With a one-sample t-test, if we compute 95% confidence limits on the mean as 112.5 & 118.4, we can conclude that

Question 46 options:

a)	the probability is .95 that the population mean lies between 112.5 and 118.4.

b)	the probability is .95 that the sample mean lies between 112.5 and 118.4.

c)	the probability is .05 that the population mean lies between 112.5 and 118.4.

d)	the probability is .05 that the sample mean lies between 112.5 and 118.4.

When we reject the null hypothesis in a oneway analysis of variance we can conclude that

a)	at least one of the group means is different from at least one other mean.

b)	all of the group means are the same.

c)	MSerror is greater than MSbetween by more than would be expected by chance alone.

d)	all of the group means are different.

How do you calculate the degrees of freedom for the chi-square analysis of contingency tables?

a)	(k-1)(N-1)

b)	(R-1)(C-1)

c)

k-1

d)

N-1

An IQ score is a standardized score, with a mean of 100 and a standard deviation of 15. Based on what you know about the percentage of scores in a normal distribution further away than 2 standard deviations above or below the mean, what is the proportion of people who took the IQ test who would have the IQ scores either below 70 or above 130?

a)	Approximately 15%

b)	Approximately 1%

c)	Approximately 5%

d)	Approximately 32%

Over the past 20 years at SCU, the university has been closed for 30 days due to inclement weather. Nine closures have happened during the Fall terms, 15 during the Winter terms, and 6 during the Spring terms. If the chi-square test performed for this analysis was not statistically significant, which of the following would represent the most proper conclusion regarding these campus closures?

a)	there has been a statistically significant departure from chance distributions of closures over the terms

b)	the closures have been equally distributed over the terms

c)	the closures have not been equally distributed over the terms

d)	closures are not independent of terms