Answer to Question #282170 in Statistics and Probability for lys

Question

2. A study was conducted to determine the level of awareness of the residents of a certain municipality on the causes of hypertension. The accompanying table shows the results with respect to the gender of the respondents. Female Weight (n=48) Male Indicator (N3D32) Mean X1 X2 W 2.8 2.8 1. Intake of food high in salt 2. Intake of food high in saturated fats. 2.4 3.0 3. Caffeine consumed in more than 4 2.5 2.4 cups of coffee a day 4. Drinking too much alcohol 5. Smoking 2.9 3.0 3.2 3.2 2.8 2.8 6. Intake of drugs or medications that may increase blood presssure. 7. Lack of physical exercise 3.0 3.1 8. Stress 2.8 3.0 9. Family history of hypertension 10 Present medical conditions like kidney 2.6 2.7 2.5 2.5 disorder, diabetes mellitus and others Average a) Find the average for the male and for the female. b) Find the weighted mean, xw, for each indicator letting the n as the weight.

Answer 1

a)

average for the male:

$\mu_1=\frac{2.8+2.4+2.5+2.9+3.2+2.8+3.0+2.8+2.6+2.5}{32}=0.86$

average for the female:

$\mu_2=\frac{2.8+3.0+2.4+3.0+3.2+2.8+3.1+3.0+2.7+2.5}{48}=0.59$

b)

weighted mean = $\frac{x_1n_1+x_2n_2}{n}$

$n=32+48=80$

$\overline{x}_1=\frac{32\cdot2.8+48\cdot2.8}{80}=2.8$

$\overline{x}_2=\frac{32\cdot2.4+48\cdot3}{80}=2.76$

$\overline{x}_3=\frac{32\cdot2.5+48\cdot2.4}{80}=2.44$

$\overline{x}_4=\frac{32\cdot2.9+48\cdot3}{80}=2.96$

$\overline{x}_5=\frac{32\cdot3.2+48\cdot3.2}{80}=3.2$

$\overline{x}_6=\frac{32\cdot2.8+48\cdot2.8}{80}=2.8$

$\overline{x}_7=\frac{32\cdot3+48\cdot3.1}{80}=3.06$

$\overline{x}_8=\frac{32\cdot2.8+48\cdot3}{80}=2.92$

$\overline{x}_9=\frac{32\cdot2.6+48\cdot2.7}{80}=2.66$

$\overline{x}_{10}=\frac{32\cdot2.5+48\cdot2.5}{80}=2.5$