Surname1 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 is19.40±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 χα 2 2≤σ2≤(n−1)s2 χ1−α 2 2 For a sample size ofn = 41, we will havedf = n – 1 = 40degrees of freedom. For a 95% confidence interval, we haveα = 0.05, which gives 2.5% of the area at each end of the chi-square distribution. We find values of χ0.975 2=13.844andχ0.025 2=41.923. (n−1)s2 χα 2 2=40x1.902 41.923=3.444and (n−1)s2 χ1−α 2 2=40x1.902 13.844=10.4305. Therefore, the 95% confidence interval for the population variance is: 3.444°≤σ2≤10.4305°
Surname2 Question 2 (i)The test is one sample t – test since the sample size = 8 less than 30. H0:μ>60dB¿the true mean noise level is greater than 60dB) Ha:μ≤60dB¿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>2.3650we 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
Surname3 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.
End of preview
Want to access all the pages? Upload your documents or become a member.