**Problem 13:**Listed below are the lead concentrations in µg/g measured in different traditional medicines. Use a 0.05 significance level to test the claim that the mean lead concentration for all such medicines is less than 16 µg/g. Assume that the sample is a simple random sample.- Data: 16, 14.5, 21.5, 22.5, 11, 3.5, 3, 22, 22.5, 3**a. Identify the null and alternative hypotheses.****b. Identify the test statistic for this hypothesis test.****c. Identify the P-value for this hypothesis test.****d. State the final conclusion that addresses the original claim.** In a large clinical trial, 400,237 children were randomly assigned to two groups. The treatment group consisted of 198,320 children given a vaccine for a certain disease, and 40 of those children developed the disease. The other 201,917 children were given a placebo, and 126 of those children developed the disease. Consider the vaccine treatment group to be the first sample. Identify the values of \( n_1, \hat{p}_1, \hat{q}_1, n_2, \hat{p}_2, \hat{q}_2, \bar{p}, \) and \( \bar{q} \).

As per our guidelines we are suppose to asnwer only one question kindly postthe remaining questions in next post13)Given data is16,14.5,21.5,22.5,11,3.5,3,22,22.5,3sample size(n)=10x x2 16 256 14.5 210.25 21.5 462.25 22.5 506.25 11 121 3.5 12.25 3 9 22 484 22.5 506.25 3 9 ∑x=139.5 ∑x2=2576.25 Mean(x)=∑xnMean(x)=139.510Mean(x)=13.95standard deviation(s)=∑x2-(∑x)2nn-1s=2576.25-(139.5)21010-1s≅8.3681a)claim is that the mean lead concentration for all such medicines is less than 16 μg/gThe null and alternative hypothesis isH0:μ=16Ha:μ<16b)Test statistic(t):-t=x-μsnt=13.95-168.368110t=-0.7747The test statistic for this hypothesis test is -0.7747c)Degrees of freedom(df)=n-1=10-1=9P-value:-P(t<-0.7747)=0.2292 (from excel formula)P-value for this hypothesis test is 0.2292d)Final conclusion:-P-value is greater than significance level(α)=0.05Hence Fail to reject H0There is no sufficient evidence to support the claim that the mean leadconcentration for all such medicines is less than 16 μg/g