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Rachel B.
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Psychology
TutorMe
Question:

Your company is developing a new drug that is claimed to lower heart rates. To test the effectiveness of this drug you collect a simple random sample of adults and randomly assign each to either a placebo group or the drug group. What are the null and alternate hypotheses in this study? Identify the independent and dependent variables.

Rachel B.
Answer:

I know that the null hypothesis states that there is no difference between the groups. Applying that to this specific study, I would say that the null hypothesis states, "The average heart rate of the drug group is not significantly different than the average heart rate of the placebo group." The alternate hypothesis, on the other hand, states what I predict will happen. Again, applying this to the specific study, I would say that the alternate hypothesis states, "The average heart rate of the drug group is lower than the average heart rate of the placebo group." An independent variable is the variable that is being manipulated or changed between the two groups. In this case, one group receives a placebo pill while the other group receives a drug. Therefore, my independent variable is the medication (or placebo) received. A dependent variable "depends" on the independent variable, or tells me what is being measured. In this example, I am comparing the average heart rate for each group. I expect that the heart rate depends on whether or not the participant received a placebo or the drug. Therefore, my dependent variable is heart rate.

Statistics
TutorMe
Question:

You predict that your class performed better than average on the last test you took. The average grade on this test is an 80, with a standard deviation of 5. Your class average was 88. Is your prediction correct? Use a 95% confidence level and assume it's a normal distribution.

Rachel B.
Answer:

Our first step is to identify our variables, or write down what we know. This can help guide you on where to start if you feel lost or confused. First, I am told that the average grade on the test is 80. In this scenario, I am going to assume that is the "population" average, which we denote with mu. So, mu = 80. Next, I am told that the standard deviation is 5. Again, assuming this is a population standard deviation, I am going to assign this value to my variable sigma. So, sigma = 5. Now, I know that my class average was 88. If the population is everyone that has taken the test, then my class would be a sample of that population. The average of a sample is called the sample mean, which we notate as x-bar. So, x-bar = 88. The final bit of information I am given tells me what I want my cutoff, or critical value to be. We'll come back to that in just a minute. Using the information I'm given, I'm going to sketch a normal, bell-shaped curve, labeling my mean in the middle and my critical region in the tail. In this problem, we are predicting that the class performed "better", this means that I'm predicting a direction and will be using a one-tailed test. Since I said "better" I am specifically going to shade the right tail of the distribution as my critical region. If my confidence level is 95%, that means only 5% of the data can fall into that shaded region. My task is to determine if my sample mean is in that 5%, or if it's in the 95%. Now that I have a visual of the problem, I can start plugging my values into my formula. Since we are told this is a normal distribution, I know I will be using z-scores to test this hypothesis. The formula for finding a z-score is: z = (x-bar - mu)/sigma. Basically, I want to subtract the population mean from my sample mean, then divide the difference by the population standard deviation. Plugging in the values I labeled at the beginning that becomes: z = (88-80)/5. Simplifying this equation I get: z = 8/5 = 1.60. I make a habit of rounding z-scores to two decimal places because the z-table uses two decimal places. When I look this z-score up in the table, I find the p-value to be p = .9452. An important thing to remember about the z-table is that it reports the area under the curve to the left of the z-score. Looking back at my drawing of the normal distribution, I see that I shaded the area to the right of the critical z-score. Finding the area to the right is simple when I remember that all probabilities add up to 1. If the area to the left of my test statistic is .9452, then I can find the area to the right by subtracting that value from 1: 1-.9452 = .0548. All that's left to do is compare our test value to the cutoff. There's a few ways to do this, but we'll go with the simplest one. We know that our critical region represents the upper 5% of the distribution. Rewriting 5% as a decimal I get .0500. In order for my prediction to be correct, my p-value needs to be less than .0500, which means it falls inside that critical region. The p-value we got from our calculations was .0548, which is greater than .0500. That means that my prediction is not correct, my class did not do better than average on the test because an average score of 88 falls within the 95% of scores we would expect to obtain by chance.

Algebra
TutorMe
Question:

Solve for x: 2x-10 = -x+2

Rachel B.
Answer:

Our goal when solving for the x variable is to isolate the x (get it by itself) on one side of the equals sign and the integers (numbers) on the other side. We do this by moving terms across the equals sign through inverse functions (that's a fancy way of saying "do the opposite"). While it doesn't matter which side we choose to put the "x" variable on, I like to choose the side that results in the fewest negative values. Traditionally, we write equations with the variable on the left, so let's go with that option. This means I want to move all of my x-terms to the left of the equals sign and all of my numbers to the right side. Let's move the x-terms first. In this problem, there is already one x-term (2x) on the left. We're going to leave that there. There is also an x-term (-x) on the right side. We need to move that across the equals sign by "making 0" or performing the opposite function. Since it is a negative x, that means it is being "subtracted" or we could make 0 by adding "x" to it (-x+x=0). We need to keep our equation balanced, so if we add "x" to one side, we need to add "x" to the other side too. So, just looking at the x-terms on the left side we have "x+2x". These are "like terms" which means we can add them. Any time we see "x" by itself we know it really means "1x," so this really reads "1x+2x". We can combine this to be one term: 3x. Now that we have all the x-terms on the same side, let's move the integer terms. There is one integer term on the left (-10) that we want to move to the right. Just like with our "-x" term, we need to perform the opposite function, or make 0, to move that 10. Adding "10" to both sides (again to keep our equation balanced) will move the "-10" to the right side of the equation. When we do this we are left with 3x on the left side and "2+10" on the right side, or 3x = 2+10. Again, we can combine like terms, so 2+10 can be rewritten as 12. So, we have 3x = 12. Our final step is to isolate the "x", or get it by itself, by moving the "3" to the other side of the equals. We will do this by performing the opposite function. We know that "3x" means "3 times x", so our current function is multiplication. The opposite of multiplication is division, that means I need to divide both sides by 3 to move the 3 to the other side. 3x divided by 3 becomes 1x, or just x. 12 divided by 3 reduces to 4. Therefore, x = 4.

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