I think the answer to this question is False but I would like to confirm. Thank you for your help.
True or False: Suppose the population mean is 50 and the population variance is 100 for a population. In a sample where n = 100 is randomly taken, 90% of all possible sample means will fall between 49 and 51.
Let's see:
CI90 = mean + or - 1.645 (sd/√n)
With the data given:
CI90 = 50 + or - 1.645 (10/√100)
Note: standard deviation is the square root of the variance.
Do the calculation for the confidence interval and see if your answer is correct.
I hope this will help.
To answer this question, we can use the concept of the sampling distribution of the sample mean.
The sampling distribution of the sample mean refers to the distribution of all possible sample means that can be obtained from samples of a given size taken from a population. The properties of the sampling distribution are determined by the population mean and the population variance.
In this case, you have been given that the population mean is 50 and the population variance is 100. For a sample size of n = 100, the central limit theorem states that the sample mean will follow a normal distribution with the same mean as the population mean (50) and a standard deviation equal to the population standard deviation divided by the square root of the sample size.
The standard deviation of the sample mean (also known as the standard error) can be calculated as follows:
Standard Deviation of Sample Mean = Square Root of (Population Variance / Sample Size)
= Square Root of (100 / 100)
= 1
Since the standard deviation of the sample mean is 1, we can conclude that approximately 68% of all possible sample means will fall within 1 standard deviation of the population mean. In this case, that means approximately 68% of the sample means will fall between 49 and 51.
However, you are interested in determining if 90% of the sample means will fall between 49 and 51. Since 90% is well beyond one standard deviation from the mean, we can say that the statement is false.