Answer to Question #207333 in Statistics and Probability for JACKY

Question

The weights (lb) of discarded plastic from a sample of households is listed and the summary statistics are n=62, x̅=1.911 lb, and s=1.065 lb. Use a 0.05 significance level to test the claim that the mean weight of discarded plastics from a population of households is greater than 1.800 lb. What null hypothesis can be formulated for this claim?

Answer 1

Given that,

Hypothesized population mean(u)=1.800

Sample standard deviation (s)=1.065

Sample size (n)=62

Sample mean (x̅)=1.911

Significance level $(\beta)$ =0.05

claim $μ>1.800$

$Hypothesis\\H_o:μ=1.800\\ H_a:μ>1.800$

$t_c=1.670$

degree of freedom $df=n-1=62-1=61$

t_critical:

$p(t_{df}\text{\textgreater}t_c)=0.05\\p({t_{61}}\text{\textgreater}t_c)=0.05$

Test statistic: $t=\frac{\bar x-\mu}{\frac{s}{\sqrt{n}}}=\frac{1.911-1.800}{\frac{1065}{\sqrt{62}}}=0.8207$

$t=0.8207$

p-value;

$p(t_{df}\text{\textgreater}0.8207)\\p(t_{61}\text{\textgreater}0.8207)\\=0.2075$

here p-value $\text{\textgreater}$ significance level

It is then concluded that the null hypothesis is not rejected.

There is no enough evidence to claim that the population mean is greater than 1.800 at the 0.05 significance level.