Each part of each exercise is worth 3 points. 1. A normal probability plot appea

Question

Each part of each exercise is worth 3 points. 1. A normal probability plot appears in Figure 6-23 on page 231 of the textbook. a. Use R to write the following function. usage npp (data, x label, y label, title) arguments data vector of observations in a sample x label: a label for the horizontal axis y label: a label for the vertical axis title title of the plot npp0 produces a normal probability plot for the observations in data. You may not use qqnorm to write npp b. Use npp to produce a normal probability plot for the data in the second column of Table 6-6 on page 230 of the textbook. The plot should be similar to the plot in Figure 6-23. 2. After you flip a coin, the probability that a head appears on the upward facing side is p. Consider the following experiment. You continuously fip the coin until k consecutive heads appears. Assume that k >0. If k 3 and p-0.7, then a possible outcome of this experiment is the following outcome, HH THTTTHTHTHHH where H represents a head and T represents a tail. a. Determine the algebraic expression for the expected number of tails that appear for k 0 and 0

Explanation / Answer

A.3) Given,

SD=3.6

n=15

H0: µ=75.2

H1: µ>75.2

=0.05

Z(/2)=1.65(as it is a one tail test)

A.a) for H0 to be rejected

(x- µ )/{SD/sqrt(n)} should be greater than 1.65

i.e, x-75.2/{3.6/sqrt(15)}>1.65

solving we get, x>76.73

hence for H0 to be rejected, the x values should be ranging from 76.73 to infinity

A.b) Given, mean(µ)=77.0

H1 is true

to find the power of hypothesis when H1 is true

first we need to find X(critical value) for which type 2 error can be committed

we fail to reject null if Z statistic is less then 1.65

1.65=x-75.2/{3.6/sqrt(15)}

solving, we get

x=76.73

now we need to calculate the probability of committing type 2 error (fail to reject null)

Z={76.7337-77/3.6/sqrt(15)}

solving, we get

Z=-0.28649

type 2 error is given as, P(Z<-0.28649)=0.3897

power of hypothesis is 1-0.3897=0.6103

A.c) probability of type 2 error is 0.3897 as explained above

A.4) Given,

values 7.9 11.3 6.9 12.7 13.2 8.8 9.3 10.6

sample mean(x)=10.0875

SD=2.250992

to estimate confidence interval for population mean

CI is given as

CI=X±Z/2*SD/sqrt(n)

10.0875±1.96*2.25/sqrt(8)

solving we get

8.52833<mean<11.64667

hence the confidence interval for mean is given as (8.52833, 11.64667)

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