Answer with explanation:
Given : Sample mean =[tex]\overline{x}=\text{48.47 pounds}[/tex]
Standard deviation : [tex]\sigma=\text{ 3.1 pounds.}[/tex]
Sample size : n = 36
Claim : [tex]\mu\neq50[/tex]
∴ [tex]H_0:\mu=50[/tex]
[tex]H_1:\mu\neq50[/tex]
Since the alternative hypothesis is two tail , then the test is two tail test.
By using a z statistic and a 0.05 level of significance. Reject [tex]H_0[/tex] if z < -1.960 or is z> 1.960.
Then , the test static for population mean is given by :-
[tex]z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]
[tex]z=\dfrac{48.47-50}{\dfrac{3.1}{\sqrt{36}}}\approx-2.96[/tex]
We reject [tex]H_0[/tex] since [tex]-2.96\leq -1.96[/tex]. We have statistically significant evidence at [tex]\alpha=0.05[/tex] to show that the mean amount of garbage per bin is not different from 50.
