41. The weight of adults in a certain city has a mean of 130 lbs, with standard deviation of 41
1bs. A sample of 25 adults living in one of the kebeles of the city is found to have a mean
weight of 129 Ibs. Does this mean the residents of the kebele weigh significantly less on the
basis of thel% level of significance? (Assume normal population.)
a State the null and the alternative hypothesis.
b. Based on the information provided would you use Z or t as the distribution of the test
statistic?
C. Give the critical region
d. What is your decision and conclusion ?
a. Null hypothesis: The residents of the kebele weigh the same as the rest of the city's adults (mean weight = 130 lbs)
Alternative hypothesis: The residents of the kebele weigh significantly less than the rest of the city's adults (mean weight < 130 lbs)
b. Since the population standard deviation is known and the sample size is large (n > 30), we can use the Z-distribution.
c. The critical region will be on the left side of the distribution since we are testing if the mean weight is significantly less. We need to find the critical Z-value corresponding to the l% level of significance. Let's assume the level of significance is 5% (α = 0.05). The critical Z-value can be found using a Z-table or Z-distribution calculator.
d. To make a decision, we need to calculate the Z-score for the sample mean weight of 129 lbs. The formula for calculating the Z-score is:
Z = (sample mean - population mean) / (standard deviation / √sample size)
Z = (129 - 130) / (41 / √25)
Z = -1 / 8.2
Z ≈ -0.122
Compare the calculated Z-score to the critical Z-value. If the calculated Z-score falls in the critical region (i.e., less than the critical Z-value), we reject the null hypothesis. If not, we fail to reject the null hypothesis.
Based on the decision, we can conclude that there is not enough evidence to suggest that the residents of the kebele weigh significantly less than the rest of the city's adults at the l% level of significance.