Question

Using the data set (link below), please calculate three separate one-way chi-square tests for Presidents' George W. Bush, Obama and Trump's approval rating for their first years in office. 1) Specifically run chi-square on the "Approving column" for W. Bush (2/1/2001-12/14/2001), Obama (1/23/2009-12/30/2009) Trump (1/23/2017-12/25/2017). Hint: It might be easier to delete all other values that are not used before analysis. 2) Once your analysis is completed, political ideology aside, compare the chi-square values and report the analysis comparison.

Answer 1

We take Approving data for W. Bush We want to test whether the hypothesis that there is a significance difference between the observed and Expected approving Null Hypothesis(H): There is no significance difference between the observed and Expected approving. Alternative Hypohtesis (H,): There is a significance difference between the observed and Expected approving. Level of Signficance (a ) = 0.05 Test statistic: Under H, Test statistic value can be defined as 0,- E (0:- E,) (0:- E,) E 85 85 18.94 18.94 20.94 17.94 19.94 20.94 21.94 87 84 86 87 88 89 87 89 85 51 55 22.94 20.94 22.94 57 57 55 55 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 66.06 358.7236 358.7236 438.4836 321.8436 397.6036 438.4836 481.3636 526.2436 438.4836 526.2436 358.7236 226.8036 122.3236 82.0836 82.0836 122.3236 122.3236 82.0836 197.6836 257.9236 197.6836 122.3236 122.3236 101.2036 170.5636 16.4836 49.8436 197.6836 64.9636 16.4836 25.6036 82.0836 82.0836 5.430269452 5.430269452 6.637656676 4.871989101 6.01882531 6.637656676 7.286763548 7.966145928 6.637656676 7.966145928 5.430269452 3.433297003 1.851704511 1.242561308 1.242561308 1.851704511 1.851704511 1.242561308 2.992485619 3.904383893 2.992485619 1.851704511 1.851704511 1.531995156 2.581949743 0.249524675 0.754520133 2.992485619 0.983402967 0.249524675 0.387580987 1.242561308 1.242561308 108.8386134 18.94 -15.06 -11.06 -9.06 -9.06 -11.06 -11.06 -9.06 -14.06 -16.06 -14.06 -11.06 -11.06 -10.06 -13.06 -4.06 -7.06 -14.06 -8.06 -4.06 -5.06 -9.06 -9.06 57 52 50 52 55 55 56 53 62 59 52 58 62 61 57 57

Expected Approving value per day = Total of Approving coloum Number of observations in Approving Coloumn 2180 33 = 66.06 z* ={(0,FED' ) -108.84 Degrees of Freedom (df )= n - 1 = 33 -1 = 32 We find p value usign Excel function P value =CHIDIST (108.84,32 ) = 0.0000 Decision: Here we observe that p value (0.0000) is less than the level of significance value (0.05).So we reject the null hypothesis (H.). Conclusion:Therefore we conclude that there is a sufficient evidence to support that There is a significance difference between the observed and Expected approving.

We take Approving data for Obama We want to test whether the hypothesis that there is a significance difference between the observed and Expected approving Null Hypothesis(H): There is no significance difference between the observed and Expected approving. Alternative Hypohtesis (H,): There is a significance difference between the observed and Expected approving. Level of Signficance (a ) = 0.05 Test statistic: Under H.,Test statistic value can be defined as 2: - E) x2 = E Total of Approving coloum Expected Approving value per day Number of observations in Approving Coloumn 19001 330 = 57.58 Simlarly we find Test statistic value -5 (0,- E,)" =187.30 E Degrees of Freedom (df )= n-1 = 330 - 1 = 329 We find p value usign Excel function P value =CHIDIST (187.30,329) = 0.99999999998 Decision: Here we observe that p value is greater than the level of significance value (0.05).So we fail to reject the null hypothesis (H.). Conclusion:Therefore we conclude that there is no sufficient evidence to support that there is a significance difference between the observed and Expected approving.

We take Approving data for Trump We want to test whether the hypothesis that there is a significance difference between the observed and Expected approving Null Hypothesis(H): There is no significance difference between the observed and Expected approving. Alternative Hypohtesis (H): There is a significance difference between the observed and expected approving. Level of Signficance (a) = 0.05 Test statistic: Under H.Test statistic value can be defined as 0,-E, (0,-E, (0,-E) 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 SH.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 38.51 0.19 1.51 -3.51 -2.51 -3.51 -1.51 -0.51 -0.51 -0.51 -3.51 -2.51 -1.51 0.51 -1.51 0.51 -0.51 -1.51 -2.51 -3.51 -1.51 -2.51 -1.51 -0.51 -1.51 0.49 0.51 0.19 0.49 -0.51 -1.51 -0.51 2.49 -0.51 -0.51 3.49 2.49 0.2401 2.2801 12.3201 6.3001 12.3201 2.2801 0.2601 0.2601 0.2601 12.3201 6.30X01 2.2801 0.2601 2.2801 0.2601 0.2601 2.2801 6.3001 12.3201 2.2801 6.3001 2.2801 0.2601 2.2801 0.2401 0.2601 0.2401 0.2401 0.2601 2.2801 0.2601 6.2001 0.2601 0.2601 12.1801 6.2001 6.2001 2.2201 2.2201 0.2601 0.2401 2.2201 12.1801 20.1601 12.1801 2.2201 6.2001 20,1601 42.1201 0.006234744 0.059207998 0.319919501 0.163596468 0.319919501 0.059207998 0.006/5409 0.00675409 0.006/5409 0.319919501 0.163596468 0.059207998 0.00675409 0.059207998 0.00675409 0.00675409 0.059207998 0.163596468 0.319919501 0.059202998 0.163596468 0.059207998 0.00675409 0.059207998 0.006234744 0.00675409 0.006234744 0.006234744 0.00675409 0.059207998 0.00675409 0.16099974 0.00525409 0.00675409 0.316284182 0.16099974 0.16099974 0.057649961 0.057649961 0.00675109 0.006234744 0.057649961 0.316284082 0.523502986 0.316284082 0.057649961 0.16099974 0.523502985 1.093744482 6.550114256 A1 41 1.49 42 -0.51 0.49 1.49 3.49 4.49 3.49 1.49 2.49 45 6.49 1887

Expected Approving value per day = Total of Approving coloum Number of observations in Approving Coloumn 1887 = - 49 = 38.51 Test statistic value 22-5 0,- ) = 6.55 E Degrees of Freedom (df )= n-1 = 49 - 1 = 48 We find p value usign Excel function P value =CHIDIST (6.55.48) = 1.0000 Decision: Here we observe that p value is greater than the level of significance value (0.05).So we fail to reject the null hypothesis (H.). Conclusion:Therefore we conclude that there is no sufficient evidence to support that There is a significance difference between the observed and Expected approving.