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A sample $X_1, X_2, X_3, X_4$ comes from a normal distribution $N(m, 2^2)$. To verify $H_0: m=4$ against $H_1: m = 1$ one uses a test with a critical region: $K = \{4X_1 − 2X_2 − 2X_3 + X_4 < −2\}$. Compute the probability of type I and type II error.

Ok, so type I error is the probability of rejecting null while it is true, so I need to obtain $\mathbb P_{m=4}(T(\mathbb X)\in K)$. So I guess my test statistics $T$ is just a mean, so $T=\frac{X_1+X_2+X_3+X_4}{4}$, or maybe I am completely wrong?

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I will assume the data are independent. Let $T(\mathbf{X}) = \mathbf{X}^T\beta$ where $\mathbf{X} = [X_1,X_2,X_3,X_4]^T$ and $\beta = [4, -2, -2, 1]^T$.

Note that

$$ \mathbb{E}(T(\mathbf{X}) \vert H_0) = 4$$

and $$ \operatorname{Var}(T(\mathbf{X})\vert H_0) = 100$$

using properties of the expectation operator and the variance operator (namely that the expectation operator is a linear operator, that $\operatorname{Var}(aX) = a^2 \operatorname{Var}(X)$, and that the covariance between random variables is 0 if they are independent).

Since $X_i$ are iid normal, then $T(\mathbf{X}) \sim \mathcal{N}(4, 10^2)$.

The probability we reject the null hypothesis when it is true is thus

$$ P(T(\mathbf{X})<-2) = .274$$

Some simulation verifies this is the case

set.seed(0)
nreps = 100000

# represent T(X) as an inner product.
b = c(4, -2, -2, 1)

#data under the null
# rows are samples.  Rerun the experiment nreps times.
x = matrix(rnorm(nreps*4, 4, 2), nrow=nreps)

# Compute the test statistic
test_stats = x%*%b

mean(-test_stats>2)
>>>0.27195 # Close enough

The type II error is failing to reject the null when the alternative is the case. You can use similar logic as I have to construct that probability.

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  • $\begingroup$ Amazing. Thank you a lot! $\endgroup$
    – amal
    May 26, 2021 at 7:41

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