Lab three 4-6

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Feb 20, 2024

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STAT 151 C1 GROUP # LAB 3 Vogan, Gabriela Yujie, Cai
1. This is an example of an observational study because the independent variables cannot be controlled by the researcher. The data cannot generalize a broader population because this study has very specific circumstances. This study discusses the natural selection involved with the snow disaster in the Sierra Nevada Mountains. This data comes from the people involved with this accident and since 40 of the 87-people died, only 87 can participate in this statistic. If females in this study showed to be more apt to survive in this situation against men, it would not be valid to use this as an argument that females can withstand harsh conditions. Other factors to considered consists of age, health, and physical dimensions, this study was also related to famine as well as freezing 2. There are 87 cases from this study. Category Categorical or Numerical? Name Categorical Gender Categorical Family Categorical Position Categorical Age Numerical Child Categorical Survival Categorical Order Categorical Alone Categorical Group Size Numerical ***Identifier Variable** 3. a)
Contingency table results: Rows: Survival Columns: Gender Female Male Total Died 10 30 40 Survived 24 23 47 TOTAL 34 53 87 Chi-Square test: Statistic DF Value P-Value Chi-square 1 6.1659327 0.013 The overall survival rate was 54%. b) The top 3 families that lost the most members were the Eddy family, Keseburg family, and the Wolfinger family. c) The survival rate of people travelling alone was 18.8%. The survival rate or people not travelling alone was 62%. 3d) 70.6 % of females survived the ordeal ad 43.4% of males survived.
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The graph clearly shows that more men died than women and that more women survived than men did. 3e) The survival rate for children was 67.4%. The survival rate for adults was 39%. . The survival rate was much higher for children.
f) Category Sample Size Mean Median Std Dev IQ range Age 87 19.666667 16 15.251128 23 STILL NEED TO COMMENT ON CENTRE, SPREAD, SHAPE OF TWO DISTRIBUTIONS f) Gender Sample Size Mean Median Std Dev IQ range Female 34 17 13.5 13.615944 19 Male 53 21.377358 23 16.105692 22
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4. In this question, you will examine the relationship between survival and gender. a) Contingency table results: Rows: Survival Columns: Gender Cell format Count (Row percent) (Column percent) (Percent of total) Femal e Male Total Died 10 (25%) (29.41%) (11.49%) 30 (75%) (56.6%) (34.48%) 40 (100%) (45.98%) (45.98%) Survived 24 (51.06%) (70.59%) (27.59%) 23 (48.94%) (43.4%) (26.44%) 47 (100%) (54.02%) (54.02%) Total 34 (39.08%) (100%) (39.08%) 53 (60.92%) (100%) (60.92%) 87 (100%) (100%) (100%) Chi-Square test: Statistic DF Value P-value Chi- square 1 6.1659327 0.013 b)
H 0 : There is no association between gender and survival rate H a : There is an association between gender and survival rate X2= 6.1659327 ~ X2 has a DF = 1 P-value = P (X2 > 6.1659327) = 0.013 < 0.05 P-value < α = 0.05 reject H0 The evidence demonstrates that there is an association between survival counts and gender at α = 0.05. (c) The percentage of the male survivors is 60.92%. The percentage of the female survivors is 39.08%. (d) P 1 : Proportion of successes (Success = Survived) for Survival where gender = "Female" P 2 : Proportion of successes (Success = Survived) for Survival where gender = "Male" P 1 - P 2 : Differences in proportions H 0 : P 1 - P 2 = 0 H A : P 1 - P 2 ≠ 0 Results: Differen ce Count 1 Total 1 Count 2 Total 2 Sample Diff. Std. Err. Z-Stat P- value P1 - P2 24 34 23 53 0.271920 09 0.109507 01 2.483129 6 0.013
H 0 : P1 - P2 = 0 H A : P1 - P2 0 p pooled = 0.5402 s pooled = 0.1095 Z* = 0.7059-0.4340.5402(1-0.5402)(1/34+1/53)= 0.0066 ~ p pooled = 0.5402 with s pooled = 0.1095 P-value = P(z <-1.3573)*2 = 0.0066*2 = 0.0132 P-value < α = 0.05 Therefore, there is evidence shows that there is a different between the percentage of female and male survivors at α = 0.05. (e) According to the part A, it shows that there is a difference between the survival counts between genders. The degree of difference doesn’t show, so it is hard to find whether the two categories are independent or not. From the part B, we used the Chi-square test to get the P value. It clearly shows that when α = 0.05, there is an association between genders and survival numbers. (f) s p1-p2 = 0.7059(1-0.7059)34+0.434(1-0.434)53 =0.1036 C95% = (p1-p2) 1.96*0.1036 = [0.068844, 0.474956]
