Solutions to Assignment

Surname 1
Student’s Name
Professor’s Name
Course
Date
Solutions to Assignment
Question 1
(i) Since the data is normally distributed and n = 41 > 30, we use
x ± Zα/ 2
S
√n
Where: α = 5% = 0.05, Z0.025=1.96, and √n= √41=6.4031
Therefore, 95% CI is 19.40 ± 1.96 ( 1.90
6.4031 )
( 19.4° ± 0.5816° )
18.8184° ≤ μ ≤19.9816°
(ii) The original population is normally distributed then the standard deviation is
estimated as follows:
( n−1 ) s2
χ α
2
2 ≤ σ2 ≤ ( n−1 ) s2
χ1− α
2
2
For a sample size of n = 41, we will have df = n – 1 = 40 degrees of
freedom. For a 95% confidence interval, we have α = 0.05, which gives 2.5%
of the area at each end of the chi-square distribution. We find values of
χ0.975
2 =13.844 and χ0.025
2 =41.923.
( n−1 ) s2
χ α
2
2 = 40 x 1.9 02
41.923 =3.444 and
( n−1 ) s2
χ1− α
2
2 = 40 x 1.9 02
13.844 =10.4305. Therefore,
the 95% confidence interval for the population variance is:
3.444° ≤ σ2 ≤10.4305°

Surname 2
Question 2
(i) The test is one sample t – test since the sample size = 8 less than 30.
H0: μ>60 dB ¿the true mean noise level is greater than 60dB)
Ha: μ ≤60 dB¿the true mean noise level is than or equal to 60dB)
(ii) The test statistic is t-tests defined as follows:
tcal .= √ n ( μ−x )
s
tcal .= √ 8 ( 60−72 )
10 =−3.3941
The decision, reject null hypothesis if |tcal .|≥ , t0.025 ,(7)
Now, t0.025 ,(7)=2.3650
Since |tcal .|=3.3941>2.3650 we reject the null hypothesis and conclude
that there is sufficient evidence to suggests that the true mean noise level is
greater than 60dB at 95% significance level. Thus, the residences have
managed to disapprove the government report.
(iii) Solution to part (iii)
(a) The R-code for determining sample size:
## Install the pwr R-package
install.packages("pwr")
library(pwr)
delta <- 0.5
sigma <- 10
d <- delta/sigma
pwr.norm.test(d=d, sig.level=.05, power = .99)
## The results

Surname 3
Mean power calculation for normal distribution with known variance
d = 0.05
n = 7348.988
sig.level = 0.05
power = 0.99
alternative = two.sided
The residents will need at least 7349 measurements.
(b) Since the population standard deviation is unknown, we use one
sample t as follows:
## R- Code for the calculation
pwr.t.test(d = d, sig.level = .05, power = .99, type = "one.sample")
One-sample t test power calculation
n = 7350.909
d = 0.05
sig.level = 0.05
power = 0.99
alternative = two.sided
The residents will need at least 7351 measurements.