Download Chapter 6 HW Solutions 6.12 a) In this problem, both x and the

Chapter 6 HW Solutions
6.12
a) In this problem, both x and the margin of error is given. Since a 95% confidence interval is of the
form
x ± margin of error,
we can write down the confidence interval directly:
73 ± 8.
b) If you wanted more confidence, your margin of error would need to be larger. (For a given sample
size, larger margins of error translate to more confidence.)
6.13 Recall that the formula for the margin of error is
σ
z∗√
n
where σ is the population standard deviation, n is the sample size, and z ∗ is the number such that the
area under a N (0, 1) curve between −z ∗ and z ∗ is exactly C. For a 95% confidence interval, z∗ = 2,
and σ is fixed at 28. So as n ranges from 10 to 20, 40, and 80, what changes is just the denominator
of this expression. The resulting margins of error are
28
2√ ,
10
28
2√ ,
20
28
2√ ,
40
28
2√ ,
80
which reduce to
17.7,
12.52,
8.85,
6.26.
Note that these margins get smaller as n gets bigger.
6.14 The margin of error formula for a C−level confidence interval is
σ
margin of error = z ∗ √ ,
n
where σ is the population standard deviation, n is the sample size, and z ∗ is the number such that the
area under a N (0, 1) curve between −z ∗ and z ∗ is exactly C. In this problem, σ and n are given as 28
and 49, respectively, so the margins of error become
28
28
margin of error := z ∗ √ = z ∗
= z ∗ · 4.
7
49
The values of z ∗ that correspond to 80%, 90%, 95%, and 99% confidence intervals are 1.28, 1.645, 1.96,
and 2.575. The diagram (which I do not include here but will sketch on the board in class) shows that
the larger the desired confidence, the larger the confidence interval.
√
6.15 a) She needs to use σ/ n as the standard deviation, not σ itself. In other words, the confidence
interval should have been
2.2
x ± 1.96 √
= 8.6 ± 0.2.
500
b) Really, she means that the method she used to compute the interval would, if repeated again and
again with different samples, contain the true mean 95% of the time.
c) In fact, the interval she computed either contains the true mean or it doesn’t: there is 95% chance
that any interval constructed this way will contain the true mean, but we can’t say that there is
a 95% chance that this particular interval contains the true mean.
d) The population distribution does not depend on the sample size. What changes as the sample
size gets large is the distribution of x.
6.16
a) Cary forgot the square root: he should have reported
2.9
3.8 ± 1.96 √
100
b) The confidence interval is a range within which we hope to find the true mean of the population. If
the original population is normal with standard deviation 2.9, we know that 95% of the population
spend time in the range
µ ± 1.96 · 2.9,
but we don’t know what the true µ is, so we don’t know where this interval is located.
c) Remember that the margin of error is given by
σ
M oE = z ∗ √
n
If we wish to double the denominator by increasing n, then we need to increase n four-fold, as
√
√
4n = 2 n.
6.52
a) x is always know (it’s based on the data!) The hypothesis has to do with µ, the unknown
population mean. The null hypothesis should be
H0 : µ = 23.
b) The sample deviation of the sample mean should be
5
√
30
c) It doesn’t make sense to report statistical significance for a test when the data is completely at
odds with the alternative hypothesis. Since the alternative here is that the mean is bigger than 50,
and the sample mean was actually lower than 50, no statistical significance should be reported.
d) The logic of the hypothesis test is that we gather evidence against the null: if there is a lot of
evidence, we reject the null, if there is not, we fail to reject the null. But we never prove the null.
6.53
a) The null hypothesis is about the population mean, not the sample mean.
b) The null hypothesis is always of the form “no difference”. In this case, the appropriate null would
be H0 : µ = 21.2, with a one-sided alterntive Ha : µ > 21.2.
c) P −values are only meaningful when they are small. In general, any P −value larger than about
.1 is not small enough to cause us to reject the null hypothesis. (The value .1 is a bit arbitrary,
but not uncommon.)
d) We are not interested in the z−test statistic itself, but rather the P −value that is associated
with this statistic. (Remember that the P −value is the area under some appropropriate tail of a
N (0, 1) distribution.)
6.54
a) H0 : p = .88, Ha : p > .88.
b) H0 : µ = .75, Ha : µ > .75.
c) H0 : µ = 0, Ha : µ > 0.
6.55
a) H0 : µ = .77, Ha : µ 6= .77.
b) H0 : µ = 20, Ha : µ < 20.
c) H0 : µ = 880, Ha : µ < 880.
6.57
a) H0 : µ = 42800, Ha : µ > 42800.
b) H0 : µ = 0.4, Ha : µ 6= 0.4.
6.58
a) P = .0384
b) don’t report
c) P = .0768
6.59
a) don’t report
b) P = 0.0455
c) P = 0.0910
6.60 The phrase “lost less weight” suggests that both groups (the early eaters and the late eaters) lost
weight. Perhaps both groups were involved in a weight loss program? We don’t know. But we do
know that the probability that a difference in weight loss between the two groups was significant, in
the sense of it probably didn’t happen by chance. Specifically, there is only a 0.2% probability that a
weight loss difference this large could have been due to chance alone.
6.62 A P −value of 0.01 means that there is only a 1% probability that the differences in wealth between
graves with pig skulls and those without could have been caused by random chance.
6.73
a) Here, H0 : µ = 0 and Ha : µ 6= 0, where µ is the difference in the means of the computer
calculations and the driver calculations.
b) The mean of the data is x = 2.73. The z statistic is
z=
x−0
√ = 4.07
3/ 20
The P −value for this z−statistic is essentially 0 (it’s so small that it’s not on the N (0, 1) table.)