Question

1. Group of non-smokers expose to cigarette, and non-smokers not exposed to cigarette. N=50 N=50 X=16.35 ng/mg x=60.58 ng/mg S=138.08 ng/mlg S= 62.53 ng/mlg Use a 0.05 significance level to test the claim that nonsmokers exposed to tobacco smoke have higher mean cotinine level than nonsmokers not exposed to tobacco smoke. a) State the hypothesis b) Find the test statistics c) Find the p value d) Find the critical value and draw the bell curve and compare the test statistic value e) Decision on the null hypothesis and write the conclusion. f) Construct the confidence interval appropriate for the hypothesis. Extra credit 4 points.

Answer 1

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The provided sample means are shown below: X1 = 60.58 X2 = 16.35 Also, the provided sample standard deviations are: $1 = 138.08 S2 = 62.53 and the sample sizes are ni = 50 and n2 = 50. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = P2 Ha: Mi > M2 This corresponds to a right-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 = 98. 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 right-tailed test is to = 1.661, for a= 0.05 and df = 98. The rejection region for this right-tailed test is R = {t:t > 1.661}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: t= X1 - X2 (n-1)s +(n2-1)s ni+n2-2 G + a) 60.58 – 16.35 (50-1)138.082 +(50-1)62.532 (50 + 30) 50+50-2 = 2.063 (4) Decision about the null hypothesis Since it is observed that t = 2.063 > tc = 1.661, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0209, and since p = 0.0209 0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean mi is greater than um, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 1.69<Mi - M2 < 86.77. Graphically. T-Test: t = 2.063, p = 0.0209 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0 -2.5 -2.0-1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

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