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?
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.
a) The null hypothesis is that the residents of the kebele weigh the same as the overall population in the city, i.e., the mean weight in the kebele is equal to 130 lbs. The alternative hypothesis is that the residents of the kebele weigh significantly less than the overall population, i.e., the mean weight in the kebele is less than 130 lbs.
b) The critical region for a one-tailed test at the 1% level of significance is in the left tail of the distribution, where the z-score is less than -2.33 (calculated from the critical value for a 1% significance level).
c) To make a decision, we need to calculate the test statistic (z-score) and compare it to the critical value. The test statistic can be calculated using the formula:
z = (sample mean - population mean) / (standard deviation / √sample size)
Plugging in the values:
z = (129 - 130) / (41 / √25)
z = -1 / (41/5)
z = -0.122
Since the test statistic (-0.122) is not in the critical region (-2.33), we fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the residents of the kebele weigh significantly less than the overall population at the 1% level of significance.
a) The null hypothesis (H0) states that the residents of the kebele weigh the same as the rest of the city, while the alternative hypothesis (Ha) states that the residents of the kebele weigh significantly less than the rest of the city.
Null hypothesis (H0): Residents of the kebele weigh the same as the rest of the city (μ = 130 lbs)
Alternative hypothesis (Ha): Residents of the kebele weigh significantly less than the rest of the city (μ < 130 lbs)
b) The critical region for a one-tailed test with a 1% level of significance is defined by the critical value of the test statistic. Since the population standard deviation is known, we can use a z-test. The critical value for the 1% level of significance is -2.33 (obtained from the standard normal distribution table).
Critical region: z < -2.33
c) To make a decision, we need to calculate the test statistic (z-score) and compare it to the critical value.
The formula to calculate the z-score is:
z = (sample mean - population mean) / (population standard deviation / √n)
Given:
Sample mean (x̄) = 129 lbs
Population mean (μ) = 130 lbs
Population standard deviation (σ) = 41 lbs
Sample size (n) = 25
Using the given values, we can calculate the z-score:
z = (129 - 130) / (41 / √25)
z = -1 / (41 / 5)
z = -5/41
The calculated z-score is -0.121.
Since the calculated z-score of -0.121 is not less than -2.33 (the critical value), it falls outside the critical region.
Decision: We fail to reject the null hypothesis. This means that there is not enough evidence to conclude that the residents of the kebele weigh significantly less than the rest of the city at the 1% level of significance.
a) The null hypothesis (H0): The residents of the kebele weigh the same as the average weight of adults in the city (μ = 130 lbs).
The alternative hypothesis (Ha): The residents of the kebele weigh significantly less than the average weight of adults in the city (μ < 130 lbs).
b) The critical region for a one-tailed test at the 1% level of significance is determined by the z-score. Since we are testing if the mean weight of the kebele residents is less than the city average, we look at the left tail of the standard normal distribution.
To find the critical value, we need to find the z-score that corresponds to a 1% level of significance. Since this is a one-tailed test, we want to find the z-value that cuts off the lowest 1% of the distribution. Using a standard normal distribution table or a statistical calculator, we find that z = -2.33 corresponds to a 1% significance level.
c) To make a decision, we compare the test statistic (z-score) with the critical value.
The test statistic (z-score) can be calculated using the formula:
z = (x̄ - μ) / (σ / √n)
Where:
x̄ is the sample mean (129 lbs),
μ is the population mean (130 lbs),
σ is the population standard deviation (41 lbs), and
n is the sample size (25).
Plugging in the values, we get:
z = (129 - 130) / (41 / √25)
z = -1 / (41 / 5)
z ≈ -0.122
Since the test statistic (-0.122) is larger (less negative) than the critical value (-2.33), it falls within the acceptance region.
Therefore, we do not reject the null hypothesis. There is not enough evidence to conclude that the residents of the kebele weigh significantly less, at the 1% level of significance, based on the given sample data.