Repeat Exercise 26 for samples of size 72 and 108. What happen...

Question

Repeat Exercise 26 for samples of size 72 and 108. What happens to the mean and the standard deviation of the distribution of sample means as the sample size increases?

Accepted Answer

Hello there. Today we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. The heights of adult male giraffes in a certain region are normally distributed with a mean of 5.4 m and a standard deviation of 0.6 m. Random samples of size 16 are taken and the mean height of each sample is recorded. Find the mean and standard deviation of the sampling distribution of sample means. Repeat the calculation for samples of size 49 and 81. What happens to the mean and standard deviation of the sampling distribution of sample means as the sample size increases. Awesome. So it appears for this particular problem we're asked to solve for Um our final answer that we're ultimately trying to solve for is we're trying to determine what happens to the mean and standard deviations of the sampling distribution of sample means as the sample size increases. And in order to do that, we must first find the mean and standard deviation of the sampling distribution of sample means, which we will need to use the information provided to us by the prom itself to do that. And then we need to take that information and repeat the calculation for the samples of size 49 and 81. And considering that information, we can then solve for our final answer. To determine what exactly happens to the mean and standard deviation of this sampling distribution as the sample size increases. Awesome. So now that we know what we're ultimately trying to solve for, let us first write down all of our known information. So we're told that our population mean, which we'll call mu, is equal to 5.4 m. We're also told the population standard deviation, which we're gonna call sigma, and sigma is equal to 0.6 m. We're also told the sample size, which we'll call N1 and N1 is equal to 16. We also know the value of N2, which is equal to 49, and we also know the value of N3 and N3 is going to be equal to 81. Fantastic. So with that in mind, we we now need to recall that that according to the central limit theorem CLT for short, the mean of the sampling distribution of sample means is the same as the population mean, which will mean that mu subscript X bar is equal to mu. Which will be equal to 5.4 m. Fantastic. At this point, we need to note that the population mean remains the same for any sample size. So with that in mind, as we should recall, the standard deviation of sampling distribution of sample means, which is also called the standard error, we could write this equation as sigma subscript X bar is equal to sigma divided by the square root of N. So with that in mind, let us consider the value when N1 is equal to 16, and let's plug in our known variables to solve for sigma X bar 1. Or rather just sigma X bar. So sigma X bar, when N1 is equal to 16 will be equal to 0.6 divided by the square root of 16, which will give us a value of 0.15 m. OK. So now we need to do the same calculations for N2 and N3. So for N2 is equal to 49, we will find that sigma. The value for Sigma subscript X will be equal to 0.6 divided by the square root of 49, which will give us a value of 0.09 m. And then finally, we need to solve for N. 3 equals 81, and when N3 equals 81, that will mean that sigma X bar is equal to 0.6 divided by the square root of 81, which will give us an answer of 0.07 m. Fantastic. So now, moving right along here. We need to note that we will observe that the standard deviation decreases as the sample size increases. This therefore means, as we should recall, that larger samples produce sample means that are, on average, closer to the true population mean. So therefore, that will mean in conclusion, That as the sample size increases, the mean of the sampling distribution remains the same as the population mean, which is 5.4 m, but the standard deviation of the sampling distribution decreases. So therefore, without a doubt, our final answer will have to be. As follows, so as the sample size increases, So once again, as the sample size increases, the mean of the sampling distribution remains the same as the population mean 5.4 m, but the standard deviation of the sampling distribution decreases, and that once and for all is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Repeat Exercise 26 for samples of size 72 and 108. What happens to the mean and the standard deviation of the distribution of sample means as the sample size increases?

Key Concepts

Central Limit Theorem

Standard Error of the Mean

Effect of Sample Size on Distribution

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