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In: Statistics and Probability

Question 7 Chi Square tests have an assumption of normality True False Question 8 Both ANOVA...

Question 7

Chi Square tests have an assumption of normality

  1. True
  2. False

Question 8

Both ANOVA and Chi square tests are always right tail tests

  1. True
  2. False

Question 9

For a Chi Square test of Association, you ask if mouse hair color (black, grey or white) is associated with whisker length (short, medium or long). The df for this analysis would be: Give answer as X

Question 10

For ANOVA, you only complete a Tukey's means comparisons test if the overall F* value was significant (i.e., p<0.05)

  1. True
  2. False

Solutions

Expert Solution

Question :7

Chi Square tests have an assumption of normality

Answer: True

Reason: The Chi square test for single variance has an assumption that the population from which the sample has been is normal. This normality assumption need not hold for chi square goodness of fit test and test for independence of attributes.

Question 8

Both ANOVA and Chi square tests are always right tail tests

Answer: True

Reason:

The tail refers to the end of the distribution of the test statistic for the particular analysis that you are conducting. ... This means that analyses such as ANOVA and chi-square tests do not have a “one-tailed vs. two-tailed” option, because the distributions they are based on have only one tail.

Question 9

For a Chi Square test of Association, you ask if mouse hair color (black, grey or white) is associated with whisker length (short, medium or long). The df for this analysis would be: Give answer as X

Answer:

Df= #( coloum-1)* (Rows-1)

Question 10

For ANOVA, you only complete a Tukey's means comparisons test if the overall F* value was significant (i.e., p<0.05)

Answer: True

Reason:  

Tukey test controls the experimentwaise error rate. A part of the cost is that individual pairwise comparisons are significant at a p-value less than 0.05. So a pairwise comparison that is significant with a t-test at a p-value of 0.05 will be non-significant with Tukey test. If you have a large number of treatments and are not interested in all pairwise comparisons, then Tukey will be too conservative.


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