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# Tutor profile: Bryan P.

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Bryan P.
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## Questions

### Subject:Spanish

TutorMe
Question:

Fill in the blanks of this sentence with the proper verb (either ser or estar) and the proper verb conjugation. Yo _______ muy cansada, pero si tú ______ listo para ir, mi hermano y yo podemos ______ listos en diez minutos, porque _______ tus mejores amigos.

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Bryan P.

### Subject:Statistics

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Question:

Maya and her 9 friends took an IQ test and scored an average of 98. After someone teased her for group for being below the average of 100, she decided to do a statistical test to see if they were really below average or if the difference was just due to chance variation. Luckily, the distribution of the IQ test they took was standardized, so the mean and standard deviation are known. The population mean is 100 and the standard deviation is 15. At alpha = 5%, was Maya's group's mean score significantly lower than the population mean or was the difference simply due to chance?

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Bryan P.

When we're asked if something is significantly lower, higher, or different, then we know we're dealing with a hypothesis test. The first thing we need to figure out is what kind of hypothesis test, and that starts with looking at what kind of data we have. These numerical test scores are a quantitative variable and we already know the mean and standard deviation of the population. And we're only testing one group, so we know we're doing a 1-sample z-test for means. We also know the question isn't simply if these scores are *different* from the average. We want to know if they're significantly less, so it's a one-sided test. Let's write our hypotheses to get us on the right track. The null hypothesis should be the less interesting or the pre-assumed answer. Almost always, it's the one that has an equal sign in it. So our null hypothesis is that Maya's group's average score is equal to the population average of 100. (H_0: Mu_group = Mu_population = 100) and the alternative hypothesis is that the group average is less than the overall population average (H_a: Mu_group < Mu_population) So we know our test statistic is z. And we can look up the formula for z in the case of a 1-sample z-test for means (or we might have it memorized by now). It is this: z = (the observed sample mean represented by x-bar minus the null hypothesis value we're testing against represented by Mu_0) divided by (the population standard deviation divided by the square root of the sample size of the group). z = 98 - 100 ------------ = -0.42 (15/sqrt(10)) What this number means is that Maya's group average is 0.42 standard deviations of the sampling distribution lower than the average group of 10 students taking the test in this population. But does that mean they were significantly lower? We don't know until we find out how probable it is to score as low or lower than they did, but the z-score is going to help us get there. At this point in the real world, we'd probably just punch it into a computer and get the answer. In school, the method you probably have to use isn't much more complex or mathematical--you just look it up on a table. When we look up -0.42 on a 1-sided z-table (because we're looking for "less than", not just "different in either direction"), we get .3372. What that means is that the probability of a group of 10 people in this population scoring a 98 or lower is 33.72%. That's our p-value. In the problem, we said our alpha is 5%. Essentially what that means is that we're not going to reject the null hypothesis in favor of the *more interesting* alternate hypothesis unless we see an outcome rarer than 5% of the time. Our p-value was 33.72% which is much higher than 5%. So our conclusion is that we don't have evidence to reject the null hypothesis. The people teasing this group for having a lower average IQ score doesn't just need to get a life--they also need to do a hypothesis test to see that the small amount that this group's average was less than the population average was possibly just due to chance.

### Subject:Algebra

TutorMe
Question:

Dan is looking at 2 potential payment plans for a subscription service. One payment plan has a start-up fee of \$45 with a \$5 monthly fee after that. The other payment plan has a lower start-up fee of \$15, but the monthly fee is \$10. Realizing that the first plan will be cheaper over the long-run, but the second plan will be cheaper up front, he wants to know what lengths of time would be best for each plan, so he can make the decision. If he plans to keep the subscription for a year, which plan should he pick?

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Bryan P.

The first thing I would naturally do, even if no one were asking me to solve a problem with this information is express the 2 subscription plans as equations. Both payment plans have a fixed portion that they pay regardless of how long the subscription lasts, so we'll call that our Y-intercept. The monthly cost is our slope--as the month increases by 1, the amount of money Dan has paid goes up \$X. So we now have the equations Y = 45 + 5X and Y = 15 + 10X. At this point, we might graph the 2 equations representing the amount of money paid after a given number of months and we'd see how they relate. The first plan starts higher but goes up slowly; the second plan starts lower but is steeper. They cross at 6 months, so we can see that, for the first 6 months, the second plan is better, but after that, the first plan is better. Graphing is an acceptable method, but not always a feasible option, especially when we get into more complex formulas The algebraic way to solve this problem is more helpful in a wider set of circumstances, so let's do it that way too. The variable Y represents the amount of money Dan has paid after X months. Notice that both of the formulas are equal to a different hypothetical version of Y. What we want to know is where those formulas are the same--where the lines cross. To do that, we set the right side of the equations equal to each other. It's a simple substitution--if Y = 45 + 5x in one equation, we can substitute it for Y in the other equation (or vice versa). Now we have 45 + 5X = 15 + 10X and we just need to solve for X. Solving for X means isolating one X on one side of the equation while keeping the equation still true by performing the same operations to both sides. So we'll start with subtracting. There are more X's on the right side, so let's subtract the lesser amount from both sides to keep things positive. Now the 45 is alone on the left side and it's equal to 15 + 5x (45 = 15 + 5x) Now, we want X to be by itself, so let's move that 15 to the other side by subtracting it from both sides. So now 30 = 5x. But remember that 5x isn't what we care about. We care about X. X is the amount of time passed where the 2 subscription plans cost the same amount. So we need to divide both sides by 5 and we're left with x = 6. Just like we saw from the graph. The equations are simple enough that we can tell that the first plan costs more before 6 months and the second plan costs more after 6 months, but if we want to be sure, we can go back to our original equations and plug in some values. When X = 3 months, what is Y in each scenario? Y = 45 + 5 (3) = \$60 Y = 15 + 10 (3) = \$45 Just like we thought. Plan 1 is more expensive when X is less than 6 months. Now let's try it at X = 9 months. Y = 45 + 5 (9) = \$90 Y = 15 + 10 (9) = \$105 Yep. More expensive in the long run.

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