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Sasha is wondering what percent of students on her campus live in the dorms. One of her friends told her that 60% of students live in the dorms at their school Sas...

Question

Sasha is wondering what percent of students on her campus live in the dorms. One of her friends told her that 60% of students live in the dorms at their school Sasha randomly samples 148 students at the school and asks each of them if they live in the dorms or not: 76 of the 148 students (51%) she sampled live in the dorms: Use this sample to test whether or not her friend seems right or not: a) Identify the null and alternative hypotheses in words and using proper notation: Also identify the v

Sasha is wondering what percent of students on her campus live in the dorms. One of her friends told her that 60% of students live in the dorms at their school Sasha randomly samples 148 students at the school and asks each of them if they live in the dorms or not: 76 of the 148 students (51%) she sampled live in the dorms: Use this sample to test whether or not her friend seems right or not: a) Identify the null and alternative hypotheses in words and using proper notation: Also identify the value of the statistic including the symbol: Solution: Null hypothesis: Ho: T = 0.5 Ha: T # 0.5



Answers

Instructions: For the following ten exercises, Hypothesis testing: For the following ten exercises, answer each question.
a. State the null and alternate hypothesis.
b. State the p-value.
c. State alpha.
d. What is your decision?
e. Write a conclusion.
f. Answer any other questions asked in the problem.
The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a standard deviation of 45 minutes. At ? = 0.01 level, is the student academic group’s claim correct?

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.

Right, we have a sample with N is equal to 49 values. That sample has mean 8.5 and standard deviation 1.5. We want to use a significant level of 0.1 or 1% to conduct the left field test for the population. Mean new equal 9.2 1st. We want to answer whether or not a this is appropriate to use the student's T distribution to solve. Yes, it is because we have any greater than equal to 30 samples. Note that we're gonna have a degree of freedom equal to n minus one or 48. Since students T tables do not use uh 48 or values typically between intervals of five for very large numbers, we're going to approximately this down to 45 so as not overestimate the p value. Next let's take the hypotheses are null hypothesis. Station on his musical 9.2 are alternative hypothesis. Is that new is less than 9.2 next computers. T statistic. Remember the T stat is defined as x minus mu, divided by estimate of the standard deviation. Then in this case that's T equals 5.6. Next use the one tail tea table to derive RP interval. We see that this test statistic T is extreme, so we have the interval P is less than 10.5 incredibly small. P value. Next we reject H not, Yes, we do, because our P value is less than equal to alpha. And we interpret this to mean that we have evidence suggesting new is actually less than 9.2.

So in this question you want to use the scientists to find out whether the data received from the Entertainment Software Association indicates that less than 50%. Yeah. Of gamers are women cameras are okay. They randomly select 100 computer and video games. I found that 38. This is a left tail. Mhm loss. Yeah. And we have the hypothesis. She's something that probably yes, you're right. Uh not caring about cyst or really less like. Okay. I want you to do that. That's right. Yes. I have outfit people to zero show one. I think the euro area tables find the cream from value in this land of significance to be negative. Mhm. Mhm. Mhm. Yeah. So now you can determine test statistics. This is a Yeah Large sample case and Speaker five. Okay. It's not. Let's see you're not both cheap. Okay. Plus your five minus all over. Straight to bed too. Shoot. Let's Yeah, that's 75 minus two. Mhm. I'm afraid fish. And about it. So. Mhm. Should be in this accident. What do you do? Who since? So you're not scared. Hi. We cannot. Yeah. With yeah, project in other places. Okay. No. Therefore there is insufficient evidence to prove that 50% of computer and video game players are okay. Yeah.

Let us read this question. A random sample of 50 medical school applicants at the university has a mineral score of 31. So x bar is 31. My end is 50 on the multiple choice portions. Medical college admission test. Okay. A student says that the main draw school for schools applicants is more than 30. Is there enough evidence to support the students? Cling. Okay, So a student says that the mean raw score for its schools applicant is more than 30. He's saying that it is more than 30. So my new is more than 30. So let this be my alternatives. So what will be Mindulle? It is less than or equal to 30 right? This is my age. Not This is my ex alternative. Okay? Yeah. Population Standard division is 2.5. So let me simply do one thing. Let me directly like the formula. Now, this is going to be 31 minus 30 upon Standard Division is 2.5. This is 2.5 upon route over 50 upon router 50. Right. This is going to be route off 50 on If I use a calculator for this, this is going to be one divided by 2.5 multiplied by route off 50. It is 2.82 My statistic. This is my Z value for my Zeev value is coming out to be 2.8 to 8. Now, this is a right tail test because I'm checking for greater than 30. So this will be a right tail test. And my Alfa in this case is 0.1 And what is my critical value? 2.32 This is 2.32 This is my critical value on my critical well on myself. Statistic that I've calculated falls in the rejection reason off. Two point it Do it. So I will say that I will reject my age, not reject h not. Which means I would say that the claim that is being made by the student is actually correct. And this would be my answer


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