6.6) a) False
b)True
c) True
d) False

6.12) a) 48% is a sample statistic because it is a value that was found from the sample data, which is a smaller group than the whole population.
b)

SE <- sqrt(.48*(1-.48)/1259)
max <- .48 + SE * 1.96
min <- .48 - SE * 1.96

The confidence interval is(0.4524028, 0.5075972).
We are 95% confident that the true proportion of US residents who think marijuana should be legal is between 45% and 51%.
c) Fewer than 10% of the population were sampled. We don’t how people were sampled. If it was a simple random sampling, then we can assume that the observations are independent. There are at least 10 people who supported legalization and 10 people who opposed so the success failure condition is met. The sample size is large so we do not need to be very concerned with skew. We can assume that the normal model is a good approximation.
d) We cannot assume that the majority of Americans think marijuana should be legalized. We cannot assume this because the lower end of the 95% confidence interval is below 50%.
6.20) margin of error = 0.02
.02 = z* SE
z = 1.96
SE = \(\sqrt{p(1-p)/n}\)
n = 2397.2   You would need to sample 2398 people. (I rounded up since we are sampling people.)

6.28) \(p_{oregon} - p_{california}\) = 0.088- 0.080 = .008
SE = \(\sqrt{p1(1-p1)/n + p2(1-p2)/n}\)
SE = \(\sqrt{.088(1-.088)/4691 + .08(1-.08)/11545}\) 

SE <- sqrt(.088*(1-.088)/4691 + .08*(1-.08)/11545)

The confidence interval is (-0.0014981, 0.0174981)
We are 95% confident that the proportion of Oregonians who are sleep deprived is .15% lower to 1.7% higher than Californians who are sleep deprived.

6.44) a) Ho: Barking deer forage in every habitat equally.
HA: Barking deer prefer to forage in some habitats.

b) We can perform a Chi Squared test.
c) Check for independence: Each case contributes a count that is independent of the other cases. Each cell count must have at least 5 cases. It fails this test.
d) Expected number of deer foraging in
woods = 20.4
grass = 62.6
forest = 168.7
other = 174.2

df = 3 

chisqr <- (20.4-4)^2/20.4 + (62.6-16)^2/62.6 + (168.7-67)^2/168.7 + (174.2-345)^2/174.2
chisqr
## [1] 276.6495

p < .01
We reject the null hypothesis. The data provides evidence that barking deer prefer some habitats over others. However because the distribution was small with fewer than 5 samples for the woods, I have some reservations about this conclusion.

6.48) a) Chi Square Test for Independence in 2 way tables
b) Ho: There is no association between a woman having depression and the amount of coffee she drinks.
HA: There is an association between a woman having depression and the amount of coffee she drinks.

c) proportion of women who suffer from depression = 2607/50739 = 0.051
proportion of women who do not suffer from depression = 48132/50739 = 0.949

d) 0.051 x 6617 = 337.5
e) df = (1)(4) = 4
p < 0.001
f) Since p is less than 0.05, we reject the null hypothesis. There is an association between a woman having depression and the amount of coffee she drinks.
g) I do not agree with the statement. This was an observational study and not an experiment, so we cannot make a conclusion about causation from the study. Just because an association is observed, that does not mean that there is causation between the amount of coffee drunk and depression in women.