H 0 : μ = 7.3.H (^) A : μ ≠ 7.3.
II . We will use a one-sample t confidence interval (at the direction of the problem—otherwise
we would most likely have chosen to do a one-sample t significance test) at α = 0.05, which,
for the two-sided test, is equivalent to a confidence level of 0.95 (C = 0.95). We will
assume that the ratings are a random sample of the population of all ratings. The sample size
is small, but we are told that the ratings are approximately normally distributed, so that the
conditions necessary for the inference are present.
III . = 6.1, s = 2.7. For df = 12 – 1 = 11 and a 95% confidence interval, t * = 2.201 A 95%
confidence interval for μ is given by:
(Note : A one-sample t test for these data yields P -value = 0.15.)
IV . Since 7.3 is in the interval (4.384, 7.816), we cannot reject H 0 at the 0.05 level of
significance. We do not have good evidence that there has been a significantly significant
change in the popularity rating of the show after its move to Wednesday night.
(a) df = min{23 – 1,27 – 1} = 22 t * = 1.717.
(b) df = 22 t * = 2.074.
(c) df = 23 + 27 –2 = 48 t * = 1.684 (round down to 40 degrees of freedom in the table).
(d) df = 48 t * = 2.021.Solutions to Cumulative Review Problems
a.
b. 2.
(The exact binomial given by the TI-83/84 is 1-binomcdf(250,0.6,160) = 0.087 . If you
happen to be familiar with using a continuity correction (and you don’t really need to be) for the
normal approximation, normalcdf (160.5,1000,150,7.75) = 0.088 , which is closer to the
exact binomial.)- . (On the TI-83/84, this is equivalent to 1-
binomcdf(5,0.3,2).)- P (the worker is an administrator | the worker is female)