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Suppose that researcher is interested in estimating the mean systolic blood pressure_ 4l, of executives of major corporations: He plans to use the blood pressures o...

Question

Suppose that researcher is interested in estimating the mean systolic blood pressure_ 4l, of executives of major corporations: He plans to use the blood pressures of random sample of executives of major corporations to estimate U. Assuming that the standard deviation of the population of systolic blood pressures of executives of major corporations is 25 mm Hg, what is the minimum sample size needed for the researcher to be 990 confident that his estimate is within mm Hg of 4?Carry your intermedi

Suppose that researcher is interested in estimating the mean systolic blood pressure_ 4l, of executives of major corporations: He plans to use the blood pressures of random sample of executives of major corporations to estimate U. Assuming that the standard deviation of the population of systolic blood pressures of executives of major corporations is 25 mm Hg, what is the minimum sample size needed for the researcher to be 990 confident that his estimate is within mm Hg of 4? Carry your intermediate computations to at least three decimal places_ Write your answer as whole number (and make sure that it is the minimum whole number that satisfies the requirements). (If necessary, consult list of formulas:)



Answers

Blood pressure screening Your company markets a computerized device for detecting high blood
pressure. The device measures an individual’s blood pressure once per hour at a randomly selected time throughout a 12-hour period. Then it calculates the mean systolic (top number) pressure for the sample of measurements. Based on the sample results, the device determines whether there is significant evidence that the individual’s actual mean systolic pressure is greater than 130. If so, it recommends that the person seek medical attention.
(a) State appropriate null and alternative hypotheses in this setting. Be sure to define your parameter.
(b) Describe a Type I and a Type II error, and explain the consequences of each.
(c) The blood pressure device can be adjusted to decrease one error probability at the cost of an increase in the other error probability. Which error probability would you choose to make smaller, and why?

We're going to start with what we know in this problem. We know that the mean systolic blood pressure of normal adults is 120 millimeters of mercury, and we know that standard deviation is 5.6. We also know that we're going to assume normal distribution. So again, we're going to use our bell shaped curve part A. If an individual is selected, so only one individualised selected. We want the probability that that individuals systolic blood pressure is between 1 20 and 1. 21.8. So we're gonna start by placing the 1 20 the mean on our bell, and we want to be between 1 20 and 1 21.8. So we will need our Z score for both, because 1 20 is right at the peak of the bell. We know that 1 20 would correspond with the Z score of zero and to calculate Z score for 1 28 we'll do 1 21 0.8 minus 120 divided by the standard deviation, which was 5.6, and we will get a Z score of 0.32 So when we talk about the X value between 1 21 21.8 were also talking about the Z value being between zero and 00.3 to. So we're going to solve that by changing this into its equivalents attraction problem. So the probability that Zuse less than 3.2, minus the probability that he's less than zero We'll use our standard normal table from the back of your textbook, and we find the probability that Z's list in 3.2 is 0.6255 We know that the probability that C is less than zero is 00.5000 because that is half the bell, so therefore we get a value of 0.1255 So the probability of us selecting an individual, um, with a pressure between 121 21.8 will be 0.1255 Now, let's go on to Part B. We're just gonna scoop this up a little bit and far for Part B. It's asking us now we're going to select a sample of 30. So let's still talk about what the population waas. The population had a mean systolic blood pressure of 1 20 with a standard deviation 5.6. And now we want to talk about the sample and the sample size. In this case, we're going to select 30 adults at random, so we now need the average of the sample means, and we need the standard deviation of the sample means, and the average of the sample means is the same as the average of the population, which is 1 20. And the standard deviation of the sample means is equal to the standard deviation of population divided by the square root event. So in this case, it would be 5.6 divided by the square root of 30. So again we're going to set up our bell shaped curve. We're going to put the average in the center and the average was 1 20. And let's look at the problem that we're trying to solve. And the problem we're trying to solve is a sample of 30 adults is selected at random. What's the probability that the sample mean is between 1 20? So this time we're gonna have an X bar in our statement and 1 21.8 so we're gonna need the 1 21.8 on our bill, and we want to be an average of the 30 adults to be between 1. 21 21.8. So, again, we're going to need to calculate the Z scores for each of those. And this time the formula is gonna change slightly. It's gonna be X minus or Sorry, X bar minus Musa of X bar over Cygnus X bar. So here we have an average of 1 21.8, minus 1 20 divided by 5.6 over the square root of 30. So the Z score turns out to be 1.76 again, we know that the Z score associated with 1 20 is zero because it's at the peak of the bell. And if you think about the fact that being between 1 21 21.8 is the same as when our Z score is between zero in 1.76 so we could rewrite that as the probability that Z is less than 1.76 minus the probability that is less than zero, and we would rely on our standard normal table and we would find that the probability that C is less than 1.76 is 0.9608 and the probability of Z being less than zero is 00.5000 and we get an answer of point for six zero eight. So for Part B, the question was, What's the probability that when we select 30 people at random there, blood pressure is between 1 21 21.8? And I should say their average is between 1 21 21.8 and that probability would be point for 608 and then finally, for part, C Part C is asking us why is the answer to part a so much smaller than part B? So let's just recap with the part A and part B were in part A. We were talking about an individual persons blood pressure, and we had an answer of 0.1255 and in part B. We were talking about taking a sample of 30 adults and their average blood pressure being between 1 21 21.8, and we found it to be 0.4608 So the question is saying, Why is this one so much smaller? And the reason would be that a sample mean will be closer to the true mean than an individual value, so the true mean is more likely to be the 0.46 several eight.

