Question

Listed below are the lead concentrations in gg measured in different traditional medicines. Use a 0.10 significance level to test the claim that the mean lead concentration for all such medicines is less than 12 ugg Assume that the sample is a simple random sample. 20.5 5.5 3.5 12 В 4.5 3.5 14 2.5 7 Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses? A Ho: -12d H1: >12 y 9/9 B HO: "12 yolg H1: 12 v glo С. HO: >12 go H1: <120 D HO: 12 H1: 12ygg Identify the test statistic? Pwania

Answer 1

Solution:- Given that samples : 20.5,5.5,3.5,12,8,4.5,3.5,14,2.5,7
x = 8.1, sd = 5.7581, n = 10

=> option D.

=> Test statistic = −2.142

=> P-value = 0.0304

  

The provided sample mean is X = 8.1 and the sample standard deviation is s = 5.7581, and the sample size is n = 10. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: A = 12 Ha: u<12 This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.1, and the critical value for a left- tailed test is te = -1.383. The rejection region for this left-tailed test is R=t:t< -1.383 (3) Test Statistics The t-statistic is computed as follows: X - μo 8.1 - 12 t= 2.142 s/n 5.7581/ 10 (4) Decision about the null hypothesis Since it is observed that t = -2.142 <te= -1.383, it is then concluded that the null hypothesis is rejected Using the P-value approach: The p-value is p = 0.0304, and since p=0.0304 <0.1, 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 the nonulation mean is less than 12, at the 0.1 significance level.