Question

0.00
0.20
0.60
0.43
0.75
0.27
1.45
0.19
0.26
0.04
0.61
0.26
0.80
0.36
0.26
1.30
0.01
0.10
0.06
0.19
0.51
1.74
0.01
0.19
0.06
0.19
0.17
0.25
0.24
0.36
0.15
0.36
0.07
1.18
0.83
0.23
0.06
0.01
0.05
0.78
1.14
0.41
0.38
0.40
1.46
0.07
0.63
0.91
0.21
0.75
0.19
0.59
0.12
1.12
0.97
0.02
0.04
0.38
0.56
0.33
0.16
0.16
0.29
0.24
0.02
0.98
0.44
0.19
0.63
0.59
0.50
0.40
1.56
0.12
3.17
0.13
0.41
0.11
0.05
0.59
1.74
0.03
0.09
0.42
0.40
0.09
0.44
0.04
1.20
1.06
0.28
0.46
0.24
0.07
0.40
0.68
0.73
0.88
0.00

It turns out that the "Pocket Change" data set was taken from a random variable X that was exponentially distributed with mean, μ = 0.50 .

In other words, the random variable X is given by

X = amount of change in pocket

X ∼ Exp ( 2 )

(a) What is the PDF for X ∼ Exp ( 2 ) ?

(b) How do the shapes of your histogram in Problem 1 and the PDF for X ∼ Exp ( 2 ) compare?

(c) How does the expected value ( $LaTeX: \mu$ μ )  and your sample average ( $LaTeX: \bar{x}$ x ¯ ) in Problem 1 compare? Do you expect these to be close? Why or why not?

(d) How does the standard deviation of X and your sample standard deviation ( s ) in Problem 1 compare ? Do you expect these values to be close? Why or why not?

(e) Let $LaTeX: \bar{X}$ X ¯ be the random variable of sample averages of size n = 30 .

How is$LaTeX: \bar{X}$ X ¯ distributed? $LaTeX: \bar{X}\sim$ X ¯ ∼ _____?______

(f) Use your mean $LaTeX: \bar{x}$ x ¯ and standard deviation s from Problem 1 to construct a 99% confidence interval on the true location of the population average.

(g) Write a complete sentence that explains what a confidence interval means in this context.

X ~Exp(2) 30

Answer 1

a) The PDF is ${\color{Blue} f_X(x)=2e^{-2x};x>0}$

b) The histogram and the theoretical distribution are plotted below.

Histogram & Exponetial distribution 寸 Ev (0 门 寸 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5

The shapes matches.

c) The theoretical expectation is
E(X) = 1.2 = 0.5
. The sample mean is
X 0.4671717
.

Yes. Since the distributions matches, the means are also close.

d)The theoretical  standard deviation is
TX = 1/2-0.5
. The sample standard deviation is
s = 0.4978957
.

Yes. Since the distributions matches, the standard deviations are also close.

e) According to Central Limit Theorem (CLT),

X ~N(0.5,0.5/30)

f) The 99% confidence interval is

0.4978957 0.4671717 ±Z1-0.01/2 (0.3383,0.5961)

g) We are 99% confident that the mean lies in the interval

(0.3383,0.5961)

Complete R code below.

X <- c(0,   0.2,   0.6,   0.43,   0.75,   0.27,   1.45,   0.19,   0.26,   0.04,   0.61,   0.26,   0.8,
0.36,   0.26,   1.3,   0.01,   0.1,   0.06,   0.19,   0.51,   1.74,   0.01,   0.19,   0.06,   0.19,
0.17,   0.25,   0.24,   0.36,   0.15,   0.36,   0.07,   1.18,   0.83,   0.23,   0.06,   0.01,   0.05,   0.78,
1.14,   0.41,   0.38,   0.4,   1.46,   0.07,   0.63,   0.91,   0.21,   0.75,   0.19,   0.59,   0.12,   1.12,
0.97,   0.02,   0.04,   0.38,   0.56,   0.33,   0.16,   0.16,   0.29,   0.24,   0.02,   0.98,   0.44,   0.19,
0.63,   0.59,   0.5,   0.4,   1.56,   0.12,   3.17,   0.13,   0.41,   0.11,   0.05,   0.59,   1.74,   0.03,
0.09,   0.42,   0.4,   0.09,   0.44,   0.04,   1.2,   1.06,   0.28,   0.46,   0.24,   0.07,   0.4,   0.68,
0.73,   0.88,   0)
plot(1:1)
dev.new()
hist(X, col="skyblue",main="Histogram & Exponetial distribution",xlab = "X",ylab = "Density",probability = TRUE)
curve(dexp(x,rate=2),lwd=2, col="blue",add = TRUE)

mean(X)
sd(X)
length(X)