Z / T Statistic Tools in Comparing Means - Case Study Example

Z / T Statistic Tools in Comparing Means The population involved a random sample of 75 men and 50 women blood donors and it involved an analysis of a particular trace element in their blood. The concentration mean of the trace elements is 28 ppm (parts per million) for the men and 33 for the women. The concentration of the trace element in the blood was calculated and found to be 14.1 ppm for the men and 9.5 ppm for the women. In the two formulas and are the means of the two samples used while Δ is the assumed variation between the sample means while σ 1 and σ 2 are the computed standard deviations of the two samples and n 1and n 2 are the sample sizes.

The calculated z-value was negative because the bigger mean for the women donors was deducted from the smaller value for the males. However, the final figure can be given as 2.37 and not -2.37 because the hypothesized difference between the two samples was zero. The order of the samples thus did not matter and thus the male mean value can be subtracted from the female mean value. An extreme z-score in the analysis whether plus or minus can result to the rejection of the null hypothesis of having no difference (Ted et al).

The area of the standard normal curve culminating from the computed z-score value of -2.37 is 0.0089. Since this test is two-tailed, the figure is multiplied by two to give a probability of 0.0178 implying that the population means of the two samples are equal or the same. Had the test been conducted in at a pre-specified significance levels of say α < 0.05 then the null hypothesis of similar means could have been rejected.