CE 344 Homework Assignment No 5-2023 (1) (1)

CIVIL ENGINEERING ANALYSIS I CE 344 Assignment No. 5 Due: October 31, 2023 (Max Points: 50) 1. (20 points) 256 students were surveyed to find out their zodiac sign. The results were: Aries (29), Taurus (24), Gemini (22), Cancer (19), Leo (21), Virgo (18), Libra (19), Scorpio (20), Sagittarius (23), Capricorn (18), Aquarius (20), Pisces (23). Test the hypothesis that zodiac signs are evenly distributed across students. Assume α=0.10. Expected - NX P1 1 12 = 256 × = 21.333 The expected frequency for the Taurus zodiac sign is, Expected = NX P2 1 12 = 256 × = 21.333 The frequencies are evenly distributed, so the expected frequency is the same for each Gemini zodiac sign to Pisces zodiac sign categories and each has an equal expected frequency which is 21.333. (Observed - Expected)2 Expected (29-21.333)^2/21.333 +(24-21.333)^2/21.333 +(22- 21.333)^2/21.333…………= 5.094 The chi-square test statistic is 5.094. DF=k-1 12-1=11 From the Excel output, the chi-square critical value at the 10% significance level is 17.275. The chi-square test statistic, 5.094, falls below the chi-square critical value of 17.275. Therefore, at the chosen level of significance, there is no significant reason to reject the null hypothesis. This suggests there isn't enough proof to claim that zodiac signs are unevenly distributed among students. 2. (10 points) For the given table of observed values below (a) Construct the corresponding table of expected values (4 points) (b) If appropriate, perform the chi-square test for the null hypothesis that the row and column outcomes are independent. If not appropriate, explain why. (6 points) 1

Observed Values 1 2 3 A 15 10 12 B 3 11 11 C 9 14 12 Where: - Eij is the expected frequency for cell in row i and column j. - Row total, is the total of row i. - Column total, is the total of column j. - Grand total is the total number of observations. A) EA1= 37x27/97=10.28 | | 1 | 2 | 3 | |---|-------|-------|-------| | A | 10.28 | 10.81 | 15.91 | | B | 6.96 | 7.22 | 11.81 | | C | 9.76 | 10.12 | 15.12 | Where: - Oij is the observed frequency for cell in row i and column j. - Eij is the expected frequency for cell in row i and column j. B) x^2A1=(15-10.28)^2/10.28 =2.19 | | 1 | 2 | 3 | |---|-------|-------|-------| | A | 12.03| 15.67 9.30 | | B | 7.61 | 9.92 | 5.88 | | C | 7.36| 9.42 | 5.58 | I believe you reject the null hypo here 3. (20 points) Two random samples taken from two normal populations yielded the following information: Sample Sample Size Sample Variance 1 n 1 =16 s 1 2 =53 2 n 2 =21 s 2 2 =32 (a) Find the statistic F=s 1 2 /s 2 2 . (4 points) (b) Find the degrees of freedom df 1 and df 2 . (4 points) (c) Find F 0.05 using df 1 and df 2 computed above. (4 points) 2

C) The degrees of freedom are df ₁ = 15 and df ₂ = 20. qf(p,df1,df2,lower.tail=FALSE) where p is probability i.e. level of significance, lower.tail=FALSE is used for right tailed critical value and lower.tail=TRUE is used for left tailed critical value. df is degrees of freedom. significance alpha = 0.05 and we need right tailed critical value (alternative hypothesis is right tailed Ha: σ2 > σ2) Degrees of freedom are 15 and 20. R code :qf(0.05,15,20,lower.tail=FALSE) R output :2.203274 D) Here we have to test the hypotheses, Hoσ = σ2 vs. Ha : σ > σ2 : The test statistic is F = 1.6563 and the critical value is F0.05 = 2.2033 1.6563 2.2033 i.e. The value of the test statistic is less than the critical value. Explanation: If critical value is greater than the test statistic we accept or fail to reject the null hypothesis otherwise reject the null hypothesis. fail to reject the null hypothesis. Conclusion: At 5% level of significance conclude that σ = σ2. a) The test statistic F = 1.6563 b) The degrees of freedom are df 1 = 15 and df 2 = 20. c) F 0.05 = 2.2033 d) We fail to reject the null hypothesis. Conclusion: At a 5% level of significance, we conclude that σ = σ^2 4