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9) A recent study of 100 elementary school teachers in _ southern state found that their mean salary was $24,000 with population standard deviation of $2100. A simi...

Question

9) A recent study of 100 elementary school teachers in _ southern state found that their mean salary was $24,000 with population standard deviation of $2100. A similar study of 100 elementary school teachers in western state found that their mean salary was 534,400 with population standard deviation of = 53200. Test the claim that the salaries of elementary school teachers in the western state more than $10,000 greater than that of elementary teachers in the southern state; Use a 0.05_ Assume Ih

9) A recent study of 100 elementary school teachers in _ southern state found that their mean salary was $24,000 with population standard deviation of $2100. A similar study of 100 elementary school teachers in western state found that their mean salary was 534,400 with population standard deviation of = 53200. Test the claim that the salaries of elementary school teachers in the western state more than $10,000 greater than that of elementary teachers in the southern state; Use a 0.05_ Assume Ihe two samples are random and independemt



Answers

Teacher Salaries A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers in a large school district. The mean of the salaries of a random sample of 26 elementary school teachers is dollar 48,256, and the sample standard deviation is dollar 3,912.40 . The mean of the salaries of a random sample of 24 secondary school teachers is dollar 45,633 . The sample standard deviation is dollar 5533 . At $\alpha=0.05,$ can it be concluded that the mean of the salaries of the elementary school teachers is greater than the mean of the salaries of the secondary school teachers? Use the $P$ -value method.

Looking at 89. We're doing an inversion normal because they're giving us percentile. So in verse Nor and O r and and I'm gonna do 0.9 a mean of 44,000 But it's damn deviation of 500. Its meaning Me. But in red, 52,000 330. And then for part B, we're doing inverse Norm White nine. £44 6500. No, not 1000 100. Divided by this were of the 10 teachers and that equal 46 6 30 four.

Right in this problem. We have two parts, even a and A B part, and both parts are centered around the fact that we have an average teacher salary. So we're talking about the population of teachers in Connecticut having an average salary of 57,000 $337 and the standard deviation is $7500. So we're going to use that in both huts. So let's focus on part A and in part A. We want the probability that a teacher one teacher has an A salary that's less than $52,000 a year. So with that, we're going to have to draw a bell curve. And with the bell curve, we always put the average in the center. So we're gonna have 57,000 337 in the center, and we're looking for on average or s aureus salary less than 52,000. So the next thing we're gonna have to do is we're going to have to calculate the Z score associated with 52,000, So we're going to 52,000 minus 57,337 divided by the standard deviation of 7500 and the Z score associated with the 52,000 is approximately negative 0.71 So what we can do is we can go back to our picture, and we could say that 52,000 is equivalent to a Z score of negative 0.71 So we can also rewrite our problem to say the probability that the Z score is less than negative 0.71 sear, then going to use your standard normal distribution chart in the back of your textbook. And when we look up negative 0.71 we're going to get a value of point 2389 So for part a, the probability that a teacher selected has a salary of less than $52,000 would be 0.2389 Now let's go to Part B and in Part B. We're going to use the same information about the population, and we know that the average salary of the teacher in Connecticut was 57,337 and the standard deviation was 7500. But in part B, we're now going to select a sample of teachers and in this instance were selecting 100 teachers. So our sample size is 100 and we now want to find out what the probability is that the sample mean of those 100 is less than $56,000. So since we have a sample of 100 we're talking sample means the central limit theorem is going to go into play. And according to the Central Limit Theorem, the average of the distribution of means will be the same as the average of the population, which in this case is 57,300 37. And the standard error of the means is going to be the standard deviation of the population divided by the square root of the sample size. And again, in this case, it's going to be 7500 divided by the square root of when, So we're going to want to draw another bell curve, and again we're gonna put the average in the center, and 56,000 is gonna be over here. And we're looking for, on average to be less than 56. That was so we need to find busy score, associate it with 56,000. So our Z score is going to be 56,000 minus 57,300 37 divided by the standard deviation of those means with standard error of the mean, which is our Sigma, or the 7500 divided by the square root of 100. And that turns out to be approximately negative 1.78 So we can now go back to our picture. And we could say that the Z score associated with 56,000 is negative 1.78 so we can rewrite our problem instead of us talking about the probability that the average is less than 56,000. We can talk about the probability that the Z score is less than negative 1.78 So we're then going to go to our standard normal distribution chart in the back of the book, and we're going to find the probability to be 0.375 So in summary, Part B was asking us what's the probability that when we select 100 teachers, their average salary is less than 56,000 and the answer would be 0.3 75

