Question

A sample of 62 households was taken to measure the amount of discarded plastics. The sample mean was 1.911 lbs. and the standard deviation was 1.065 lbs. Test the hypothesis that the amount of discarded plastics per household was different from the population mean 1.800 lbs. at a significance level 0.01.

Answer 1

Null and Alternative Hypotheses

Ho: u = 1.8 Ha: u * 1.8

Rejection Region

The significance level is a = 0.01 and the critical value for a two tailed critical value is to = 2.659

Test Statistics

X - μo 1.911 – 1.8 t= t 0.821 s/n 1.065/V62

Decision about the null hypothesis

= 0.821 <te= 2.659, it is then concluded that the null hypothesis is not Since it is observed that t rejected. Using the P-value approach: The p-value is p = 0.415, and since p= 0.415 > 0.01, it is concluded that the null hypothesis is not rejected.

Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean u is different than 1.8, at the 0.01 significance level.

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