The third box options are greater than, not, equal to, and less than. I need in an hour pls help Thankyou. Listed below are the lead concentrations in ug/g 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 16 ug/g. Assume that the sample is asimple random sample.1113.510.56.18.56.55.51912.5BEREEAssuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses?O A. Ho: H=16 ug/gH:H<16 ug/gΟ Β. Ho μ 16 μοH: p> 16 ug/gOC. Ho:H>16 ug/gO D. Ho: H=16 ug/gH: u<16 ug/gH: u#16 ug/gDetermine the test statistic.(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.Ho There isevidence to conclude that the mean lead concentration for all such medicines is16 ug/g.sufficientnot sufficient Listed below are the lead concentrations in ug/g 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 16 ug/g. Assume that the sample is asimple random sample.1113.510.518.56.55.51912.5Assuming all conditions for conducting a hypothesis test are met, what are the null and alternative hypotheses?O A. Ho: H= 16 µg/gO B. Ho: u= 16 ug/gH1: µ> 16 µg/gH,: µ<16 µg/gOC. Ho: H> 16 µg/gH: u<16 µg/gO D. Ho: =16 µg/gH: u#16 ug/gDetermine the test statistic.(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.Ho. There isevidence to conclude that the mean lead concentration for all such medicines is16 ug/g.RejectFail to reject

Given data is11,13.5,10.5,9,18.5,6.5,5.5,4,19,12.5 sample size(n)=10α=0.10From given data Mean(x)=sum of the data settotal no.of data setMean(x)=11+13.5+10.5+9+18.5+6.5+5.5+4+19+12.510Mean(x)=11010Mean(x)=11standard deviatio(s)=∑(x-x)2n-1s=(11-11)2+.....+(12.5-11)210-1s=5.08The null and alternative hypothesis isH0:μ=16μg/gH1:μ<16μg/gThe correct option is "A"Test statistic(t):-t=x-μsnt=11-165.0810t=-3.11Test statistic(t)=-3.11P-value:-Degrees of freedom(df)=n-1=10-1=9P(t<-3.11)=0.0063 (from t table)P-value is less than significance level(α)=0.10Reject H0.There is sufficient evidence to conclude that the mean lead concentration for such medicines is less than 16μg/g