Alright, salaries for teachers in an elementary school are normally distributed, so that means we're going to be modeling our data in the bell shape. It has a mean of $44,000. We have a standard deviation of $6500 and we're going to randomly survey 10 teachers, so our sample size is 10. So let's tackle this problem one part at a time. So party in words. What would you describe? X as well. X is a single piece of data, and in this case, our data represents one teacher's salary. Part B X can be approximated by and because they've already said that the salaries are normally distributed, we could say that Ex could be approximated by a normal distribution. And when we do that, we always follow it up with what's that average of that normal distribution and the standard deviation. So in our problem, X is approximated by a normal distribution with an average of $44,000 with a standard deviation of $6500 part C. What does Sigma X stand for inwards? Well, the symbol sigma in statistics means some of so we're summing up all of the exes. So in this case, X will stand for the some of the salaries of 10 elementary school teachers parts D. The sum of X can be approximated by well, since our original data is approximated by the normal distribution, we could say that the some of the 10 teachers salaries will also be approximated by the normal, um, distribution. But this time our average and our standard deviation are going to be modified. So our average is going to get multiplied by how many is in our sample size. And with our standard deviation, we're going to have to apply our central limit serum because we're selecting 10 teachers from that population. And the central limit theorem tells us that the standard deviation will be Sigma divided by the square root of n. But then, because there's 10 teachers, we're gonna have to multiply that by end. So in this instance, the sum of X is approximated by a normal distribution with an average or a means of 44,000 times 10 and a standard deviation of 6500 divided by the square root of 10 times 10. And if we put all of that into our calculator, we will get an average of 440,000 and a standard deviation off $20,554.80. Let's go and tackle Part E in Part E. We're trying to find the probability that the teachers earn a total of over 400,000. So by them saying a total of we're talking The sum of X is greater than $400,000 and the best way to tackle this is to draw a picture or a graph of what's going on. And we're going to use the information from Part D. That said, the average of the some is $440,000 and the standard deviation of that sum is $20,554.80. And what we're trying to do is we're trying to find the probability that that sum is greater than 400,000. So we can do this one of two ways. Um, we can do it just by putting the data right into our calculator, and by putting it right into our calculator, we would be saying normal CDF and then when you used the cumulative density function you need to provide four different pieces of information the lower boundary of your shaded area, which would be the 400,000 the upper boundary, which would be a very, very high number. Because as we go into that right tail, we get extremely large numbers. So we're going to use 10 to the 99th Power. Then we need the average of our data which was 440,000 and the standard deviation, which was $20,554.80. Now the other approach, we could standardize this data and we could find the Z score that goes along with this boundary. And when we find a Z score, we always take the piece of data which in this case was $400,000. We subtract the average, which was $4,440,000 and we divided by our standard deviation, which was 20,550 $4.80. And when you calculate that out, we will get a Z score of approximately negative 1.946 So the other way we can put the information into our calculator is to still use that normal a cumulative density function. But this time, our low end on the Z scale is negative. 1.946 The upper end. Very seldom do we go much past five standard deviations. So we're gonna say five. The average of the standard normal curve is zero. And the standard deviation is what? So we could type either of those into our calculator and end up with the appropriate probability. Now I'm going to elect to put this one in because it's got fewer numbers. It's got less zeros, less room to make errors as I'm typing it in. So I'm gonna bring in my graphing calculator, and we're going to use the normal cumulative density function. You're gonna find it under the second there's which is number two. We're gonna type in our low bound negative 1.946 are upper boundary, five hour average zero in our standard deviation one, and we get approximately 0.9742 So 0.9742 is the area to the right of $400,000. And that would be the probability that the some of the 10 teacherssalaries is greater than $400,000 part F in part f. You are asked to calculate the 90th percentile for an individual teacher salary. So 90th percentile means that 90% of the information is less than that value. So since we've already said that the teacher's salary is approximated by a normal distribution with a mean of $44,000 and a standard deviation of $6500 we condone Aw, the bell shaped curve putting the average in the center and 90th percentile would be referring to this boundary line right here, where 90% of the data is lower than that. So what we're gonna do is we're going to first find our Z score associated with that boundary. And to find our Z score, we're going to use our inverse norm feature on our calculator. And then we're going to put in the value or the area to the left, into that left tail, the average of the standard normal curve and the standard deviation of the standard