The weight of adults in a certain city has a mean of 130 lbs, with a 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 weight significantly less on the basis of the 1% level of significance and construct 99% confidence interval of sample weight? (Assume the normal population.)
To determine whether the residents of the kebele weigh significantly less compared to the overall population, we can conduct a hypothesis test and construct a confidence interval.
Hypothesis Test:
1. Null Hypothesis (H0): The residents of the kebele weigh the same as the overall population.
2. Alternative Hypothesis (H1): The residents of the kebele weigh significantly less than the overall population.
To conduct the hypothesis test, we can use the t-test since we have a sample size of 25 (which is less than 30) and we do not know the population standard deviation, but we do know the sample standard deviation.
Calculate the t-statistic:
t = (sample mean - population mean) / (sample standard deviation / sqrt(sample size))
t = (129 - 130) / (41 / sqrt(25))
t = -1 / (41 / 5)
t = -0.122
To determine if the t-statistic is significant at the 1% level of significance, we need to compare it to the critical t-value.
The critical t-value can be found using a t-table or a statistical software. For a one-tailed test at a 1% level of significance with 24 degrees of freedom (sample size - 1), the critical t-value is approximately 2.492.
Since the absolute value of the calculated t-statistic (-0.122) is less than the critical t-value (2.492), we fail to reject the null hypothesis. This means that there is insufficient evidence to conclude that the residents of the kebele weigh significantly less than the overall population.
Confidence Interval:
To construct a 99% confidence interval for the sample weight, we can use the formula:
Confidence interval = sample mean ± (critical value * standard error)
The critical value can be found using a t-table or a statistical software. For a 99% confidence level and 24 degrees of freedom, the critical value is approximately 2.797.
The standard error can be calculated as the standard deviation of the population divided by the square root of the sample size:
Standard error = 41 / sqrt(25)
Standard error = 8.2
Substituting the values into the formula:
Confidence interval = 129 ± (2.797 * 8.2)
Confidence interval = 129 ± 22.924
Confidence interval = (106.076, 151.924)
Therefore, we can be 99% confident that the true population mean weight lies between 106.076 lbs and 151.924 lbs based on the sample data.