Problem. 13. That's a still a blood pressure of others follows a normal distribution with mean 112 and standard division of 15. A person is considered pre hypertensive. Three. Khyber Offensive if it's still a blood pressure, is between 121 130 120 and 130. We want to find the probability that the blood pressure of a randomly selected person is pre high. Burton's Let's recall the normal distribution, the normal distribution Isabelle shape centered about the mean she means we have here the mean 112 and they have a standard division of 15 for the region. We want to calculate we have the blood pressure between 121 130 which lies here about these two points 121 130. Using the normal distribution and getting the area under the curve between 121 130 we can find the required probability. We can do so by transforming these values into the standard values. Standard values have the bill shapes under at zero, and we have here that one and they took. We can calculate it using the formula which equals X minus mu, divided by sigma, where X represents the Khyber or the blood pressure. And that is the normal values, the standard values of the normal distribution. Then we can calculate that one as X one minus mu, divided by Sigma X one is 120 minus mu, which is given as 112 divided by Sigma, which is 15. And she means that one equals eight, divided by 15.53 as that one. Similarly, you can get the two x two minus meal funded by segment equals the two. X two is 130 minus 112 divided by 15 equals 18. Divided by 15 equals 1.2. Now we can use the standard normal distribution tables to get this area. But the table gives us the area to the left of the school, which means now we can get this area to the left, 120 get this area to the left of 130 and by getting carry on minus area to we can get the required area or that required probability. Let's get area one and area to from the standard tables. We want to get the value of the area to the left of the set of 4.53 We have here that 1.5 and we can get over and 53 by entering this table or this column. Then the value is 0.7 oh 194 And for the two we get 1.2 and to get 1.2 Oh, we entered the first column. Then the area is a 0.88 493 which means the required probability, the probability to have X between 121 130 equals all 0.88493 which is area too minus Oh, 0.7 oh 194 which is area one. And the answer is 4.18 to 99 or in percentage approximately equals 18.23 person. And this is the final answer of our problems.

The following is a solution for number 27. This looks at the mean systolic blood pressure being less than 100 and 30 for a certain floor for like a nursing student. And first we look to see if the conditions for inference have been met and you're given a normal probability plot. Um Mhm. Uh huh. I did not mean to do that. Sorry, normal probability plot. So uh this is what it kind of looks like in your books. You don't have to do that, but as long as it's linear, then we can assume normality. So that's one condition for inference and then the other one is a box plot and that kind of tells us if there are any outliers or skew nous. So this is what it kind of looks like for you and this definitely does not have any outliers. There are no stars on the outside. And then that really looks pretty darn good to me. That median is basically smack dab in the middle, maybe a little bit left skewed, but for the most part, the conditions for inference of Metz of the normal probability plot, it's linear. So we can assume normality and the box plot, it's free of outliers or really any type of skew as well. So they actually did a mini tab print out where the test statistic is negative 1.41 and a p value of 0.88. So it asked us what are the hypotheses and you don't have to use the context clues here in your original prompt, but h not is mu equals 1 30. And it says less than for the alternative. Some use less than 130 because that's what this nursing patient patient. Uh Thanks. And then it says identify the P value. There it is .088. That's just it's his pee pee. And that's the p value. So .088. And will the nurse rejected? 5? no P value is greater than alpha so failed to reject. Okay. So the conclusion here would be that there is not sufficient evidence to say that The blood pressure the average blood pressure is less than 1:30 on our floor. So not enough evidence to suggest the average. What pressure? All right. Blood pressure on her floor is less than 130.


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