Suppose for a population.In a sample where n = 100 is randomly taken, 95% of all possible sample means will fall above 76.08.

Question

Suppose  for a population.In a sample where n = 100 is randomly taken, 95% of all possible sample means will fall above 76.08.

Quizplus · Accepted Answer

This statement is true because the sample mean follows a normal distribution with a standard error equal to the population standard deviation divided by the square root of the sample size. Using a standard normal probability table, we can find that the z-score corresponding to 95% of the area above a certain value is 1.645. Therefore, we can calculate the minimum value for the sample mean to be above 76.08 as follows:

76.08 = population mean + 1.645 * (population standard deviation / sqrt(sample size))
76.08 = 11ebc114_1253_a3a1_8acb_d58f1283ca35_TB8562_11 + 1.645 * (1253 / sqrt(100))
76.08 = 11ebc114_1253_a3a1_8acb_d58f1283ca35_TB8562_11 + 206.29
Sample mean > 282.37

Therefore, we can say with 95% confidence that the sample mean will be above 76.08 if it is at least 282.37. This statement is an application of the Central Limit Theorem, which states that as sample size increases, the distribution of sample means approaches a normal distribution with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of the sample size.

Given that we know the population mean and standard deviation are both unknown and that only a sample of size n = 100 has been taken, it is reasonable to assume that the statement is referring to the T-distribution rather than the normal distribution.

Using a T-distribution with 99 degrees of freedom (df = n - 1), we can calculate the value of t using a t-table or calculator such as Excel. For a 95% confidence interval with df = 99, t = 1.984.

We can then calculate the margin of error (ME) using the formula:

ME = t * SE

where SE is the standard error of the mean, equal to the population standard deviation divided by the square root of the sample size:

SE = sd / sqrt(n)

We do not know the population standard deviation, so we can estimate it using the sample standard deviation (s). Assuming that the sample is a simple random sample and that the sample is large enough to justify using the sample standard deviation as an estimate of the population standard deviation, we can calculate:

s = 11

SE = s / sqrt(n) = 11 / sqrt(100) = 1.1

ME = t * SE = 1.984 * 1.1 = 2.1824

Finally, we can calculate the lower bound of the confidence interval:

LB = sample mean - ME

We do not know the sample mean, so we cannot calculate this value, but we do know that it must be above 76.08, which is the lower bound of the confidence interval. Therefore, the statement is true.

Suppose for a Population