normal curve. So again, I'm gonna bring in my calculator, we're gonna hit second. There's and it's number three in this menu. We're going to type in our 90%. That's in the left tail, the average of the standard normal and the standard deviation of the standard normal curve. And we get a Z score off approximately 1.28155 So now we need to turn that Z score into raw data. We need to find this X score. Now we know it's gonna be larger than $44,000 just because of the placement on the belt. And if we take our traditional Z score formula and we rewrite that and transform that into solving for a value of X. If I cross multiply, I would get X minus mu equals E times sigma. And then, if I add mu to both sides, I would get mu plus zig MMA or sorry you plus Z time Sigma is equivalent to that ex score, so I'm going to put my information in an X equals my average teacher's salary. Plus the Z score 1.28155 multiplied by the standard deviation which was 6500, and I find out that my ex score or my raw piece of data at the 90th percentile would be 52,000 300 $30 and seven cents. Now, depending on how far out you carry this Z score, the sense they're gonna change ever so slightly. So what we're saying here is that 90% of the teachers have a salary that are less than $52,330.7 Part G. As we tackle part G, we're trying to find the 90th percentile of the some of the 10 teachers salaries. So again, let's draw a picture representing the some of those salaries. And in a previous step, Part D, we said that the sum of X was approximated by a normal distribution with an average of $440,000 and a standard deviation of 20 1000 $554.80. So putting that information on our curve, we would have $4,440,000 and a standard deviation of $20,554.80. And just like the previous problem, the 90th percentile would be up here, so we would want 90% in that left part of the curve. So again, we're gonna find a Z score and to find the Z score again, we're going to take the, um, inverse norm with 0.90 And the standard deviation is or sorry, the mean is zero the standard deviation of one. And we get the same Z score that we got in the previous step. The Z score again is 1.28155 But this time our ex formula has to change a little bit. So in the previous problem, we had the ex formula as being this But because one we're doing a sample of 10 and to we are modifying both e average and the standard deviation are ex formula is going to be this, um you times n plus z times and our standard deviation Because of our central limit, theorem will be sigma over the square root of n times n or we know we've got these values here. So we're going to say that X equals our average, which is 440 $1000 which was the mu times the end, plus our Z score, which is 1.28155 and then this answer turned out to be our 20,000 $554.80. So let's bring our calculator in and calculate that out. So we have $440,000 plus 1.28155 multiplied by the standard deviation of this data, which is 20,554 and 80 cents. We get an X value of 46 $466,342 and I guess we would say no sense. And again, based on how far out you carry your Z score, the sense might change. So what we're saying is the sum of the teacher salaries 90% of the some of the teacher salaries will be less than $466,342. Part H in Part H. We want to know if we were surveyed and we're sorry if we survey 70 people instead of 10 people. What would happen to this problem? How would it change the distribution in Part D. So in part d, we were talking about the some of X was approximated by the normal distribution and we used mu times n and we used Sigma over the square root of N times n so our information would change too. Our average 44,000 times 70 and our standard deviation is going to change to 6500 divided by the square root of 70 times 70. And in doing so, this is definitely causing our mean to move upward. So if I were to multiply 440 or sorry, 44,000 times 70 I'm getting 308,000. No, try again. Three million 80,000. And for a standard deviation, I'm getting $54,382 and 90 cents. So in the previous, just looking at the picture, the previous picture had an average here of 440,000. So now our picture has an average of 3,080,000. So the average has definitely slid to the rights or whole bell slides to the right. In the previous problem, the standard deviation Waas what was it? 20,000 $554.80. Now our standard deviation is $54,382.90. So what it's saying is our bell is going to be spread out further as well. So that is how by changing from a sample size of 10 to a sample size of seven, the graph is going to again shift the bell and stretch out of the bell summarizing it. This might have been the first one, the sum of 10. The second one, The peak is going to be somewhere higher up, and the bell is going to be very stretched out. Finally, part I of this problem. If each of the 70 teachers received a $3000 raise, how would that change the distribution in Part B? So in part B, we said that X was normally approximated, with an average of $44,000 and a standard deviation of $6500. So this is what the Bell would have looked like by giving each teacher a $3000 raise. Well, then, that average is going to change to be 47,000. And it's not going to affect the standard deviation at all, because the spread will remain the same. So, summarizing those two pictures on the same graph, we would have one bell right here with 44,000 as our peak and then the other bell would be the same shape and size. Just the peak would move to the right by $3000. So the difference between H and I with H not only does the peak move, but the spread changes with I, the peak just moves further to the right, and that concludes this problem.


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