Week Five Quiz >>> In the jury version of hypothesis testing the alternative hypothesis is >>> Probability and Statistics: To p or not to p?
1.
Question 1
In the jury version of hypothesis testing, the alternative hypothesis is
H1: not guilty
H1: guilty
5.
Question 5
If the pp-value for a test was 0.07 and \alpha = 0.05α=0.05, the appropriate decision is:
reject H0 at the 5% significance level
accept H0 at the 5% significance level
not reject H0 at the 5% significance level
reject H1 at the 5% significance level
8.
Question 8
Under the central limit theorem, when sampling from non-normal distributions the distribution of \bar{X}Xˉ tends to:
a normal distribution with mean \muμ and variance \sigma^2σ2, as n \to \inftyn→∞
a normal distribution with mean \muμ and variance \sigma^2/nσ2/n, as n \to \inftyn→∞
9.
Question 9
When sampling from a Bernoulli distribution, the sample proportion, PP, has:
E(P)=π and Var(P)=π(1−π)
E(P)=π and Var(P)=π(1−π)/n
10.
Question 10
When performing a hypothesis test of a single proportion, we use the test statistic:
\frac{P – \pi}{\sqrt{p\, (1-p)/n}} \sim N(0,\, 1)p(1−p)/nP−π∼N(0,1)
\frac{P – \pi}{\sqrt{\pi\, (1-\pi)/n}} \sim N(0,\, 1)π(1−π)/nP−π∼N(0,1)
1.
Question 1
In the jury version of hypothesis testing, the null hypothesis is:
H0: not guilty
H0: guilty
2.
Question 2
In hypothesis testing, a Type I error is:
rejecting H0 when H0 is true
not rejecting H0 when H0 is false.
4.
Question 4
In general, we test at the 100\alpha100α% significance level, for \alpha \in [0,\, 1]α∈[0,1] such that we control for the:
probability of a Type I error
probability of a Type II error
7.
Question 7
If a hypothesis test does not reject H0, then which of the following might have occurred?
A Type I error
A Type II error
9.
Question 9
The sampling distribution of the sample proportion, PP, is:
N(\pi,\, \pi\, (1-\pi)/n)N(π,π(1−π)/n) for any nn
N(\pi,\, \pi\, (1-\pi)/n)N(π,π(1−π)/n) as n \to \inftyn→∞

3.
Question 3
In hypothesis testing, the conditional probability P(H0 not rejected|H0 is true)=
\alphaα
\betaβ
1−α
1−β
4.
Question 4
In general, we control for the probability of a Type I error by choosing: