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Point) To test the efficacy of a new cholesterol-lowering medication; I0 people are selected at random: Each has their LDL levels measured (shown below as Before), ...

Question

Point) To test the efficacy of a new cholesterol-lowering medication; I0 people are selected at random: Each has their LDL levels measured (shown below as Before), then take the medicine for 10 weeks, and then has their LDL levels measured again (After) . Subject Before( B) After( A) 192 186 164 143 164 134 134 99 189 185 118 107 138 123 184 189 155 121 10 127 122Let D = B -A Find the 80% confidence interval for the average difterence botween LDL levels before and after taking the medicatlon_Kp

point) To test the efficacy of a new cholesterol-lowering medication; I0 people are selected at random: Each has their LDL levels measured (shown below as Before), then take the medicine for 10 weeks, and then has their LDL levels measured again (After) . Subject Before( B) After( A) 192 186 164 143 164 134 134 99 189 185 118 107 138 123 184 189 155 121 10 127 122 Let D = B -A Find the 80% confidence interval for the average difterence botween LDL levels before and after taking the medicatlon_ Kp



Answers

Ten subjects with borderline-high cholesterol levels were randomly recruited for a study that involved taking a nutrition education class. Cholesterol readings were taken before the class and 3 months after the class. Let $d=$ preclass cholesterol $-$ postclass cholesterol. Excel was used to find the $95 \%$ confidence interval for the mean amount of reduction in cholesterol readings after taking the nutrition education class. Verify the results shown on the output by calculating the values yourself. Assume normality.

Look at this question In a test of the effectiveness of garlic for lowering cholesterol, 49 subjects were treated with raw garlic and is equal to 49 the first thing to notice and is equal to 49. Therefore, my degree of freedom is going to be in minus one, which is 40 in next. Cholesterol levels were measured before and after the treatment. The change in their levels off LDL cholesterol had a mean, off zero point for the change had a mean of zero point forward. X bar was 0.4 and standard deviation off 21 under standard deviation off 21 Really standard deviation off 21. Okay, construct a 95 98% confidence interval 98%. Which means if my Alfa my Alfa is 0.2 my Alfa by two will be 0.1 construct a 98% confidence interval estimates off the main again. We do not know sigma for the population. This 21 is the standard deviation for the sample, so we will use the tea distribution. What does the confidence interval suggest about effectiveness off garlic and reducing LDL cholesterol Okay, let's start to solve this again. I will just simply substitute these values in this formula. And the formula gives me a confidence interval. It is going to be a lower limit. Comma, upper limit. This will give me two values. One will come from the minus sign and when will come from the plus sign. So what exactly is the in double that I'm getting in? This case is minus 6.836 minus 6.836 minus 6.836 to 7 point something. This does not seem to be effective. 7.6367 point 636 and zero does lie in this interval. So the change that is Zito just lie in this interval. So it is quite possible that there is absolutely no change. And before and after. So looking at this interval, we cannot say anything conclusive. We do not have enough evidence to say that this is effective

Hello. Hi. Hearing this question, we have to determine the chi square value first. Okay, So the critical value at 98% of confidence, we had to find out on when you Skype sky skirt table first V c. A number off samples given here is 49. Okay, so you can find a decrease of freedom. Degrees of freedom will be getting and minus one that is equal to 48. You can see that 48 value is not there in the table on you can see 48 is greater than 40 from The terrible will be using the value 40 in order to get the values off particular values. Okay, now see values 98% so we can write physical 2.98 Then you can obtain Alfa values. I'll physical 21 minus e one minus three is one minus 10.981 minus 0.98 You can write 0.0 toe, then left. Detailed critical value. Left tail critical value that is chi square one minus alphabet will be K square one minus. Alfa by. To hear, it will be 0.99 from the table. We get it is the call to 22 point 164 Next to write tailed critical value. This guy square alphabet. Do you get square off? 0.1 from the table. We obtained the value 63.691 Okay, once we get chi square values, the next thing we can find out, we can find the confidence in trouble. Estimate four Standard deviation. We know the formula. The formula is given by what is square root off and minus one by Chi Square Alfa by two times standard deviation less than sigma. Less than squared off and minus one by chi. Squared one minus Alfa by two times standard deviation. Okay, so once we know these values, we just start substitute all these values given and minus one value. You know that is equal to 48 Chi square. Well, we found out that is equal toe 63 point on 691 times. The standard deviation value we know it is given in the question is is equal to 21. We get next to less than Sigma less than here on the right. Answer with how again? 48 by the chi square value. We know that is able to convict 2.164 times the standard deviation again. Two D one. So we just started simplified these values. Okay, when you simply find you get the value 18 point 2306 less than standard deviation value less than 13 point 9041 Okay, so from here, you can see this value does not give it any information about effectiveness. Right? You can see this value does not give any information about them effectiveness. While the value we can still stigma values between 18.23 on 13.9. Right? So that's why we cannot have any information about effectiveness off particular value. Here. I hope that this answer your question. Thank you.

All right, that's a good question. 34. So here were given a, um up a table computed by excel in the question so we can use the information from the table directly to, um make decision off whether we're going to reject no high positives or not. So first, I'm going to stay. The no hypothesis and the alternative deposits is it's going this question are on. No hypothesis is on. The difference meant me no difference. Sequel to zero and the alternative hypothesis says me no differences, um is positive. So which is greater than their own? I hear d is indicating the score for prey class my nose opposed to close All right. And our confidence level off Kulka is equal to zero point both five. So in the fulling table were given that if we look at the third last roll, we can know that our T star is equal to three point three night and P value iss. They're opined arrow there Oh 397 which is prayer? A very small and obviously it is smaller than the over here. So why does we're going? Teoh reject No high pauses is and we can also compare our T star with the tea critical value you hear, the degree of freedom is nigh Alfa a their own point of a fire and al Qaisi area for one toe. So this critical battle is equal to 1.833 So because our key star is greater than this 1.83 straight So which minty star fought into the rejected Rijn? So we're going to reject the no hypothesis as well.

These are the statistics that we have. We have a sample size of 49. Our sample standard deviation is 21 our sample means 0.4. Now, what exactly is the claim? If I look at the question? Okay. We are testing the claim that with the garlic treatment, the mean change in LDL cholesterol is greater than zero, which means are null hyper. Yeah, are null hypothesis and our alternative hypothesis. Okay, so we are checking for me to be greater than zero. And what does this mean exactly? This is the difference in before and after. So this will become our alternative hypothesis. So what will be a null hypothesis? It will be mu is less than or equal to zero. Okay, now we're going to use a T test since we do not know the population standard deviation. And what is the formula to calculate the T statistic? It s t is equal to expert. This is the value that we already have minus immune. That is the hypothesized me which happens to be zero in this case, upon s sample standard deviation, we already have divided by root end the verbal root end on. What is n n is 49. So if I put in all the values over here mighty value turns out Toby 0.133 Mighty statistic turns out to be This turns out to be 0.133 Okay, Our Alfa is 0.5 We are going to perform a test and we need to know Alfa. And this is going to be our Alfa for this question. And what is the degree off freedom? Degree of freedom is given by n minus one. Degree of freedom is given by end minus one. What is N sample size, which is 49. So what is it? Minus 1 49 minus one, which is 48. This becomes my degrees of freedom. Now, in order to calculate the p values, we need only two things. One is the T statistic, and one is the degrees off freedom. Okay, so what is a P value that I'm getting the p value? If I use a online tool, then I get my P value somewhere around 0.4472 So my p value my P value is 0.4472 And this is much greater than Alfa. Right? So what I can say is, since my P value is greater than Alfa, I fail to reject each. Not I fail to read, Shake age not. Okay. Now let us look at the question. What was it? Not h not waas. That are mean. That is the difference of the mean Sorry. That is the mean difference is less than or equal to zero. So we do not have enough statistical evidence to say that arm you is greater than zero R In medical terms, we say that we do not have sufficient evidence to suggest that the treatment with raw garlic actually reduces the cholesterol levels. Okay, and these would be my answers.


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