Answer to Question #337399 in Statistics and Probability for John Pilarte

Question

The weights (lb) of discarded plastic from a random sample of 62 household has sample mean of

$\bar{x}=1.911$

and a sample standard deviation of

$s=1.065$

.The sanitation department claims that the mean discarded plastic from all households is greater than 1.8 lb.Use a 0.05 significance to test this claim.What is the claim written in symbolic form?

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu\le 1.8$

$H_a:\mu>1.8$

This corresponds to a right-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.05,$ $df=n-1=61$ degrees of fredom, and the critical value for a right-tailed test is $t_c =1.67022.$

The rejection region for this right-tailed test is $R = \{t: t >1.67022\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{1.911-1.8}{1.065/\sqrt{62}}\approx0.82067$$

Since it is observed that $t = 0.82067< 1.67022=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for right-tailed, $df=61$ degrees of freedom, $t=0.82067$ is $p =0.207515,$ and since $p=0.207515>0.05=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$

is greater than 1.8, at the $\alpha = 0.05$ significance level.