SolutionHW5 - STAT500 HW#5_solutions STAT500 HW#5 Solutions...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
STAT500 HW#5_solutions STAT500 HW#5 Solutions 1) Chicago Title Company problem T 0 = 0.831; because nT 0 = 2544*(0.831) = 2114.1 > 5, n*(1 - T 0 ) = 2544*(1 - 0.831) = 429.9 > 5, thus the one-proportion z-test can be used. Ho: π = 0.831; Ha: π 0.831 α = 0.024; two-sided test π ˆ = 2081/2544 = 0.818 Test statistic z = 4 0.169)/254 * (0.831 0.831 - 0.818 = -1.749 P-value= 2*P (z > |-1.749|) = 2*(1-0.9599) = 2*0.0401 = 0.0802 > α =0.024 We fail to reject Ho (since p-value > α = 0.024). Based on the observed data, there is not sufficient evidence to conclude that this year’s percentage of home buyers purchasing single-family houses is different from the 1995 figure of 83.1%. 2) a. N=18<30, use normal probability plot to see if the data is normal distribution. 2100 2000 1900 1800 1700 1600 1500 1400 1300 99 95 90 80 70 60 50 40 30 20 10 5 1 Data Percent Normal Probability Plot for Volume (cu.f ML Estimates Mean: StDev: 1718.33 133.884 As the plot suggests, we can observe that all points fall inside the confidence bands, so there is no evidence to suggest that the data do not come from a normal distribution. We can use one-sample t-test in this problem. Ho: µ 1600; Ha: µ>1600 α = 0.05; one-sided right-tailed test y =1718.3, s=137.8 t = 18 / 8 . 137 1600 - 1718.3 / = - n s y μ = 3.64 P-value= P( t > 3.64) = 1 – P( t < 3.64) = 1-0.999=0.001.
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
STAT500 HW#5_solutions Cumulative Distribution Function Student's t distribution with 17 DF t P(T <= t) 3.6400 0.9990 Since P-value = 0.001 < α = 0.05, we can reject H 0 at level 0.05. Based on the observed data, there is sufficient evidence to conclude that the average amount of recycled paper is greater than 1600 cubic feet per 2-week period. b. A 95% confidence interval for µ is given by y + t α/2 *(s / n ) = 1718.3 ± (2.11*137.8/ 18 ) = (1649.725, 1786.875). 3) We have independent samples. Furthermore, the two populations (Type I devices and Type II devices) appear to be normally distributed (see below): Type I devic Type II devi 1.4 1.3 1.2 1.1 1.0 0.9 99 95 90 80 70 60 50 40 30 20 10 5 1 Data Percent 1.259 1.311 AD* Goodness of Fit Normal Probability Plot for Type I devic...Type II devi ML Estimates - 95% CI Also, the standard deviations for both populations are reasonably different (s 1 is twice the size of s 2 ; s 1 / s 2 = 0.06360/0.03057 = 2.08). Therefore, we will use a two-sample t- test with non-pooled variances (in Minitab). Two Sample T-Test and Confidence Interval (at 99% C.I.) Two sample T for Type I vs Type II N Mean StDev SE Mean Type I 10 1.2360 0.0636 0.020 Type II 10 0.9970 0.0306 0.0097
Image of page 2
STAT500 HW#5_solutions 99% CI for mu Type I - mu Type II: (0.171,0.3072) T-Test mu Type I = mu Type II (vs >): T = 10.71 P = 0.0000 DF = 12 We want to test the following hypotheses (at α = 0.01): H o : μ I - μ II = 0 ; H a : μ I μ II > 0. From Minitab, we obtain P-values = 0 (with t = 10.71; df = 12) < α = 0.01. So we can reject H o and conclude that the mean level of emission for Type I devices (μ 1 ) is greater than Type II devices (μ 2 ). 4) a. The value of the pooled-variance t-statistic, t = -4.04.
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern