StatisticsProblems

Question1

Theaverage for group A =6.3

Theaverage for group B = 5.777778

Thestandard deviation of group A=0.2739

Thestandard deviation of group B=0.2108

Thevariance for group A= 0.075

Thevariance for group B=0.044444

Valueof the t-test =3.955

Ifthe t critical,for df =16, sig.=0.05 and two tail test = 2.120 then the t statisticin this study which is 3.955 is greater than the t-critical and hencewe reject the null hypothesis and conclude that group A and group Bare different(Chong et al., 2008).

Question2

a).Z score=(X-U)/ Sqrt (Variance) where X is the test sample mean, U isthe population mean.

Zscore=(123.5-120)/sqrt (121)

Zscore=0.3182

b).I use Z critical=1.96 (two tails) for my comparison.

TheZ score of 0.3182 is less than the Z critical of 1.96 hence we failto reject the null hypothesis and conclude that the value of 123.5mg/dL is not significantly different from the limit for prediabeteshistorically expected of 120 mg/dL.

C)Ifthe obtained results is higher than the historically expected valuethen the Zscore had to be greater than the Z critical for one toreject the null hypothesis and conclude that value is different fromthe historically expected value.

d).The group should be not considered prediabetes and deserves a propertreatment since the groups average glucose is not higher than thehistorically expected limit for prediabetes of 120mg/dl.

References

Chong,T. T. L., Hinich, M. J., Liew, V. K. S., &amp Lim, K. P. (2008).Time series test of nonlinear convergence and transitionaldynamics.&nbspEconomicsLetters,&nbsp100(3),337-339.