4 Listed below are the lead concentrations in ug/g measured in different traditional medicines. Use a 0.01 significance levelrandom sample.test the claim that the mean lead concentration for all such medicines is less than 15 pg/g. Assume that the sample is a simple162216.5 9.5 6 3.5 3.5 8 11.518.5 DAssuming all conditions for conducting a hypothesis test are met, what are the null and altemative hypotheses?O A. Ho = 15 pg/gO B. H, u> 15 ug/gHμ< 15 μggH1 ut 15 pg/gO C. H, μ= 15 μggO D. H, u= 15 pg/gH, u> 15 pg/gH, p< 15 jug/gDetermine the test statistic.O (Round to two decimal places as needed.)Determine the P-value.(Round to three decimal places as needed.)State the final conclusion that addresses the original claim.V Họ. There isVevidence to conclude that the mean lead concentration for all such medicines is7 15 pg/g.

The provided information are: The observations: 16,22,16.5,9.5,6,3.5,3.5,8,11.5,18.5 First calculate the mean and Statndard deviation: X X2 16 256 22 484 16.5 272.25 9.5 90.25 6 36 3.5 12.25 3.5 12.25 8 64 11.5 132.25 18.5 342.25 ∑iXi=115 ∑iXi2=1701.25 to calculate the mean Mean=∑iXin =11510 =11.5 Standard Deviation=s=∑Xi2-∑iXinin-1 =1701.25-1152109 =3799 =6.4893 using the Mean and Standard deviation: Null and Alternative Hypothesis are: c) H0:μ=15 μg/gH1:μ<15μg/g To Determine the test Statistics: the t-statistic is computed as follows: t=X¯-μsn =11.5-156.489310 =-1.71 test statistic=-1.71 to calculate the p-value p-value=0.061 (from Statistical table) P-value Approach: If p-value ≤level of significance ;Reject the Null Hypothesis Here p-value=0.061>0.01; Do not Reject the null Hypothesis Conclusion Do not Reject H0 There is not sufficient evidence to conclude that the mean lead concentration for all such medicines is < 15 μg/g