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A researcher measures the relationship between sleep medicationuse (times used per week) and time spent working (in hours perweek). Answer the following questions b...

Question

A researcher measures the relationship between sleep medicationuse (times used per week) and time spent working (in hours perweek). Answer the following questions based on the resultsprovided.SleepMedication Use Time SpentWorking 918437717232Part (a) Compute the Pearson correlation coefficient. (Round your answerto three decimal places.)Part (b) Multiply each measurement of drug use times 3 and recalculatethe correlation coefficient. (Round your answer to three decimalplaces.)Part (c) Divide eac

A researcher measures the relationship between sleep medication use (times used per week) and time spent working (in hours per week). Answer the following questions based on the results provided. Sleep Medication Use Time Spent Working 9 18 4 37 7 17 2 32 Part (a) Compute the Pearson correlation coefficient. (Round your answer to three decimal places.) Part (b) Multiply each measurement of drug use times 3 and recalculate the correlation coefficient. (Round your answer to three decimal places.) Part (c) Divide each measurement in half for time spent working and recalculate the correlation coefficient. (Round your answer to three decimal places.) Part (d) True or false: Multiplying or dividing a positive constant to one set of scores (X or Y) does not change the correlation coefficient. Note: Use your answers in (a) to (c) to answer true or false. True or False



Answers

A study of the effects of smoking on sleep patterns is conducted. The measure observed is the time, in minutes, that it takes to fall asleep. These data are obtained: $\begin{array}{lllll}\text { Smokers: } & 69.3 & 56.0 & 22.1 & 47.6 \\ & 53.2 & 48.1 & 52.7 & 34.4 \\ & 60.2 & 43.8 & 23.2 & 13.8 \\ \text { Nonsmokers: } & 28.6 & 25.1 & 26.4 & 34.9 \\ & 29.8 & 28.4 & 38.5 & 30.2 \\ & 30.6 & 31.8 & 41.6 & 21.1 \\ & 36.0 & 37.9 & 13.9 & \end{array}$ (a) Find the sample mean for each group. (b) Find the sample standard deviation for each group. (c) Make a dot plot of the data sets $\mathrm{A}$ and $\mathrm{B}$ on the same line. (d) Comment on what kind of impact smoking appears to have on the time required to fall asleep.

Mhm. This is a template off one of the past questions that we did. We're going to use this same template again. This all this question as well. Let's go through the question. Ah, clinical trial was conducted to test the effectiveness off a drug Zopa clone for treating insomnia in older subjects. Before treatment with Rubicon, 16 subjects had a mean time off. 102.8. Okay, So mean off 102.8. All right. After the treatment, the 16 subjects had a mean wake time off. 98.9. Afterwards this was before and x bar. This will be after afterwards. What did they have? 98.9 98.9. Okay, under standard deviation off 43.2, 43.2 minutes and s waas 43.2 minutes. All right. Assume that the 16 sample values appeared to form Normally distributed population can construct in 98% confidence interval for the estimate off. Mean weight over the population with Zo pick long treatments with topical treatments. That is after they are treated. So this is going to be my I mean, this is going to be my sample standard deviation and in this case is equal to 16. So degree of freedom is going to be degree off. Freedom is going to be 15 now. I want a 98% confidence interval, which means my Alfa is equal to 0.2 So what is going to be my Alfa by two Alfa by 20.1 And what is the formula that we're using? It is this very same formula expert plus minus the margin of error and margin of a resident by the Alfa by two as by routine. And when I do this, the confidence interval that I'm getting is 74. Sorry. 71.421 26.4. 71.4 21 26.4. 71. Excuse me. 71.421 26.4. This is the confidence interval that I'm getting. What does the results suggest? About the mean time of 10 to before the treatment? Well, what was the treatment? Uh, to test the effectiveness for treating insomnia. Okay, so 102 does appear to fall in this region. So we can say that there is not enough statistical evidence to conclude anything because it does fall in this region. So we do not have enough evidence to say that the sleep time has actually increased or decreased. We cannot say anything so the drug does not seem to be very effective or does not seem toe change mute by much before New Waas 102 pointed and after that newest 98.9. So that is it does not seem to be very effective.

So if you let d stand for the time without the drug minus with the drug and looking at the number of hours of sleep then we can put all those values in and find what R. D. Bar is. And let's quickly do that calculation And I put all 16 of those values in so and is 16. Yeah and let's see what we get there. We end up getting that. The mean is negative 1.5-5 and our standard deviation is .54-2. And then there's more decibels on it. So we're going to take for a 90 confidence interval. We need to figure out what those t. Values are. And so are degrees of freedom is 15 And so 15° of freedom or the 90 confidence interval which means there's going to be five in each tail That corresponds with. And let me get to the value looks like this is 1.7 5 3. So we're going to take our negative 1.5-5, We're going to take it plus or minus and then our T. Star value which is at 1.753. And then we're going to take that standard deviation .54-2 Over the square root event, which is 16. Uh huh. And then when we do our calculation we can go to our task and let me get those values. And I find out that this low value is down here at negative 1.7626 And the upper is up at negative 1.28 74 and so are mean difference where 90% confident confident that that mean differences somewhere between those two numbers.

We want to construct a confidence interval about the population mean mu given a sample, the sample mean X bar equals 2.33 sample size and equals 10 sample standard deviation S equals 2.2 at the confidence level of 95% Construct this confidence interval. The first thing we need to do is identify the T score associated with this degree of freedom and confidence level. So our degree of freedom is nine. Our confidence level is the probability or rather the area under the students distribution between the critical t values .95. And we can use a tea table to map for these two variables. The exact T score needed. Doing so. Using a tea table from google or a textbook, we obtained TC was 2.262 Next we want to identify the margin of error. E given by the formula left equals two times as of a root. And plugging in these variables for this problem we obtain equals 1.43 Plugging into our confidence interval, we can compute the conference interval as experiment E is less than new and less than X Y plus E. During so we obtained 0.9 is less than you as less than 3.76. We can interpret these findings to suggest that we can say with 95 confidence That the mean is between 0.9 and 3.76 and that therefore the mean is always positive.

So I'm letting d stand for taking the number of hours without medicine minus with the medicine, so I'm subtracting top minus bottom how a bunch of our negative and I get that the D Bar is negative .43 uh 57 Okay. And I get the standard deviation of those values to come out to the .67665. And we had a sample size of 14, so we're going to take that devalue, mhm plus or minus, and we have to find the T. Star value For 13° of freedom and we're finding a 95 confidence interval. Just double check and make sure I have that correct? Yes, I do. So 95 confidence interval. And when we look up the 95 confidence interval for 14° of freedom, We end up finding out for 95 14, I'm sorry, 13° of freedom Comes out to be 2.160. So we would put our 2.160 right here. And then we take our standard deviation Over the square root event. And when we do that calculation out, we find that the using the subtraction sign, we get negative .8264 And the -4 .4357 Plus. That product ends up giving me negative 0.45 So we're 95% confidence that that means differences. Somewhere in here.


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