Question #143897
(3.2.23) Listed below are pulse rates​ (beats per​ minute) from samples of adult males and females. Does there appear to be a​ difference? Find the coefficient of variation for each of the two​ samples; then compare the variation.
Male 88,69,63,71,71,52,65,54,83,70,64,63,99,56,65
Female 67,84,82,70,76,85,86,85,87,88,94,72,86,79,80
1
Expert's answer
2020-11-17T16:21:29-0500

Coefficient of variation:

CV=sxˉ100CV=\frac{s}{\bar x}\cdot100


xˉ=xn\bar x=\frac{\sum x}{n}

s=(xxˉ)2n1s=\sqrt{\frac{\sum{(x-\bar x)^2}}{n-1}}


xˉmale=88+69+63+71+71+52+65+54+83+70+64+63+99+56+6515=68.867\bar x_{male}=\frac{88+69+63+71+71+52+65+54+83+70+64+63+99+56+65}{15}=68.867

smale=(8868.867)2+...+(6568.867)2151=12.755s_{male}=\sqrt{\frac{(88-68.867)^2+...+(65-68.867)^2}{15-1}}=12.755

CVmale=12.75568.867100=18.52CV_{male}=\frac{12.755}{68.867}\cdot100=18.52


xˉfemale=67+84+82+70+76+85+86+85+87+88+94+72+86+79+8015=81.4\bar x_{female}=\frac{67+84+82+70+76+85+86+85+87+88+94+72+86+79+80}{15}=81.4

sfemale=(6781.4)2+...+(8081.4)2151=7.42s_{female}=\sqrt{\frac{(67-81.4)^2+...+(80-81.4)^2}{15-1}}=7.42

CVfemale=7.4281.4100=9.16CV_{female}=\frac{7.42}{81.4}\cdot100=9.16


CVfemale<CVmaleCV_{female}<CV_{male} therefore the variation for male pulse rates appears to be greater than the variation for female pulse rates.


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