The totals of the individual weights of garbage discarded by 62 households in one week have a mean of 27.553 lb. Assume that the standard deviation of the weights is 12.458 lb.  Use a 0.05 significance level to test the claim that the population of households has a mean less than 30 lb, which is the maximum amount that can be handled by the current waste removal system. What is your final conclusion. Could someone help me solve this?

## Your answer

 Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: To avoid this verification in future, please log in or register.

## 1 Answer

SAMPLE STATISTICS: mean weight of garbage (x bar)=27.553lb, size (n)=62 households.

POPULATION STATISTIC: standard deviation (σ)=12.458lb.

CLAIM: Mean weight of garbage is less than 30lb.

COUNTERCLAIM: Mean weight of garbage is greater than or equal to 30lb.

SIGNIFICANCE LEVEL: ɑ=0.05 (95% confidence level)

Null hypothesis, H₀: µ=30 (from counterclaim, which permits µ=30)

Alternative hypothesis, H₁: µ<30 (claim)

This is a left tail test.

TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(27.553-30)/(12.458/√62)=-1.547, corresponding to a P-value of 0.061.

ɑ=0.05 and sample P-value 0.061>0.05, which means that H₀ fails to be rejected. We conclude that there is insufficient evidence to decide with 95% confidence whether or not the mean weight of garbage is less than 30lb. Although the sample weight was less than 30lb, it wasn’t significantly less.

by Top Rated User (660k points)

1 answer
1 answer
1 answer
1 answer
1 answer
1 answer
1 answer
0 answers
1 answer
0 answers