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Therefore, we are 95% confident that the confidence interval [0.068844, 0.474956] can capture the true proportion. If there is no difference between p1 and p2, then p1-p2 will equal to 0. However, 0 is not captured in the range [0.068844, 0.474956], so we can conclude that the difference between p1 and p2 is greater than 0. We can say that there is a difference between p1 and p2. 5. (a) Contingency table results: Rows: Bin(Age) Columns: Survival Cell format Count (Row percent) (Column percent) (Percent of total) Died Survived Total 1 to 7 12 (57.14% ) (30%) (13.79% ) 9 (42.86%) (19.15%) (10.34%) 21 (100%) (24.14%) (24.14%) 7 to 13 2 (15.38% ) (5%) (2.3%) 11 (84.62%) (23.4%) (12.64%) 13 (100%) (14.94%) (14.94%)
13 to 19 1 (8.33%) (2.5%) (1.15%) 11 (91.67%) (23.4%) (12.64%) 12 (100%) (13.79%) (13.79%) 19 to 25 2 (28.57% ) (5%) (2.3%) 5 (71.43%) (10.64%) (5.75%) 7 (100%) (8.05%) (8.05%) 25 to 31 12 (70.59% ) (30%) (13.79% ) 5 (29.41%) (10.64%) (5.75%) 17 (100%) (19.54%) (19.54%) 31 to 37 4 (57.14% ) (10%) (4.6%) 3 (42.86%) (6.38%) (3.45%) 7 (100%) (8.05%) (8.05%) 37 to 43 0 (0%) (0%) (0%) 1 (100%) (2.13%) (1.15%) 1 (100%) (1.15%) (1.15%) 43 to 49 3 (75%) (7.5%) (3.45%) 1 (25%) (2.13%) (1.15%) 4 (100%) (4.6%) (4.6%) 49 to 55 0 (0%) (0%) (0%) 1 (100%) (2.13%) (1.15%) 1 (100%) (1.15%) (1.15%) 55 to 61 3 (100%) (7.5%) (3.45%) 0 (0%) (0%) (0%) 3 (100%) (3.45%) (3.45%) 61 to 67 1 0 1
(100%) (2.5%) (1.15%) (0%) (0%) (0%) (100%) (1.15%) (1.15%) Total 40 (45.98% ) (100%) (45.98% ) 47 (54.02%) (100%) (54.02%) 87 (100%) (100%) (100%) Chi-Square test: Statistic DF Value P-value Chi-square 10 25.90810 3 0.0039 Highest survival rate are: [7,13] and [13,19] Lowest survival rate are: [26,31) with survival rate 29.41% , [1,7) and [31,37) both have the survival rates of 42.86% (b) H0 : There is no association between age and survival. Ha: There is an association between age and survival X2= 25.908103 ~ X2 with degree of freedom of 10 P-value = P(X2>25.908103) = 0.0039 P-value < α =0.01-reject H0 The evidence demonstrates that there is an association between survival and rate. 6.
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(a) Contingency table results: Rows: Group Size Columns: Survival Cell format Count (Row percent) (Column percent) (Percent of total) Died Survived Total 1 13 (81.25% ) (32.5%) (14.94% ) 3 (18.75%) (6.38%) (3.45%) 16 (100%) (18.39%) (18.39%) 2 2 (50%) (5%) (2.3%) 2 (50%) (4.26%) (2.3%) 4 (100%) (4.6%) (4.6%) 3 1 (33.33% ) (2.5%) (1.15%) 2 (66.67%) (4.26%) (2.3%) 3 (100%) (3.45%) (3.45%) 4 5 (62.5%) (12.5%) (5.75%) 3 (37.5%) (6.38%) (3.45%) 8 (100%) (9.2%) (9.2%) 7 0 (0%) (0%) (0%) 6 (100%) (12.77%) (6.9%) 6 (100%) (6.9%) (6.9%)
9 0 (0%) (0%) (0%) 9 (100%) (19.15%) (10.34%) 9 (100%) (10.34%) (10.34%) 12 5 (41.67% ) (12.5%) (5.75%) 7 (58.33%) (14.89%) (8.05%) 12 (100%) (13.79%) (13.79%) 13 6 (46.15% ) (15%) (6.9%) 7 (53.85%) (14.89%) (8.05%) 13 (100%) (14.94%) (14.94%) 16 8 (50%) (20%) (9.2%) 8 (50%) (17.02%) (9.2%) 16 (100%) (18.39%) (18.39%) Total 40 (45.98% ) (100%) (45.98% ) 47 (54.02%) (100%) (54.02%) 87 (100%) (100%) (100%) Chi-Square test: Statistic DF Value P-value Chi-square 8 22.073269 0.0048 According to the data, the group size with the lowest survival rate is 16 members and the group sizes with the highest survival rate are 7 and 9 members. Therefore, we can predict that there is an association between
group sizes and survival rates. When the survival rate decreass, the group size will increase. 8 males and 8 females separate the group size with lowest survival rate, so the percentage of male and female is half and half. (b) H0: There is no relationship between survival rate and group size. Ha: There is a relationship between survival rate and group size. X2= 22.073269 ~ X2with degree of freedom of 8 P-value = P(X2> 22.073269) = 0.0048 P-value < α = 0.01-rejeft H0 The evidence and data shows that there is a relationship between survival rate and group size when α = 0.01.
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