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Section 7.1 Hypothesis Testing: Hypothesis: Null Hypothesis (H 0 ): Alternative Hypothesis (H 1 ):

Section 7.1 Hypothesis Testing: Hypothesis: Null Hypothesis (H 0 ): Alternative Hypothesis (H 1 ):. a statistical analysis used to decide which of two competing hypotheses should be believed (analogous to a court trial).

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Section 7.1 Hypothesis Testing: Hypothesis: Null Hypothesis (H 0 ): Alternative Hypothesis (H 1 ):

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  1. Section 7.1 Hypothesis Testing: Hypothesis: Null Hypothesis (H0): Alternative Hypothesis (H1): a statistical analysis used to decide which of two competing hypotheses should be believed (analogous to a court trial) a statement often involving the value of a parameter in a distribution that we want to decide whether or not to believe; the hypothesis is called simple if it specifies only one possible value and is called composite if it specifies multiple possible values a statement assumed to be true at the outset of a hypothesis test (comparable to “innocence” in a court trial) a statement for which sufficient evidence is required before it will be believed (comparable to “guilt” in a court trial)

  2. Critical (Rejection) Region: Type I Error: Type II Error:   p-value (probability value): a set describing the observed results of data collection that will lead to rejecting H0 rejecting H0 and accepting H1 when H0 is true (in a court trial, saying that the defendant is guilty when the defendant is really innocent) failing to reject H0 when H1 is true (in a court trial, saying that the defendant is innocent when the defendant is really guilty) denotes the probability of making a Type I error and is called the significance level of the test denotes the probability of making a Type II error the probability of observing data as supportive or more supportive of H1 than the actual data, calculated by assuming H0 is true

  3. Tables 7.1-1 and 7.1-2 summarize hypothesis tests about one or two proportions (for sufficiently large sample size(s)). 1. Do Text Exercise 7.1-2 three different ways: First Way (a) Let C0 = {x | x 0} be the critical region. (b)  = P{x : x 0; p = 3/5} = P{X = 0; p = 3/5} = (2/5)4 = 16/625 = 0.0256  = P{x : x> 0; p = 2/5} = P{X > 0; p = 2/5} = 1 – P{X = 0; p = 2/5} = 1 – (3/5)4 = 1 – 81/625 = 544/625 = 0.8704

  4. Second Way (a) Let C1 = {x | x 1} be the critical region. (b)  = P{x : x 1; p = 3/5} = P{X = 0; p = 3/5} + P{X = 1; p = 3/5} = (2/5)4 + (4)(3/5)(2/5)3 = 16/625 + 96/625 = 0.1792  = P{x : x> 1; p = 2/5} = P{X > 1; p = 2/5} = 1 – [P{X = 0; p = 2/5} + P{X = 1; p = 2/5}] = 1 – [(3/5)4 + (4)(2/5)(3/5)3] = 1 – 297/625 = 328/625 = 0.5248

  5. 1.-continued Third Way (a) Let C2 = {x | x 2} be the critical region. (b)  = P{x : x 2; p = 3/5} = P{X = 0; p = 3/5} + P{X = 1; p = 3/5} + P{X = 2; p = 3/5} = (2/5)4 + (4)(3/5)(2/5)3 + (6)(3/5)2(2/5)2 = 16/625 + 96/625 + 216/625 = 328/625 = 0.5248  = P{x : x> 2; p = 2/5} = P{X > 2; p = 2/5} = P{X = 3; p = 2/5} + P{X = 4; p = 2/5} = (4)(2/5)3(3/5) + (2/5)4 = 96/625 + 16/625 = 0.1792

  6. 2. (a) A bowl contains 2 red balls and 3 other balls each of which is either red or white. Let p denote the probability of drawing at random a red ball from the bowl. Consider testing H0: p = 2/5 vs. H1: p > 2/5. Drawing ten balls from the bowl one at a time at random and with replacement, we let X equal the number of red balls drawn, and we define the critical region to be C = {x | x 6}. Calculate .  = P{x : x 6; p = 2/5} = P{X 6; p = 2/5} = 1 – P{X 5; p = 2/5} = 1 – 0.8338 = 0.1662 (from Table II in Appendix B)

  7. 2.-continued (b) Calculate  if p = 3/5. If p = 3/5,  = P{x : x 5; p = 3/5} = P{X 5; p = 3/5} = P{10 – X 5; 1 –p = 2/5} = 1 – P{10 – X 4; 1–p = 2/5} = 1 – 0.6331= 0.3669 (from Table II in Appendix B)

