the mean.

Th original aim of the teacher is to check the effects of the particular added unit on the performance of students in the subject.

The teacher goes about this by testing the students before and after learning the unit.

The best way to compare of course, is to take a difference of the test scores for different samples. This first gives the idea of whether the newly introduced unit affects performance.

This set of differences is then checked for normality.

So, the best manner to make a plot of these differences. Like we mentioned earlier, a normal distribution is bell shaped. So, the plot of these differences would be a bell shaped curve if the distribution was normal and we wouldn't get a bell shaped curve if the distribution wasn't normal.

Checking for outliers help to eliminate part of data that can totally scatter the regular behaviour of the data distribution.

So, the best way for the teacher to check for normality is to find the difference between the two scores for a number of sample distributions. Make a plot of the differences and check for outliers.

Hope this Helps!!!

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bearhunter [10]
<h3>Answer:</h3>

x=2

<h3>Solution:</h3>
  • In order to isolate x, we should first of all take the square root of both sides.
  • If we take the square root of the left-hand side, we will get
  • x-\displaystyle\frac{1}{2}
  • How about the right-hand side? Well, we should take the square root of the numerator (9) and the denominator (4)
  • So we have
  • x-\displaystyle\frac{1}{2} =\frac{3}{2}}
  • Move -1/2 to the right:
  • x=-\displaystyle\frac{3}{2} +\frac{1}{2}
  • x=\displaystyle\frac{4}{2}
  • x=2

Hope it helps.

Do comment if you have any query.

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2 years ago
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tangare [24]

Answer:

X>0, so the second one

Step-by-step explanation:

Since the point on 0 is OPEN that means the number (0) is NOT part of the solution. Therefore, X is greater than 0.

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