Problem #2
x=c(1.8, 2, 2.1, 1.7, 1.2, 2.3, 2.5, 2.9, 1.6, 2.2)	#Daily DO (mg/l)
day = 1:10

#histogram
hist(xx)
#QQ plot
qqnorm(x)
qqline(x)
> ks.test(x, "pnorm", mean=mean(x), sd=sd(x))

        One-sample Kolmogorov-Smirnov test

data:  x
D = 0.089, p-value = 1
alternative hypothesis: two.sided
--
All of these indicate that the data is Normally distributed
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(a)t.test(x)  # provides a 95% confidence interval
Equation 8-18
        One Sample t-test

data:  x
t = 13.2235, df = 9, p-value = 3.354e-07
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 1.682726 2.377274
sample estimates:
mean of x
     2.03
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(b) N = (t_alpha/2 * s/delta)^2
delta = desired half interval
To reduce the interval by 10%, 
delta = 0.9*(2.377274-1.682726)/2
delta
[1] 0.3125466
s = sd(x) = 0.4854
N = (qt(0.025, 9)*sd(x)/delta)^2
12 additional observations are needed
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(c)95% confidence interval on the true variance
(Equation 8-21)

N=length(x)
alpha=0.05
alpha2=alpha/2
(N-1)*var(x) / qchisq(alpha2, N-1)
(N-1)*var(x) / qchisq(1-alpha2, N-1)
{0.11114980, 0.7854422}
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(d) The 95% CI of mean DO from (a) is (1.682726 2.377274)
If the 95% CI for another set of 10 days is (2.5, 3.2)
Clearly the two CIs do not overlap - which means that the
mean DO is significantly (at 95% confidence) differently between 
the two 10 day periods.