  8. (c) (d) Calculate  if p = 4/5. If p = 4/5,  = P{x : x 5; p = 4/5} = P{X 5; p = 4/5} = P{10 – X 5; 1 –p = 1/5} = 1 – P{10 – X 4; 1–p = 1/5} = 1 – 0.9672= 0.0328 (from Table II in Appendix B) Calculate  if p = 1. If p = 1,  = P{x : x 5; p = 1} = 0

  9. 3. Do Text Exercise 7.1-6 and calculate  if p = 0.15. (a) (b) Let Y = the number of ones (1s) observed in 8000 rolls The test statistic is z = The one-sided critical region with  = 0.05 is y / 8000 – 1/6 ————————  (1/6)(5/6) / 8000 zz0.95 = – z0.05 = – 1.645 1265 / 8000 – 1/6 ———————— =  (1/6)(5/6) / 8000 – 2.05 With y = 1265, we have z = Since z = – 2.05 < – z0.05 = – 1.645, we reject H0. We conclude that the probability of rolling a one (1) is less than 1/6.

  10. (c) Find a two-sided 95% confidence interval, instead of the one-sided interval. Since H0 is rejected. we expect the hypothesized proportion 1/6 not to be in the confidence interval: 1265 / 8000 1.960 (1265 / 8000)(6735 / 8000) / 8000 0.15013 , 0.16612 If p = 0.15,  = P{z : z– 1.645; p = 0.15} = y / 8000 – 1/6 P ———————— – 1.645 ; p = 0.15 =  (1/6)(5/6) / 8000 y / 8000 –0.15 + (0.15 – 1/6) P ———————————— – 1.645 ; p = 0.15 = (1/6)(5/6) / 8000

  11. y / 8000 –0.15 + (0.15 – 1/6) P ———————————— – 1.645 ; p = 0.15 = (1/6)(5/6) / 8000 y / 8000 – 0.15 1/6 – 0.15 P ————————  ———————— – 1.645 ; p = 0.15 =  (1/6)(5/6) / 8000  (1/6)(5/6) / 8000 y / 8000 – 0.15 1/6 – 0.15 P ————————  ———————— – 1.7169 ; p = 0.15 =  (0.15)(0.85) / 8000  (0.15)(0.85) / 8000 P(Z 2.46) = 1 – (2.46) = 1 – 0.9931 = 0.0069

  12. 4. Do Text Exercise 7.1-16. (a) (b) (c) Let Y = number of yellow candies observed in n random candies The test statistic is z = The two-sided critical region with  = 0.05 is p– 0.2 ——————  (0.2)(0.8) / n |z| 1.960 H0 is rejected when p = 5/54, but H0 is not rejected for each of the other 19 samples. If H0 is true, then when this hypothesis test is performed repeatedly, H0 will be rejected 5% of the time in the long run.

  13. (d) (e) If 95% confidence intervals for p are repeatedly obtained, then 95% of the intervals in the long run will contain p. 219 —— 1124 If the 20 samples are pooled, then, p = 219 / 1124 – 0.2 ———————— =  (0.2)(0.8) / 1124 – 0.43 We have z = Since z = – 0.43, and |z| < z0.025 = 1.960, we fail to reject H0. We conclude that the proportion of yellow candies produced is not different from 0.2.

  14. 5. Do Text Exercise 7.1-20. (a) (b) p1– p2 —————————  p(1 – p)(1/n1 + 1/n2) The test statistic is z = The one-sided critical region with  = 0.05 is z1.645 p1= 135 / 900 = 0.15 p2= 77 / 700 = 0.11 p= 212 / 1600 = 0.1325 z = + 2.341 Since z = 2.341, and zz0.05 = 1.645, we reject H0. We conclude that the proportion of babies with low birth weight is higher for developing countries in Africa than for developing countries in the Americas.

  15. (c) (d) The one-sided critical region with  = 0.01 is z 2.326. Since z = 2.341 z0.01 = 2.326, we would still reject H0. The p-value of the test is P 2.341 ; p1 =p2 = p1– p2 —————————  p(1 – p)(1/n1 + 1/n2) P(Z 2.341) = 1 – (2.341) = 1 – 0.9904 = 0.0096

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