mayafuller10
contestada

. A National Institute of Health survey of 40467 U.S adults found that 3156 of them experienced at least one major depressive episode in the previous year. Make a 95% confidence interval for the proportion of all U.S. adults who experienced depression in that year. (Source: https://www.nimh.nih.gov/health/statistics/major-depression)

Respuesta :

fichoh

Using the confidence interval for sample proportion formula, the C.I is (0.075; 0.081)

Using the relation :

[tex]p ± z_{crit} \sqrt{ \frac{p(1 - p)}{n} } [/tex]

  • Sample size, n = 40467

  • Sample proportion, p = 3156/40467 = 0.078

  • Zcrit at 95% = 1.96

  • Margin of Error = [tex] 1.96 \sqrt{ \frac{0.078(0.922)}{40467} } [/tex]

  • Margin of Error = 0.0026128

Lower boundary = 0.078 - 0.0026128 = 0.075

Upper boundary = 0.078 + 0.0026128 = 0.081

Hence, the confidence interval is (0.075; 0.081)

Learn more :

Otras preguntas