Question

4. (10 pts) A study is conducted to estimate the difference in the mean occupational exposure to radioactivity in computer science students in the years 2000 and 2008. These data based on independent samples of students for the 2 years are obtained: 2000 nj=16 X=94 rem Sı?=.040 2008 nz=16 X 25.62 rem S22=.028 An initial test confirms that the variances of the two populations can be assumed equal so that the pooled variance Sp can be used, where

a) Find a 95% confidence interval for estimating the difference in mean exposure to radioactivity for the two years.

b) Perform a hypothesis test to test the claim that radioactivity has been reduced. What are your hypotheses? What is your test statistic?

Answer 1

Solution: We need to construct the 95% confidence interval for the difference between the population means Mi - 2, for the case that the population standard deviations are not known. The following information has been provided about each of the samples: Sample Mean 1 (X1) = 0.94 Sample Standard Deviation 1 (81) = 0.040 Sample Size 1 (N) = 16 Sample Mean 2 (X2) 0.62 Sample Standard Deviation 2 (52) = 0.028 Sample Size 2 (N2) = 16 Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = ni + n2 - 2 = 16+ 16 - 2= 30. The critical value for a = 0.05 and df = 30 degrees of freedom is te = t1-a/2;n-1 = 2.042. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: SP (ni - 1)3 + (n2 - 1)s ni + 12 - 2 1) < 0.0282 (16 - 1) x 0.042 + (16 16+ 16 - 2 0.035 Since we assume that the population variances are equal, the standard error is computed as follows: 1 1 se + SPV ni n2 1 1 0.035 x + 16 16 0.012 Now, we finally compute the confidence interval: CI (X1 - X2 - te x se, X1 - X2 + te x se) (0.94 - 0.62 - 2.042 x 0.012, 0.94 - 0.62 +2.042 x 0.012) (0.295,0.345) Therefore, based on the data provided, the 95% confidence interval for the difference between the population means Hi - M2 is 0.295<H1 - H2 <0.345, which indicates that we are 95% confident that the true difference between population means is contained by the interval (0.295, 0.345).

Solution: The provided sample means are shown below: X1 = 0.94 X2 = 0.62 Also, the provided sample standard deviations are: 81 = 0.040 82 = 0.028 and the sample sizes are n = 16 and n2 = 16. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 12 Ha: Mi < u2 This corresponds to a left-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 30. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this left-tailed test is to = -1.697, for a = 0.05 and df = 30. The rejection region for this left-tailed test is R = {t:t< -1.697}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - X2 (14 –1)s{+(12-1)3 (+) n+12-2 0.94 -0.62 = 26.215 (16--1)0.0402+(16-1)0.0282 16+16-2 (16+1) (4) Decision about the null hypothesis Since it is observed that t = 26.215 > t = -1.697, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=1, and since p=1 > 0.05, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean ji is less than 12, at the 0.05 significance level.