Convert Barbie’s chest, waist, and hips measurements to z-scores. Do these z-scores provide evidence to justify the claim that the Barbie doll is too thin of a representation of adult women? Justify your response with an appropriate statistical argument.

I have the z-scores:
Chest=(82.3-90.3)/5.5 = -1.455
Waist=(40.7-69.8)/4.7 = -6.191
Hips=(72.7-97.9)/5.4 = -4.667

I also know that this data does justify the claim, but I don't understand what the statistical argument to support this is? Help?

Question

Convert Barbie’s chest, waist, and hips measurements to z-scores. Do these z-scores provide evidence to justify the claim that the Barbie doll is too thin of a representation of adult women? Justify your response with an appropriate statistical argument. 

I have the z-scores:
Chest=(82.3-90.3)/5.5 = -1.455 
Waist=(40.7-69.8)/4.7 = -6.191 
Hips=(72.7-97.9)/5.4 = -4.667 

I also know that this data does justify the claim, but I  don't understand what the statistical argument to support this is? Help?

anonymous · Answer

It depends on the level of significance of the test. If we assume a 0.05 significance level (the usual standard).

The null hypothesis is that, presumably, the doll's dimensions are roughly proportional to the average woman's chest/waist/hips sizes; the alternative is that at least one (or perhaps all, if you're comparing them each individually) is not equal.

For a two-sided test like this, you need to find the critical values for a 95% confidence (i.e. 5% significance) test. Graphically, it would mean the same as finding the dashed cutoff values that give the desired probabilities (areas under the curve).
|dw:1413928556751:dw|