Use the following information to answer Questions 42 – 44. Ten healthy adult sub
ID: 3393873 • Letter: U
Question
Use the following information to answer Questions 42 – 44.
Ten healthy adult subjects were asked to walk on a treadmill while connected to monitors. One of two kinds of disturbances was applied during the walk, and the number of steps until adaptation was recorded. Each subject experienced both kinds of disturbance, in random order. Here are the summary data for these 10 subjects:
N Mean StDev SE Mean
Disturbance A 10 7.320 2.10016 0.66413
Disturbance B 10 5.170 1.86073 0.58842
Difference A–B 10 2.150 2.68711 0.84974
The null and alternative hypotheses to test whether there is significant evidence of a different mean adaptation time (in steps) for the two disturbances are:
a)H0 : mA = 7.32 and mB = 5.17 vs. Ha : mA 7.32 and mB 5.17
b)H0 : mA = mB vs. Ha : mA mB
c)H0 : mDifference A-B = 0 vs. Ha : mDifference A-B 0
d) H0 : mA = mB vs. Ha : mA > mB
The test statistic for the appropriate test of hypothesis is
a) z = 2.53.
b) t = 2.42.
c) t = 2.53.
d) z = 2.42.
The experimenter stated that the assumptions for the appropriate test are satisfied. Based on these data,
a) we would reject H0 at significance level 0.10, but not at 0.05.
b) we would reject H0 at significance level 0.05, but not at 0.02.
c) we would reject H0 at significance level 0.02, but not at 0.01.
d) we would reject H0 at significance level 0.01.
Explanation / Answer
1.
b)H0 : mA = mB vs. Ha : mA mB [ANSWER]
*********************
2. As n = 10, we use t.
If ud = the difference in means,
Formulating the null and alternative hypotheses,
Ho: ud = 0
Ha: ud =/ 0
As we can see, this is a two tailed test.
df = n - 1 = 9
Getting the test statistic, as
X = sample mean = 2.15
uo = hypothesized mean = 0
n = sample size = 10
s = standard deviation = 2.68711
Thus, t = (X - uo) * sqrt(n) / s = 2.5301893 [ANSWER, C]
***********************************
d)
Also, the p value is
p = 0.032225361
Thus,
OPTION B: we would reject H0 at significance level 0.05, but not at 0.02. [ANSWER]