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Stats 242.3(02) Review Assignment Solutions 1. Notice that from the table we see that in the sample of n1 = 1708 Catholics, 541 of their father's reached the High school graduate level of education, while for the sample of n 2 = 1358 Protestants, 406 of their father's reached the High school graduate level of education a) Determine 95% confidence limits for the proportion of Catholics whose fathers reach the High School graduate level of education. ˆ 1 ˆ 0.3167 1 0.3167 541 0.3167, ˆ .0113 1708 n 1708 ˆ z0.025 ˆ or 0.3167 1.960 .0113 Thus 95% confidence limits are: 0.2947 to 0.3388 ˆ b) Determine 99% confidence limits for the proportion of Protestants whose fathers reach the High School graduate level of education. ˆ 1 ˆ 0.2990 1 0.2990 406 0.2990, ˆ .0124 1358 n 1358 ˆ z0.025 ˆ or 0.2990 2.576 .0124 Thus 95% confidence limits are: 0.2670 to 0.3310 ˆ c) Test to see if the proportion of Catholics whose fathers reach the High School graduate level of education is the same as the proportion of Protestants. (use = 0.05 and = 0.01) The Test statistic ˆ1 ˆ2 x x 541 406 947 z where ˆ 1 2 0.3089 n1 n2 1708 1358 3066 1 1 ˆ 1 ˆ n1 n2 Thus z 0.3167 0.2990 0.0178 1.058 0.0168 1 1 0.3089 1 0.3089 1708 1358 Using = 0.05 , z0.025 = 1.960, Since -1.960 < z < 1.960, H0 is accepted. Using = 0.01 , z0.005 = 2.576, Since -2.576 < z < 2.576, H0 is accepted. d) Estimate the difference between the proportion of Catholics and the proportion of Protestants whose fathers reach the High School graduate level of education with a 95% confidence interval. 95% confidence interval for 1 - 2 ˆ 1 ˆ1 ˆ 2 1 ˆ 2 ˆ1 ˆ 2 z0.025 1 , n1 n2 0.0178 1.960 0.3167 1 0.3167 1708 0.2990 1 0.2990 1358 Page 1 or -0.0151 to 0.0506 2. A civil engineer wishes to measure the compressive strength of two different types of concrete. A random sample of 10 specimens of the first type yielded the following data (in psi) Type I: 3250, 3268, 4302, 3184, 3266, 3297, 3332, 3502, 3064, 3116 10 x i 1 i 33581, 10 x i 1 2 i This value was in error 1241571690 whereas a sample of 10 specimens of the second type of concrete yielded the data Type II: 3094, 3106, 3004, 3066, 2984, 3124, 3316, 3212, 3380, 3018 10 x i 1 i 31304, 10 x i 1 2 i This value was in error 1078094016 A. Compute a 95 % confidence interval for the mean strength of the Type I concrete. 10 x xi n i 1 33581 3358.1 10 10 2 10 2 xi xi n i 1 i 1 s n 1 1241571690 33581 10 352.73 2 9 95% confidence intervals x t0.025 s 352.73 or 3358.1 2.262 or 3105.8 to 3610.4 10 n Compute a 95 % confidence interval for the mean strength of the Type II concrete. 10 x xi n i 1 31304 3130.4 10 10 2 10 2 xi xi n i 1 i 1 s n 1 1078094016 31304 10 133.15 2 9 95% confidence intervals x t0.025 s 133.15 or 3130.4 2.262 or 3035.2 to 3225.6 10 n B. Compute 95 percent confidence limits for the variance of strength for both types of concrete. n 1 s 2 2 0.025 to n 1 s 2 2 0.975 95% CI for variance Type I Type II 58864.9 to 414670.9 8387.8 to 59087.2 Page 2 4. An embryologist wished to estimate the mean difference in time at which eye pigmentation is first evidenced in an embryo for 2 varieties of birds. A sample of n = 33 embryos from species A revealed that it took an average of 74.32 hours for eye pigmentation to appear with a standard deviation of 2.51 hours. A second sample of m =33 revealed that it took an average of 100.21 hours for eye pigmentation to appear with a standard deviation of 19.44 hours. a) Estimate the difference in the mean tome at which eye pigmentation appears for the two varieties of birds, with a 95% confidence interval. Solution: 95% Confidence limits x1 x2 z0.025 100.21 74.32 1.960 s12 s22 or n1 n2 2.512 19.442 or 19.20 to 32.58 33 33 b) Using a 5% significance level test the Null Hypothesis H0: Variety A’s pigmentaion requires the same period of time to appear as the time required by variety B against the alternative Hypothesis HA: Variety A’s pigmentaion requires a shorter period of time to appear than the time required by variety B Solution: The test statistic z x1 x2 0 s12 s22 n1 n2 100.21 74.32 2.512 19.442 33 33 = 7.59 We will reject H0 if z > z0.05 = 1.645. Since this is true H0 is rejected for = 0.05. Page 3