Statistical Foundations DEADLINE: Monday, December 14, 2020 7:30AM (EDT) You will need to do all computations in R and submit your R file. Any explanation can be added as comments in R. Make sure to...

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Answered Same DayDec 13, 2021

Answer To: Statistical Foundations DEADLINE: Monday, December 14, 2020 7:30AM (EDT) You will need to do all...

Pooja answered on Dec 14 2021
133 Votes
#1a#
x <- c(0.6248, 0.6520, 0.6226, 0.6230, 0.6237, 0.6368, 0.6280, 0.6131,
0.6118, 0.6220, 0.6096, 0.6223, 0.6159, 0.6151, 0.6300, 0.6297,
0.6298, 0.6121, 0.6107, 0.6435, 0.6192, 0.6548, 0.6392, 0.5978,
0.6351, 0.6128, 0.6134, 0.6275, 0.6403, 0.6310, 0.6261, 0.6521,
0.6065, 0.6262, 0.6049, 0.6214, 0.6262, 0.6170, 0.6141, 0.6314)
shapiro.test(x)
#With W=0.97, p>1%, I fail to reject ho and conclude that the coefficient of restitution is normally distributed.#
#1b#
library(BSDA)
z.test(x,sigma.x=sd(x),conf.level = 0.99)
#With z=302.37, p<5%, I reject ho at 5% level of significance and conclude that mean coefficient of restitution of baseballs exceeds 0.623. mu > 0.623#
#1c p-value =2.2e-16 = 0.000#
#1d#
power=1-pnorm((mean(x)-0.63)/(sd(x)/sqrt(40)))
power
#1e#
d=abs((mean(x)-0.63))/sd(x)
d
b=1-0.80
b
# n=40#
#2a#
n1 = 10
n2 = 10
x1 = 14.8
x2 = 15.6
s1 = 1.4
s2 = 1.7
Spsq = ((n1-1)*s1^2 + (n2-1)*s2^2) / (n1+n2-2)
t= (x1 - x2)/sqrt(Spsq*(1/n1+1/n2))
t
df=n1+n2-2
t_critical = qt(0.05/2, df, lower.tail = FALSE)
t_critical
#With |t|#2b#
lower = (x1 - x2) - t_critical*sqrt(Spsq*(1/n1+1/n2))
lower
upper = (x1 - x2) + t_critical*sqrt(Spsq*(1/n1+1/n2))
upper
#Since the above confidence interval contains zero, I fail to reject ho and conclude that both production processes produce ink with...
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