Download Answers to Homework set #4

Ch 21. #11 a) one-tailed because I would like to prove that the
population proportion is too low, rather than too low or too high.
B) probability of rejecting a true null. The null hypothesis is that
the proportion is .27 and the alternative is that it is lower (smaller)
than .27. so this would be the probability of deciding based on the
(unusual/misleading) data that the proportion of minorities is too
low (i.e. below .27) when really it is truly .27.
c) probability of accepting a false null. So this would be the
probability of deciding that .27 (or 27%) are minorities when
something else is true, like perhaps only 22% are minorities or
18% or…..many beta errors.
D) power is 1-beta. So it is doing the right thing (opposite of beta
error). It is the probability of rejecting a false null. So it is the
probability of stating that less than .27 (or 27%) are minorities
when some proportion smaller (in the population) is the truth.
E) when the p-value is less than alpha we reject the hypothesis so
if alpha goes from .01 to .05 it is easier to reject the null
hypothesis. So if you can reject it more often you are more likely
to reject a false null hypothesis which is good so that power must
increase.
F) since power is a good thing and it is good to get more data then
only using 37 rather than 87 employees should result in the loss of
power for the test. G) n=37 and p-hat=.19 and 95% using 2 then
.19 +- 2* (sqrt ((.19*.81)/37))=.19+-.13=(.06, .32)
To test use a z-score of (.19-.27)/(sqrt((.27*.73)/37)) = -1.09
Using the A-51 normal tables we get a p-value of .1379
comparing this to an alpha=.05 says we find supportive evidence
for p=.27.
Ch. 23 #9 98% is there. a) original distribution normal by
histogram, independence n/N<.10
b) 98.28+-t(df=51)*(.68/sqrt(52))=98.28+-2.403 *.094=98.28+.226 = (98.03, 98.5) approximately.
C) we are 98% sure that the mean for all folks’ body temperature
is within that interval
d) if we picked 52 people again and again, say 100 times then we
would make 98 intervals that have the mean inside them
e) a test of the mean is 98.6 against it is not 98.6 would produce a
t-score of (98.28-98.6)/.094 =-3.4, look for df=52-1=51 (closest is
50) and find a number near 3.4 in A-53..that is something larger
than 2.678 so we know the p-value (looking up to the top of A-53
under two tail) is less than .01. We reject the idea that the mean is
98.6 since .01<.05 (alpha I suggested). So, this data makes us
conclude that the normal temperature is not 98.6
#11 a) the less sure we need to be the narrower we can make the
interval, so 90% is less sure so I could build a narrower interval, it
would be 98.28 +-1.676*.094 = (98.1, 98.4)
b) the more sure interval makes it more likely that we have the
true (all people’s) mean body temperature identified but at the
sacrifice of having to all more ‘wiggle room’ about where that
value really lies.
c) more data means more information meaning we can be more
sure and/or narrow our interval. If we compare this new 98%
interval made from 500 people it will be narrower than the last
based on only 52 people.
D) using the approximate formula I derived in class of
(2s/MOE)*(2s/MOE) we get ((2*.68)/.1)*((2*.68)/.1)=184.96
which you should round up to 185 people. Deveaux uses a more
exact formula to get 252.
Ch. 24 #7 from the authors’ website (via our class website or
directly
http://media.pearsoncmg.com/aw/aw_deveaux_introstats_1/data/d
ata_index.html) I downloaded the cereal data and used DDXL.
Highlighting the 2 columns and choosing ‘confidence intervals’
and then 2-sample, with 95% confidence:
the sample mean for the
children’s is 46.85 and for
the adult’s is 10.367; we
know that the stdev of the
difference of 48.65 and
10.367 (or 36.483) is
gotten by using the sqrt of
(((7.67*7.67)/27)+
((6.6*6.6)/18)) =
sqrt(2.18+2.42)=2.14 (note
the 7.67 and 6.6 were gotten by using chart/histogram in DDXL.
The df=40 are shown by DDXL to left and come from the formula
in the footnote on p. 454. So t with df=40 and confidence of 95%
is 2.021 (check this in A-53). So
36.483+-2.021* 2.14 = (32.15, 40.82)
although not shown the histogram for the
adult cereal looks somewhat triangular
rather than bell-shaped and may pose a
concern about the truth of assumptions.
To test whether the null hypothesis that
there is no difference in the mean sugar
content of children and adult cereals
versus that the mean for the children’s is
larger (greater, higher) we will need a t-score =
(36.483-0)/2.14=17.05. Looking at A-53 with df=40 we see that
the closest number is 2.704….so the p-value is less than .01. with
.01<.05=alpha we definitely reject the idea that the mean sugar
contents are the same and accept/decide that the children’s mean
content is higher than the adult’s.