SolutionHW5

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

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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.

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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
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.

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