Question

7. The weight of adults in a certain city has a mean of 130 lbs, with standard deviation of 41
lbs. A sample of 25 adults living in one of the kebeles of the city is found to have a mean
weight of 129 lbs. Does this mean the residents of the kebele weigh significantly less on
the basis of the 1% level of significance? (Assume normal population.)
a) State the null and the alternative hypothesis.
b) Give the critical region
c) What is your decision?

Answer 1

a) State the null and the alternative hypothesis.
Null hypothesis (H0): The residents of the kebele have the same mean weight as the city's general population (μ = 130 lbs).
Alternative hypothesis (H1): The residents of the kebele have a significantly different mean weight than the city's general population (μ ≠ 130 lbs).

b) Give the critical region
We are given a 1% level of significance, and since this is a two-tailed test, we will be looking at 0.5% at each tail.
For a two-tailed test with a 1% level of significance, we use the Z critical values: -2.576 and 2.576.

c) What is your decision?
To make a decision, we need to calculate the test statistic.

Test statistic (Z) = (sample mean - population mean) / (population standard deviation / sqrt(sample size))
Z = (129 - 130) / (41 / sqrt(25))
Z = (-1) / (41 / 5)
Z = (-1) / 8.2
Z ≈ -0.122

Since -0.122 is not in the critical region (-2.576, 2.576), we fail to reject the null hypothesis.

Decision: There is no significant evidence at the 1% level of significance to suggest that the residents of the kebele weigh significantly less or more than the general population of the city.