The chance of developing breast cancer is under 11% for women. 10 Ho: H = 11. He μ = 11 |ο Hoμ 211 Hp <11 Ho: s11, H₂ μ> 11 Ho: p=0.11, H₂p= 0.11 O Ho <11, H₂ μZ 11 Ho: p20.11, H₂; p< 0.11 Ho p≤0.11, H₂p> 0.11 Ho: p<0.11, H₂. pz 0.11 Incorrect. The given statement is considering a population proportion and not a population mean. Part (1) X Private universities' mean tuition cost is more than $20,000 per year. Ho: H=20,000, H₂ μ = 20,000 ọ Ho-l>20,000, Ο Ho: a s 20.000, O Ho: p=20,000, O Hops 20,000, O Ho:p> 20,000, Ο Ho: μ ε 20.000, O Ho: pz 20,000. Hg με 20.000 Hg a > 20.000 H₁: p= 20,000 H₂ p > 20,000 Hps 20,000 Hg με 20.000 H: p<20,000

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What is your conclusion using a one sided z test if you assume α=0.01 and test the hypothesis that the proportion of colic is less among infants of mothers who don’t smoke than among the infants of mothers who smoke?

(10) When estimating the difference between 2 proportions, show that when n₁ = n₂ = n, we can be at least 100(1 - a)% confident that the error that we make when using P1 P2 as an estimate of p₁ - P2 is less than e when n = a/2 2€²

Preliminary data analysis indicate that it is reasonable to use a t-test to carry out the specified hypothesis test. Perform the t-test using the critical-value approach. %3D X = 44.2 and s = 7.3 Use a significance level of a = 0.05 to test the claim that u 35.8. The sample data consists of 10 scores for which H:µ±35.8 0 Н и-35.8 Test statistics: t=3.64 Critical Value: t=+/-2.262 H:µ=35.8 Reject There is sufficient evidence to support the claim that the mean is different from 35.8. П. : и 3 44.2 0. H:µ±44.2 a. Test statistics: t=3.64 Critical Value: t= +/- 2.262 H:µ=44.2 Н, There is sufficient evidence to support the claim that the mean is different from 44.2. Reject

2. Вычислите интеграл 2 S 1 e2z e²x + 1 et.. dx.

2t 2.15 A critical part has the following failure distribution: ƒ(t) · 40 OseS 40 yr. 402, 402 » Find: (a) R(1yr) (b) MTTF

7. Is (H + G)- = H¯' + G¬'? Explain your answer.