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5.A drug _ that provides relief from headaches was tried on 18 randomly selected patients. The experiment showed that the mean time to get relief from headaches for...

Question

5.A drug _ that provides relief from headaches was tried on 18 randomly selected patients. The experiment showed that the mean time to get relief from headaches for these patients after taking this drug was 24 minutes with a standard deviation of 4.5 minutes. Assuming that the time taken to get relief from a headache after taking this drug is (approximately) normally distributed, determine a 95% confidence interval for the mean reliel time for this drug for all patients.

5.A drug _ that provides relief from headaches was tried on 18 randomly selected patients. The experiment showed that the mean time to get relief from headaches for these patients after taking this drug was 24 minutes with a standard deviation of 4.5 minutes. Assuming that the time taken to get relief from a headache after taking this drug is (approximately) normally distributed, determine a 95% confidence interval for the mean reliel time for this drug for all patients.



Answers

A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: $2.7 ; 2.8 ; 3.0 ; 2.3 ; 2.3 ; 2.2 ; 2.8 ; 2.1 ;$ and 2.4 . a. i. $\overline{x}=$ _____ ii. $s_{x}=$ _____ iii. $n=$ ____ iv. $n-1=$ _____ b. Define the random variable $X$ in words. c. Define the random variable $X$ in words. d. Which distribution should you use for this problem? Explain your choice. e. Construct a $95 \%$ confidence interval for the population mean length of time. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound. f. What does it mean to be "$95\%$ confident" in this problem?

We're in a hypothesis test to ensure that mu does not deviate from the population mean 58.4. Given a sample with data, sample size N equals 40 sample mean X bar equals 59.5 sample standard deviation as equals 8.3 Given this sample data and our population, we knew we wanted to drive the p value for this hypothesis test. So remember that our population standard deviation sigma is unknown. So we're gonna have to use a student's T distribution to solve this problem. So, using student's T distribution, the first thing we want to identify is the test statistic, T remember that T is given by the formula here, x minus mu over after the root end plugging in, we obtain T eagle, 59.5 minus 58 point 4/8 58.0.3 out of 40. Next, what we have to note is the degree of freedom for the T distribution. In this case it's n minus one equals 39 DF is always n minus one. So the P value, which is the area under the student's t distribution, is then The probability that T is greater than our T value for a degree of Freedom 39.

So we are given 200 pieces of data and further data. The meanest 12.8 in this area deviation is 1.7. And we want to make a 90% confidence interval. So we use this formula, the mean goes right here, Plus and minus the Z score for 90%,, goes right there, and then we have the standard deviation divided by the square root of the of value. And when we get that, that's what we get. Yeah.

So this is question 42. And in question 41 you had to find a 95% confidence interval. And so you were taking the X bar that was obtained, which was 1.5 hours and then plus or minus. And if we are to assume a normal population, then we're going to be using the 95% confidence interval number will be 1.96 and then we're going to use our sample size. And again we could use a T star value here if we weren't assuming assuming a normal distribution but with, uh, 69 degrees of freedom. But if we use that and then we had the standard deviation of the sample, which was 690.5 hours and then divided by the square root of 70. And so you obtain this being our margin of air. And again, this is from question 41 and you you get an interval that's going to end up being as low as, uh, let's see if we take 1.5 minus the 1.96 minus. Excuse me times 0.5, divided by the square root of 70. And that gives us 1.38 hours, as you know, and we have changing that to an addition Sign we get that the upper limit is 1.617 hours. So what does this mean? And that's what we're supposed to explain. And if we repeatedly took samples and I didn't put all the word, took samples of size 70 and then calculate an X bar in a standard deviation and then calculate our interval and use that formula of X bar plus or minus. And in our case, we would use that 1.96 times the standard deviation over the square root event. And we crank out an interval. We find that 95% of the time 95% of the time a calculated interval will contain the actual population needs. Okay, So, for instance, we calculated one here, and we got we think the mean is somewhere between 1.38 hours and 1.617 hours. Now, are we totally competent? No. Because when we use this technique, we find that 95% of the time when you get an X bar, crank out the interval and get a low limit and upper limit. We find that 95% of the time the calculated interval will contain the actual population means. So this one might. It probably does. We have a 95% chance that it does, but it might miss as well, because we know 5% of the time you calculate one of these intervals and you get quote a miss, so that would be a pretty good explanation.

In this question, we have to determine the sample sizes. The first one is at 95% confidence level at in part A. It's 95% confidence level. Which means my Alfa is zero point my Alfa 0.0 fight. This means that my Alfa by two is 0.2 fight. Okay, what is the margin of error? The margin of error e is given us. This is given other formulas e alfa by two Sigma by route and Z Alfa by two into sigma by route. And now N is what we want to find out. We know the value off the alphabet. How do we find that we use a critical value calculator and we put our value as 0.25 and the critical value for this turns out to be 1.96 This is 1.9 1.96 Now what is Sigma Sigma? Standard deviation And we are being told that we have to use a planning value for the population standard deviation off eight minutes. So I signifies it. So our margin of error has to be within two minutes, right? Yeah, this has to be within two minutes. And l z l four by two is 1.96 multiplied by sigma, which happens to be eight upon route. And now this is the value that I want to find out off route and And when I use the calculator, the value that I get is any is equal to 60 do and is equal to 62. This is the sample size that is required now when I want 99% confidence level. What is going to be my Alfa? By to my first of all, my Alfa 0.10 point 01 which means that my Alfa by two is 0.5 So what is going to be the critical value for this? Given Alfa by two it is 2.5 two points. Fight if I take it again, it is 2.575 2.575 Eat. All right, We just put in the values this is imagine, Offender, This is critical value multiplied by Sigma's eight upon Route 10. This is what we're gonna find out. What is the logic? Tell us what should happen Okay. Our sample size should increase right because we want to be more confident. So which means that our sample size should increase. So if I use a calculator for this, I'm getting my end as 107 My end is turning out to be close to 107 So this is a sample size that I required if I want to be 99% conflict.


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