Consider Mary's experiment regarding whether learning of 6th graders on a math lesson is affected by background noise level. Mary has collected her data. What is the null hypothesis for her study? What is the alternative hypothesis for her study? What are the assumptions that must be met about her data before she can correctly use an independent t-test to test the hypotheses? Why? How would she see if her data met these assumptions? How much room does she have to violate any of these assumptions and still get accurate results from the t-test? Explain and support your answers

Check the answers to the questions below

Step-by-step explanation:

a) If is the average learning rate of the 6th graders without background noise level and is the average learning rate of the 6th graders with background noise level

The Null Hypothesis is that the learning rate of the 6th graders is not affected by the background noise level.

Null hypothesis,

b) The Alternative Hypothesis is that the learning rate of the 6th graders is affected by the background noise level.

Alternative hypothesis,

c) Assumptions that must be met about the data before she can correctly use independent t-test

There must be random selection of the 6th graders

That the two groups are normally similar in their learning abilities

The division of students into the two groups should be at random

d) She has to  make these assumptions to prevent bias and inaccuracy of results. If these assumptions are not made, the outcome of the experiment may not reflect the true effect of background noise on the learning of the 6th graders.

She can still get accurate results if she include some bias in the selection to prove a particular result.

5.5

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)