Testing the Difference Between Two Means (d) find the standard...

Question

Testing the Difference Between Two Means (d) find the standardized test statistic t, Assume the samples are random and dependent, and the populations are normally distributed.

[APPLET] Migraines

A researcher claims that injections of onabotulinumtoxinA reduce the number of days per month that chronic migraine sufferers have headaches. The table shows the number of days chronic migraine sufferers suffered migraines before and after using the treatment. At , α= 0.01 is there enough evidence to support the researcher’s claim? (Adapted from Journal of Headache and Pain)

Table comparing the number of migraine days before and after treatment for chronic migraine patients.

Accepted Answer

Hello and welcome back, everyone. Our next problem says a matched pair study records the number of times participants woke up at night, before and after drinking chamomile tea. The differences, and the parenthesis before minus after for 8 participants have a mean difference, D bar equal to -1.5, and standard deviation as sub D of 1.2. Find the test statistic T for testing whether chamomile tea reduces nighttime a weakenings. And our answer choices are A, T is approximately 3.536, B, T is approximately 2.115, C, T is approximately 1.281, or D, T is approximately 0.942. So let's recall we're going to be doing a paired. Tea test. So we're going to need our equation there, which says that T. As equal to and then the numerator D bar. Minus mu sub D. All that divided by our standard deviation, so S sub d. And then divided by the square root of N. So, we have D-bar given to us, as well as our standard deviation. We know that N is 8 participants. What is mu D? Well, when we do one of these matched pair studies and we're looking for a change. We always start with a null hypothesis that the mean differencem sub D equals zero, because that status quo or baseline assumption is that there was no change. Which is what mean if we didn't see an average difference. And then it's our alternative hypothesis that would be mu D in this case, we're saying before minus after and we're looking to see if the number of wakenings decreases. So, we'd expect that to be a positive number, we'd say mucub D is greater than 0. But when we plug these into our equation. We use the null volume use of D is 0. And this is why it's so useful to have it that way, because we eliminate that term. Because we're going to be looking for in the end, the probability that the result we got, or something more extreme is different enough from what we would expect that we can reject our hypothesis. So somewhat long-winded, but. That is the reason for having this null hypothesis of the mu equaling 0. So let's look at that T equation. D bar we know is -1.5, so we have -1.5. Then you sub D is 0, so we'll just not put that in there and it gets eliminated. The standard deviation is 1.2, so now the denominator of we have 1.2. It's being divided by the square root of N, which is the square root of 8. So you could pause this for a moment if you want to check this on your calculator. But the answer we get once you put all this in is -3.536. So, our test statistic T is going to be answer A. For this paired sample test. T is approximately 3.536. Again, we use this paired T test formula, uh, with that null hypothesis being that new D is equal to zero. See you in the next video.

Key Concepts

Dependent Samples

Standardized Test Statistic (t)

Hypothesis Testing

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