Statistic Methods - Math Problem Example

1. The 95% confidence interval for the unknown mean from a Normal distribution is given by 96 where is the sample mean, is the population standard deviation and n is the sample size. Here 1.96 is the value of the standard normal curve such that P (Z 1.96) = 0.025. The confidence interval is the result from the test statistic N (0, 1). Here = 96.4, n = 25 and = 14. Hence the 95%confidence interval is 96.4 1.96(14/) 96.4 5.488 (90.912, 101.888) To test for the hypothesis H0: = 100 Vs H1: 100 using this confidence interval. Since the 95% confidence interval contains the value 100, we have insufficient evidence to reject H0 at 5% significance level. Hence we can conclude that the population mean is = 100. 2. As the population variance in unknown we have to assume that the sample means follow a t-distribution. The test statistic is tn - 1 where s is the sample standard deviation. Hence the 95% confidence interval is given by t24, 0.025 The tabulated value for t24, 0.025 is 2.064 and s = 18. Hence the confidence interval is 96.4 2.064(18 /) 96.4 7.4304 (88.9696, 103.8304) Since the 95% confidence interval contains the value 100, once again we have insufficient evident to reject H0: = 100 against H1: 100 and hence can conclude that the population mean = 100. 3. The p-value for the above hypothesis is p = 0.05 for both the situation of population variance is known or unknown. This is the level of significance of the test. The p-values can be calculated from the test statistic for appropriate test. For Q1, the standard normal distribution test statistic is. (96.4 - 100) / (14/5) = -1.286 From Z tables of standard normal distribution, P (Z -1.286) = 1 - P (Z 1.286) = 1 - 0.89973 = 0.10027 Since we are conducting two tailed test, the associated p-value is 0.10027 2 = 0.20054. Since this is much higher than 0.05, we have insufficient evidence to reject H0. For Q2, the t-distribution test statistic is (96.4 - 100) / (18 / 5) = -1 From the t-distribution tables, P (t24, 0.025 -1) = 0.15 Since this is one sided, 0.15 2 = 0.3. This is also higher than 0.05 and hence we have insufficient evidence to reject H0. The p-value for H0 is 0.05 for both questions and p-value for Q1 = 0.20054 and p-value for Q2 = 0.3. 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 interval is 5.291 1.188 (4.103, 6.479) 7. The 95% prediction interval for next Y is given by At x* = 44, the prediction interval is 4.4239 3.983 (0.4409, 8.403) At x* = 52, the confidence interval is 5.1175 3.627 (1.4905, 8.7445) At x* = 54, the confidence interval is 5.291 3.5344 (1.7566, 8.8254) 8. The assumption she make on the data are that they came from two samples are from normal distributions as the samples are less than 30. The distributions of the sample means are N and N. Here we don't know the population variance and hence t-distribution is used instead of standard normal distribution. The test statistic follows a t-distribution with degrees of freedom d, where d is Here s1 = 2.5, s2 = 3.5, n1 = 25 and n2 = 20. = Hence the rounded off degrees of freedom in 32 (No data for v=33 degrees of freedom in the table) The test statistic is = 9. H0: 1 = 2 Vs H1: 1 2 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. 10. d = . Put n1 = n2 = n and s1 = s2 = s d = 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. Read More