For an IQ test, we know the population n=100 and the ox = 16. We are interested
in creating the sampling distribution when N = 64. (a) What does that sampling
distribution of means show? (b) What is the shape of the distribution of IQ
means and the mean of the distribution? (c) Calculate ox for this distribution.
(d) What is your answer in part (c) called, and what does it indicate? (e) What is
the relative frequency of sample means above 101.5?
Is "n" the mean? Is "ox" the standard deviation? It would help if you abbreviations matched standard usage or you wrote out the terms.
Where is the data from the sample?
To answer these questions, we need to understand the concept of sampling and sampling distributions.
Sampling Distribution aims to show the distribution of sample means, given a certain population.
(a) The sampling distribution of means shows the distribution of sample means. In this case, we want to create the sampling distribution of means when the sample size (N) is 64.
(b) The shape of the distribution of IQ means can be approximated to a normal distribution, as per the Central Limit Theorem. This theorem states that when the sample size is sufficiently large (typically greater than 30), the sampling distribution of means becomes approximately normally distributed, regardless of the shape of the original population. The mean of the distribution of IQ means would be the same as the mean of the population, which is 16.
(c) To calculate the standard deviation (ox) for the distribution of IQ means, we need to use the formula: ox = standard deviation / square root of the sample size. Since we know the population standard deviation (ox = 16), and the sample size (N = 64), we can calculate the standard deviation for this distribution:
ox = 16 / square root of 64 = 16 / 8 = 2.
(d) The answer in part (c) is called the standard error of the mean. It indicates the level of error or variability we can expect in our sample means when compared to the population mean.
(e) To find the relative frequency of sample means above 101.5, we need to convert the IQ value of 101.5 into a z-score using the formula: z = (x - mean) / standard deviation.
In this case, let's assume that the mean of the distribution is 16 and the standard deviation is 2.
So, z = (101.5 - 16) / 2 = 85.5 / 2 = 42.75.
Next, we need to find the area to the right of this z-score in a standard normal distribution table. Let's assume it is 0.9999. This means that the relative frequency of sample means above 101.5 is 1 - 0.9999 = 0.0001, or 0.01% (approximately).
In summary:
(a) The sampling distribution of means shows the distribution of sample means.
(b) The shape of the distribution of IQ means is approximately normal, and the mean of the distribution is 16.
(c) The ox (standard deviation) for this distribution is 2.
(d) The answer in part (c) is called the standard error of the mean, indicating the level of error or variability in sample means.
(e) The relative frequency of sample means above 101.5 is 0.01% approximately.