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Adrical bial was conducted - testthe efiectnenas> olu drug kx treatng nsomn okter subjects Bebre tealmont 24 subjects had moan wake lime of 103.0 After Ueatmie...

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Adrical bial was conducted - testthe efiectnenas> olu drug kx treatng nsomn okter subjects Bebre tealmont 24 subjects had moan wake lime of 103.0 After Ueatmient te 74 sutjecs had = mean Wuke Itte 0/ 79 8 nn ada sndurd detulton 0 20 1ntn Assumno tat the 24 euineta vakuos appoar to bo hom _ nolmally dsstribuied papulation and constnuci 95 ** confxtence iierval estmaie olthe mean wake Ime lor population wth drug treaiments Whaf does Ihe resth sunesi about the Mean wake tme mnbelore Ihe treatm

Adrical bial was conducted - testthe efiectnenas> olu drug kx treatng nsomn okter subjects Bebre tealmont 24 subjects had moan wake lime of 103.0 After Ueatmient te 74 sutjecs had = mean Wuke Itte 0/ 79 8 nn ada sndurd detulton 0 20 1ntn Assumno tat the 24 euineta vakuos appoar to bo hom _ nolmally dsstribuied papulation and constnuci 95 ** confxtence iierval estmaie olthe mean wake Ime lor population wth drug treaiments Whaf does Ihe resth sunesi about the Mean wake tme mnbelore Ihe treatmeni? Doas Ihu drug aPaear be efectre? Construxt Uta 95 % conlxsunce intotval esenuato mejn Waka Eno k populaton Jeutt min < 4 < (Rourd decial pluce = nuaded )



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Preliminary data analyses indicate that you can reasonably apply the $t$ -interval procedure (Procedure 8.2 on page 346 ). \In $1908,$ W. S. Gosset published the article "The Probable Error of a Mean" (Biometrika, Vol. 6, pp. 1-25). In this pioneering paper, written under the pseudonym "Student," Gosset introduced what later became known as Student's $t$ -distribution. Gosset used the following data set, which gives the additional sleep in hours obtained by a sample of 10 patients using laevohysocyamine hydrobromide. $$\begin{array}{ccccc} \hline 1.9 & 0.8 & 1.1 & 0.1 & -0.1 \\ 4.4 & 5.5 & 1.6 & 4.6 & 3.4 \\ \hline \end{array}$$ a. Obtain and interpret a $95 \%$ confidence interval for the additional sleep that would be obtained on average for all people using laevohysocyamine hydrobromide. (Note: $\bar{x}=2.33 \mathrm{hr}$ $s=2.002 \mathrm{hr} .$ b. Was the drug effective in increasing sleep? Explain your answer.

So we're giving a problem that we have to find at the 5% significance level. The sample size is told to be of 10. So and is equal to 10. Which implies that are Significant value or critical value is at five and 10 and we're losing one till test. So we have our critical value of W. C. is equal to 44. Now when we take a look at this test that we're supposed to do it's one tailed. However which tail is it? We are asked to say that one is more effective than the other. So mm everything is mostly positive here and we find what are our values? Mhm. Use the following data set which gives the additional sleep in hours obtained by 10 patients who used love level high school. So I mean high bromide. So that's gonna be so the control group minus the other one or not. The control group. So what we'll just do here is calculate the both ends since I'm not sure if it's a To a left tail or right tail. So we'll go ahead and find out that right tail is at 44 and then the other tail is going to be 10 times 11 divided by two. So that's gonna be 55 minutes 44 Which is gonna be 11. Very handy. Okay. So if it's less than 11 won't reject and if it's greater than 44 will also reject. Sound good? So um we got 10 students at the positive values so 1.9 plus 0.8 plus 1.1 plus 0.1 was 4.4 plus 5.5 plus 1.6 plus 4.6 plus 3.4. And we get 23.4 for our w nut. Okay. And or critical value that just leads us straight in the middle. So at alpha equals .05 We failed to reject the null hypothesis and this was just a right tail test that are evaluation is insignificant. So now if we tried it again at the 1% significance level, mhm at point a one that changes our values to be mhm uh 50. So we got 50 right here and then we do 10 tons 11 divided by two. So 55 -50, which is going to be five. And once again 23.4 still lands us right in there. So we failed to reject the my hypothesis once more.

And this question. We have a table that gives a sample of 10 patients who have a particular disease. So 10 patients, which gives us the end n is what our sample says, which happens to be 10 in this case, the number of observations. The next one is submission of exile. There is submission of all the observations, which means X one plus x two plus X three all the way up to extend right the submission of all of them. So if I add all of these up, I'm going to use a calculator for this one. So this is going to be 1.9 plus 0.8 plus 1.1 plus 0.1 minus 0.1 plus 4.4 plus 5.5 plus 1.6 plus 4.6 plus 3.4. The addition is 23.3. The addition is 23.3. Now, what is going to be expire expire is nothing. But the sample means so everybody is given by submission of X I upon in which is nothing but 23.3 by 10, which happens to be 2.33 hours. This is our answer

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.

To calculate the standard division, we have first to calculate them in value, which equals the total sum of the four numbers divided by four equals seven plus seven plus nine plus mine divided by four equals eight hours. Then we can calculate the standard division by the formula Square root of some nation off X minus Export squared, divided by n minus one equals square root one divided boy three but the boy a boy seven minus its quit, which is one plus seven minus eight squared one blows mine minus it. Also one squared plus Lyon minus it squared. This gives square root four. The valiant boy three equals 1.1 play five hours. This means that operation off sleep claim over the mean difficult had the valley off plus or minus 1.155 hours for birth. E. We can't conclude or make any conclusions from these values because the same bill is not representative off the village


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