Statistical Methods Math Problem Example | Topics and Well Written Essays - 2000 words

Statistical Methods - Math Problem Example 4. Given data: , Sxx = 27100, Syy = 280.1 and Sxy = 2665 Cxx = Sxx - n = 27100 - 10(51.82) = 267.6 Cyy = Syy - n= 280.1 - 10(5.12) = 20 Cxy = Sxy - n= 2665 - 10 (51.8) (5.1) = 23.2 The slope a of the fitted regression line Y on X is a = Intercept b = = 5.1 - (0.0867) (51.8) = 0.6091 Estimate S for the standard deviation of the model s = = 5. H0: A = 0 Vs H1: A 0 where A is the slope of the fitted line Y on X. Test statistic: Under H0, tn - 2 distribution where a is the sample slope parameter, A is the population slope parameter, s is the sample estimate for the standard deviation. The results from Q4 are a = 0.0867, s = 1.4495 and Cxx = 267.6 and A = 0. Test statistic: Pr {-2.306 t8 2.306} = 0.95. The value 0.9786 is contained in this interval and hence we have insufficient evidence to reject the null hypothesis. The p-value is 0.3 which is higher than 0.05. Hence we can conclude that the population slope parameter A = 0. 6. The Fitted line is y = ax* + b where 'a' is the slope and 'b' is the intercept. From previous questions we have the results a = 0.0867 and 0.6091 At x* = 44, the value of the line is y = 0.0867(44) + 0.6091 = 4.424 At x* = 52, the value of the line is y = 0.0867(52) + 0.6091 = 5.118 At x* = 54, the value of the line is y = 0.0867(54) + 0.6091 = 5.291 The 95% confidence interval for mean of Y is given by the formula where a = 0.0867, b = 0.6091, x* = 44, 52 and 54, = 51.8, n = 10, s = 1.4995, Cxx = 267.6 and t8,0.025 = 2.306 At x* = 44, the confidence interval is 4.4239 1.9784 (2.4455, 6.4023) At x* = 52, the confidence interval is 5.1175 1.09428 (4.023, 6.21178) At x* = 54, the confidence... The value 0.9786 is contained in this interval and hence we have insufficient evidence to reject the null hypothesis. The p-value is 0.3 which is higher than 0.05. Hence we can conclude that the population slope parameter A = 0. The tabulated value for t32 distribution at upper 5% significance level is 1.694. Since our test statistic value is higher than this value, we reject H0 at 10% significance level. The tabulated value for t32 distribution at upper 2.5% significance level is 2.037. The test statistic is lower than this. Hence we accept H0 at 5% level of significance. The estimated p-value should be between (0.05, 0.1) excluding the upper and lower limits. The result is statistically significant and we can conclude there is a difference in the mean profit outputs. Since we can only say that the means are not equal and cannot say about which is larger, we recommend carrying out one-sided test and then choosing about which course is best. This is reasonable as the sum of n-1 values of a sample gives the other value of the statistic and there is dependence between the n terms and so the degrees of freedom of sample size n are n - 1. As we infer about 2 samples it is reasonable to use 2(n-1) as degrees of freedom when the sample sizes and variances are equal.