Question

Migraines (recurrent headaches that are moderate to severe) can be accompanied by a range of symptoms, one of which is an aura - a series of sensory disturbances, often characterized by flashes of light, blind spots, or a tingling sensation in the face or hands. Researchers gave a new medication to a random selection of 6 migraine patients and the length of symptoms (measured in minutes) was recorded for an attack before the medication (?1) and for an attack after the medication (?2). The results appear below:

1 2 6 Patient Before medication After medication 5 127 86 71 49 3 4 45 63 32 82 91 84 70 101

Calculate the value of the test statistic the researchers would use to test their claim that the new medication reduces the average length of migraines with aura symptoms. If needed, you may assume that the length of symptoms both before and after being given the medication follows a normal distribution. Give your answer accurate to 3 decimal places.

Answer 1

hypothesis

$H0 : \mu d =0$

0 < prl: DH

Calculate the difference of means by subtracting each respective element from both data sets x01 - x02 = 71 - 49 = 22 X11 - X12 = 86 - 70 = 16 X21 - 22 = 45 - 32 = 13 - X32 = 63 - 82 = -19 x41 - X42 = 127 - 101 = 26 x51 - 52 = 91 - 84 = 7 X31 We then add our of our differences of mean and divide by the number of items in the data set: d = {x1;- X2; n 10 = 22 + 16 +13 + -19 + 26 + 7 6 d = 65 6 d = 10.8333 om the mean: Calculate the sample variance s? by adding the sum of squared differences d = (22 - 10.8333)2 = 11.16672 = 124.6952 d = (16 - 10.8333)2 = 5.16672 = 26.6948 d = (13 - 10.8333)2 = 2.16672 = 4.6946 d = (-19 - 10.8333)2 = -29.83332 = 890.0258 d = (26 - 10.8333)2 = 15.16672 = 230.0288 d = (7 - 10.8333)2 = -3.83332 = 14.6942 S2 = Sum of Squared Differences n-1 s2 = 124.6952 + 26.6948 + 4.6946 + 890.0258 + 230.0288 + 14.6942 5 s2 = 1290.8334 5 s2 = 258.16668

Now calculate the sample standard deviation s: S = všŽ S = 1258.16668 S = 16.067566088241 First find degrees of freedom: Degrees of Freedom = n - 1 Degrees of Freedom = 6-1 Degrees of Freedom = 5 Calculate a: a = 1 - Confidence% a = 1 -0.95 a = 0.05 Find a spread range: = 12(a) a = 12(0.05) a = a = 0.025 Find t-score for 20.025 using 5 degrees of freedom: tscoreo.025 = 3.0029 <--- Value can be found on Excel using =TINV(0.05,5)

now test statistic can be find out as follows

= d / .sd/vn = 10.8333 / 16.067566088241/15 = 10.8333 / 16.067566088241/2.4494897427832 =

1.651529

test statistic is smaller than critical value.so we fail to reject null hypothesis.

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