Question

question 8
ch 20* you are testing $h_{0}:\mu = 100$ against $h_{a}:\mu > 100$ based on an srs of 16 observations from a normal population. the t statistic is $t = 2.13$. the p - value for the statistic
is less than 0.01.
falls between 0.05 and 0.10.
falls between 0.01 and 0.05.
question 9
ch 20 how do we turn matched pairs data into one - sample data?
by randomly choosing one value within each pair
by taking the differences within each pair
by taking the maximum within each pair
by taking the sum for each pair

Explanation:

Question 8

For a one - sided \(t\) - test with \(n = 16\) observations, the degrees of freedom \(df=n - 1=16 - 1 = 15\). Looking at the \(t\) - distribution table, for \(df = 15\), \(t_{0.025}=2.131\) and \(t_{0.01}=2.602\). Since our \(t\) - statistic \(t = 2.13\) is less than \(t_{0.025}=2.131\) but greater than \(t_{0.05}=1.753\) (for \(df = 15\)), the \(P\) - value (the probability of getting a \(t\) - statistic as extreme or more extreme than the observed one, assuming \(H_0\) is true) falls between \(0.01\) and \(0.05\).

In matched - pairs data, to convert it into one - sample data, we use the method of taking the differences within each pair. This is because we are interested in the effect or change within each pair. For example, if we have pairs \((x_1,y_1),(x_2,y_2),\cdots,(x_n,y_n)\), we calculate \(d_i=x_i - y_i\) (or \(y_i - x_i\)) for \(i = 1,2,\cdots,n\) and then analyze the single sample of differences \(d_1,d_2,\cdots,d_n\).

Answer:

falls between \(0.01\) and \(0.05\)

Question 9