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Author Question: Which of the following levels of significance would decrease the chance of a Type II error? a. ... (Read 23 times)

audragclark

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Which of the following levels of significance would decrease the chance of a Type II error?
 
  a. 0.10
  b. 0.05
  c. 0.01
  d. 0.001

Question 2

Under what condition would the mean, median, and mode be equal?
 
  a. Equal range and standard deviation
  b. Multimodal distribution of scores
  c. Normal distribution of scores
  d. Small variance



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nanny

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Answer to Question 1

ANS: A
There is a greater risk of Type II error with more extreme levels of significance and less of a risk with more conservative levels of significance.
There would be less chance of a Type II error with 0.05 than with 0.01 or 0.001, but 0.10 is even more liberal. There is a greater risk of Type II error with more extreme levels of significance and less of a risk with more conservative levels of significance.
There would be less chance of a Type II error with 0.01 than with 0.001, but 0.10 is even more liberal. There is a greater risk of Type II error with more extreme levels of significance and less of a risk with more conservative levels of significance.
This is the most restrictive answer and therefore the greatest risk of a Type II error. There is a greater risk of Type II error with more extreme levels of significance and less of a risk with more conservative levels of significance.

Answer to Question 2

ANS: C
In the normal curve, the mode, median, and mean are equal.
Equal range and standard deviation would not mean that the mode, median, and mean are equal. In the normal curve, the mode, median, and mean are equal.
With a multimodal distribution, the mean, median, and mode would not be equal. In the
normal curve, the mode, median, and mean are equal.
A small variance would not mean that the mode, median, and mean are equal. In the normal curve, the mode, median, and mean are equal.




audragclark

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  • Posts: 579
Reply 2 on: Jul 8, 2018
Excellent


JaynaD87

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Reply 3 on: Yesterday
YES! Correct, THANKS for helping me on my review

 

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