Suppose a test of musical ability has a normal distribution with a mean of 50 and a standard deviation of 5. Approximately what percentage of people are over 55?

Question

a. 6%
b. 16%
c. 45%
d. 84%

A z-score of 1.75 indicates how many standard deviation units above the mean?

a. 1.75
b. 3.5
c. 4.75
d. 2.25

If you have a confidence interval of 112 to 136,

a. All of the subjects scores fall within this range
b. You have an effect size of 12
c. You can be 95% sure that this interval includes the true population mean
d. You can be 95% sure that the results did not happen by chance.

If the significance level for a two-tailed test is .05, what is the corresponding significance level for each tail of the test?

a. .05
b. .025
c. .075
d. .1

Answer 1

Z score is your score in terms of standard deviations.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions desired.

If you are talking about a 95% confidence interval, C.

.05/2 = ?