Answer to Question #115693 in Statistics and Probability for juan

Question

Spanish: Los pesos (lb) de plástico desechado de una muestra de hogares son 62 con una media de 1,911 lb y la desviación estándar es 1,065 lb. Use un nivel de significancia de 0,05 para probar la afirmación de que el peso promedio de plástico desechado de la población es mayor de 1,800 lb.

English: The weights (lbs) of discarded plastic from a household sample are 62 with an average of 1,911 lbs and the standard deviation is 1,065 lbs. Use a significance level of 0.05 to test the claim that the population's average weight of plastic waste is greater than 1,800 lbs.

Answer 1

$H_0: \mu\leq 1800\\ H_1:\mu > 1800\\ \alpha =0.05\implies Z_{α}=Z_{0.05}=1.65,\\ Z=\frac{\bar x-\mu_0}{\frac{\sigma}{\sqrt n}}\\ Z=\frac{1911-1800}{\frac{1065}{\sqrt 62}}=0.82\\ \text{the rejection region} :\;z>1.65\\ \text{The decision is to accept}:\; H_0\\ \text{this means that we do not support the claim}\\ \text{that the population's average weight of plastic}\\ \text{ waste is greater than 1,800 lbs. }$