for each day, the conditions for the chi-square test are present.III . P- value < 0.15 (from Table C). (χ 2 cdf(7.533.1000,4) = 0.11; or, you could use the χ 2 GOF–
Test in the STAT TESTS menu of a TI-84 if you have it.)
IV . The P -value is larger than any commonly accepted significance level. We do not have strong
evidence that there are more defects produced on Monday and Friday than on other days of the
week.
Because we have a single population from which we drew our sample and we are asking if two
variables are related within that population, this is a chi-square test for independence.
I. H 0 : Type of respondent and opinion toward the legalization of marijuana are independent.H (^) A : Type of respondent and opinion toward the legalization of marijuana are not independent.
II . We will do a χ 2 test of independence. The following table gives the observed and expected values
for this situation (expected values are the second row in each cell; the expected values were
calculated by using the MATRIX function and the χ 2 –Test item in the STAT TESTS menu):
Since all expected values are at least 5, the conditions necessary for the chi-square test are present.
III. , df = (3 – 1)(3 – 1) = 4 0.025 < P -value
< 0.05 (from Table C).
(Using the TI-83/84, P -value = χ 2 cdf(10.27,1000,4)=0.036.)
IV . Because P < 0.05, reject H 0 . We have evidence that the type of respondent is related to opinion
concerning the legalization of marijuana.
Solutions to Cumulative Review Problems
The t -test statistic for the slope of the regression line (under H 0 : β = 0) is 6.09. This translates into
a P -value of 0.002, as shown in the printout. This low P -value allows us to reject the null
hypothesis. We conclude that we have strong evidence of a linear relationship between X and Y .
- For C = .99 (a 99% confidence interval) at df = n – 2 = 6, t * = 3.707. The required interval is