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00214515Using traditional methods it takes I01 hours I0 receive an has been proposed; advanced drtang " license researcher used the technique = training techni...

Question

00214515Using traditional methods it takes I01 hours I0 receive an has been proposed; advanced drtang " license researcher used the technique = training technique using Cornputer Aded Instruttion (CNI standard devlation , on 40 students and observed tiat they had known t0 be there evidence mean 0f 103 hours; Assume the population the 0.01 level that the technique lengthens the training trne? Step State the null and alter tlative hypotheses;Answer PointsTables Keypad Kryboard shortcu

00214515 Using traditional methods it takes I01 hours I0 receive an has been proposed; advanced drtang " license researcher used the technique = training technique using Cornputer Aded Instruttion (CNI standard devlation , on 40 students and observed tiat they had known t0 be there evidence mean 0f 103 hours; Assume the population the 0.01 level that the technique lengthens the training trne? Step State the null and alter tlative hypotheses; Answer Points Tables Keypad Kryboard shortcu



Answers

Doing Time. Refer to Exercise 10.115 a. Use your procedure from Exercise 10.123 to perform the hypothesis test. b. Compare your result in part (a) to the one you obtained in Exercise $10.115(\mathrm{a}),$ where you didn't use the normal approximation.

Today we need to solve a problem about hereby equals little 0.79 seven men. It is two minus X minus five point for the whole square, developed by 0.5. So this is the 0.5 point four comma 40.79 79 I have raised off normal. A distribution is the relate to maximum off the coop, which is exactly 5.4 in the normal distribution equation. X minus 5.4 part in Generally it's minus mu. Let him Ulisse, I mean, or Everett. So 5.4 I was That's the end off a question. Thank you.

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.

All right for this question they are asking us to find first the null hypothesis in the alternative hypothesis is so we have a claim that the freshman students on a campus study at least 2.5 hours per day. The null hypothesis is that claim so freshman study at least greater than equal to 2.5 hours for a day. And the alternative hypothesis is what the statistics cost. They're skeptical of this claim. So the alternative hypothesis is that freshman studied less than 2.5 hours per day. So what we need to do is we need to figure out is this null hypothesis Is this claim true? In order to do that, we need to find the p value. So in order to find the P value, the first thing we need to do is find the test statistic on that shows us how many standard deviations away from the mean is the results. So our T statistic, we find that with the formula of the, um, sample me and minus the population mean over the sample standard deviation over the square root of the sample size. So when we put that information in, Um, we're gonna go ahead and convert the minutes in the problem. Two hours. Um, to do that, you just divide by 60. So we have our 137 minutes becomes too point to a hours minus 2.5 hours over the sample standard deviation, which is, um, 0.75 hours, 45 minutes. It's 450.75 hours divided by the square root of the sample size, which is 30. And when you put that in your calculator, you get negative 1.6 07 Now that we have our t score, um, we need to find the P value. To do that, all you need to do is look up a tea table. You can even just google that and you'll find a tea table. You find negative 1.6 of seven, and it will give you the P value. The P value that you are given with that, um, to score is 0.6 now. What this means is that the probability of getting 137 minutes as the means so that's the 2.28 hours in the study is basically a 6% chance if the null hypothesis is true. So in order to decide about this study, we need to know what Alfa is. Alfa is the significance level. So Alfa is the probability of rejecting the null hypothesis when in fact it's actually true. The Alfa that they gave us is 0.1 which means that 1% of the time on the threshold is 1% of the time. You would reject the normal hypothesis when, actually it's true. Our P value, as we've already found, is 0.6 If the P value is higher than the Alfa, then that means that we do not reject the NOL. We keep the null. So go ahead and write for our d um, that we failed to rejection and all the P value. When the pea value is high, then all flies. When the people you is low, the norm must go. So that's a little saying to help you remember. So since the P value is greater than Alfa, then re failed to reject. So our conclusion in summary is that the student claim that freshman study at least two enough hours today is correct hours per day, and what it means is that we at least it's true. Um, what it means is that at least we don't have enough evidence to say that that's not true. So we're going to say that, according to this study, that the null hypothesis we cannot reject it, that it is still true that freshman students study at least 2.5 hours pretty.


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