Stat 371-003, Solutions to Homework #6
1. 7.31 (pg 245)
The estimated standard error of the difference between the sample means is
r(8.73)2
5+(7.19)2
5≈5.06
(a) Our test statistic is t= (¯y1−¯y2)/ˆ
SE(¯y1−¯y2)≈(31.72 −29.22)/5.06 ≈0.494.
For a test at significance level α= 0.10, we need the 95th percentile of a tdistribution
with 7.7 degrees of freedom. From the table at the back of the book, we get 1.86. In R,
we get qt(0.95, 7.7) ≈1.87.
Since 0.494 <1.87, we fail to reject the null hypothesis.
Note that we could calculate the P-value in R as 2*(1-pt(0.494, 7.7)) ≈0.64.
(Using the table, we would just see that the P-value is >0.4.)
(b) The result should not be surprising. Such data could reasonably ascribed to chance
variation. If the underlying averages were the same, we would get data like this most
of the time.
2. 7.36 (pg 246)
(a) True (we reject H0), since the P-value = 0.07 < α = 0.10.
(b) True (we fail to reject H0), since the P-value = 0.07 > α = 0.05.
(c) False! The P-value is the chance of getting data this extreme (or more so), if µ1=µ2.
First, it doesn’t concern a statement about ¯y1and ¯y2. Second, it concerns the chance of
data given µ1=µ2, not the other way around.
3. 7.44 (pg 255)
Yes, we would reject H0. The 95% confidence interval is the set of plausible values for
µ1−µ2, and so is the set of values, δ, for which a test of H0:µ1−µ2=δwould not
be rejected. Since 0 is not in the confidence interval, we would reject the hypothesis that
µ1=µ2.
4. 7.51 (pg 264–265)
Our estimate of the standard error of ¯y1−¯y2is
r(0.621)2
8+(0.652)2
8≈0.318
Our test statistic is t= (¯y1−¯y2)/ˆ
SE(¯y1−¯y2)≈(6.169 −5.291)/0.318 ≈2.758.
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