For the following data,
(a) justify the use of a two-proportion z -test for H 0 : p 1 – p 2 = 0.
(b) what is the value of the test statistic for H 0 : p 1 – p 2 = 0?
(c) what is the P -value of the test statistic for the two-sided alternative?Answer:
(a) n 1 1 = 40(0.3) = 12, n 1 (1 – 1 ) = 40(0.7) = 28,
n 2 2 = 35(0.4) = 14, n 2 (1 – 2 ) = 35(0.6) = 21.Since all values are at least 5, the conditions are present for a two-proportion z -test.
(b) .
(c) z = –0.91, P -value = 2(0.18) = 0.36 (from Table A). On the TI-83/84, this P -value can be found
as 2 × normalcdf(-100,-0.91) .
You want to conduct a one-sample test (t- test) for a population mean. Your random sample of size 10
yields the following data: 26, 27, 34, 29, 38, 30, 28, 30, 30, 23. Should you proceed with your test?
Explain.
Answer: A boxplot of the data shows that the 38 is an outlier. Further, the dotplot of the data casts
doubt on the approximate normality of the population from which this sample was drawn. Hence, you
should not use a t -test on these data.
Although it may be difficult to justify, there are conditions under which you can pool your estimate of
the population standard deviation when doing a two-sample test for the difference between
population means. When is this procedure justified? Why is it difficult to justify?
Answer: This procedure is justified when you can assume that the population variances (or