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Consider an x distribution with standard deviation 𝜎 = 22. (Foreach answer, enter an exact number.)(a) If specifications for a research project require thest...

Question

Consider an x distribution with standard deviation 𝜎 = 22. (Foreach answer, enter an exact number.)(a) If specifications for a research project require thestandard error of the corresponding x bar distribution to be 2, howlarge does the sample size need to be? n =(b) If specifications for a research project require thestandard error of the corresponding x bar distribution to be 1, howlarge does the sample size need to be? n =

Consider an x distribution with standard deviation 𝜎 = 22. (For each answer, enter an exact number.) (a) If specifications for a research project require the standard error of the corresponding x bar distribution to be 2, how large does the sample size need to be? n = (b) If specifications for a research project require the standard error of the corresponding x bar distribution to be 1, how large does the sample size need to be? n =



Answers

Consider an $x$ distribution with standard deviation $\sigma=12$. (a) If specifications for a research project require the standard error of the corresponding $\bar{x}$ distribution to be $2,$ how large does the sample size need to be? (b) If specifications for a research project require the standard error of the corresponding $\bar{x}$ distribution to be $1,$ how large does the sample size need to be?

So we have in our original distribution of exes and it has a standard deviation of 12 and we want to know in part a if we need that standard air, So we're going to take samples of sample sizes of end and we need that standard air to be equal to two. We want to know what sample size do we need to take? So we know the standard deviation is 12. And so that means if we divide by two and multiplied by the square root of an and we'll have to square both sides, uh huh. We can see that this becomes six and six squared is 36. So we would need a sample size of 36 to make that standard air B a size of two. Now on the other hand, if we want the standard air To only be one, then we will need whoops tried it Right that as a one. Not too. And then if we get multiply both sides by the square root of van and square both sides. That means we only to sample size a lot larger. We need a sample size of 144 and notice what happens if you cut the standard air in half, you actually have to have four times the number of samples. Yeah, yeah.

In problem 15. We have a distribution with mean equals 53 new equals 53 and the standard deviation equals 21 sigma equals 21. From this distribution or from this population, we will take a random sample of size 49 then and equals 49 as a random sample from the population and them some assembling distribution of X board. Somebody distribution will be approximately will follow a normal distribution with me and a standard division. As long as we have relatively large sample and equals 49 then assemble distribution will be approximately normal. So the sampling distribution will follow a normal distribution with a mean of the same mean of the population which is 53 and a standard division of or standard error, which equals the population Standard division divided by the square root of the sample size. Then it's 21 divided by the square root 49 and this is a normal distribution annotation. You're right here. The variance. This means the answers of our problem is the first answer is normal distribution with a block symmetry will be approximately normal, with a mean the mean of the sampling. Distribution equals the same mean of the population? 53 and the standard deviation for the standard error equals 21. Divided by square root of 49 which is 21 divided by seven will be three. These are the final answers of our problem.

All right, This question wants us to look at how the standard air changes based on our sample size. So recall that the standard air of X, it's just the population standard deviation divided by the square root of the sample size. So heart, eh has a sample of 50. So our standard air equals 25 which is our population standard deviation over the squared of our sample size. And that works out to be 3.535 Then Part B asks for an equal to 100 which again, our standard error Is there a population sigma over the square root of our sample size, which is in this case equal to 2.5. Then for an equal, the 1 50 same thing Sigma divided by the square root of our sample size, which this one works out to be. 2.412 Then finally, Part D wants us to bump up the sample size to see 100. So standard error is our population standard deviation bye bye squared of our sample size, which is in this case 1.7678 So now let's look at the trend So if we look we start at 3.5, then is our sample size goes up, goes down to 2.5. That goes down to two point. Oh, then it goes down to 1.76 So what can we say as the sample size increases the standard air decreases? Which makes sense because as you get more and more samples, you expect them to be less variable overall. And this will come into play later when we look at calculating probabilities from sampling distributions and why we like bigger sample sizes because they're more representative of the

In this problem, I can calculate the value of a and B by any is required to. I'm just writing the formula that alpha by two multiplication 10 by e. Holy Square divided by jailed alpha by two multiplication five by e Holy square. Simplifying it further, I can write the value edge, Then the square by five square which is equal to 100 by 25 which is equal to four. So four will be our final answer. I hope you understand how I just use this formula to get the value first. I just tried the formula. After that, I just put all the data which are given in the problem and solve to get four as our final answer.


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