TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW09.4

Which statement is correct?

In the situation of a (two-sided) two-sample $t$-test at 5%-significance level, assume that the null hypothesis $H_{0}:d=d_{0}$ was rejected. Comment on the following statements.

(a) The null hypothesis would also be rejected at the 7%-level.

(b) The equivalent 99% confidence interval does not contain $d_{0}$.

(c) If both sample sizes are increased by a factor 4, then the value of the $t$-statistic is halved if all other estimates remain the same.

(d) If one of the sample sizes is increased, then the width of the 95%-confidence interval is increased (if all other estimates remain the same).

(e) There is a 5% chance that the null hypothesis is true.

(f) If the null hypothesis is true, there is a 5% chance that we came to the wrong conclusion.

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Lösung von Tutorin

(a)

True; as increases our rejection region gets larger.

(b)

We don't know that. We know that the 95%-C.I. does not overlap $d_{0}$, but the 99%-C.I. is larger and might contain $d_{0}$.

(c)

False; we pick up a 2 in the denominator of the denominator, so the statistic is doubled, not halved (if all other estimates remain the same)

(d)

False; width would decrease (if all other estimates remain the same).

(e)

No, if the null hypothesis was true, we falsely rejected - we did do this - so it was just incorrect to reject (at 100%)

False; if the null is true there is a 5% chance that we reject the null.

(f)

2 interpretations:

First, if the null hypothesis is true, then there is a 5% chance that we incorrectly reject the null. This perspective comes from not fixing the result. Second, if we condition on rejecting the null, then we know that we made a wrong decision as is no longer random; i.e., we made a mistake.

Lösung von Simplex

(a)

True, because the null hypothesis was already rejected at the 5%-level.

(b)

Unknown, because even though the rejection zone decreases, we cannot say if it is small enough to not reject $H_{0}$.

(c)

False:

$$t={\frac {\mu _{1}-\mu _{2}}{\sqrt {{\frac {\sigma _{1}^{2}}{n_{1}}}+{\frac {\sigma _{2}^{2}}{n_{2}}}}}}\Rightarrow {\frac {\mu _{1}-\mu _{2}}{\sqrt {{\frac {\sigma _{1}^{2}}{4n_{1}}}+{\frac {\sigma _{2}^{2}}{4n_{2}}}}}}={\frac {\mu _{1}-\mu _{2}}{\sqrt {{\frac {1}{4}}\left({\frac {\sigma _{1}^{2}}{n_{1}}}+{\frac {\sigma _{2}^{2}}{n_{2}}}\right)}}}={\frac {\mu _{1}-\mu _{2}}{{\frac {1}{2}}{\sqrt {{\frac {\sigma _{1}^{2}}{n_{1}}}+{\frac {\sigma _{2}^{2}}{n_{2}}}}}}}=2\cdot {\frac {\mu _{1}-\mu _{2}}{\sqrt {{\frac {\sigma _{1}^{2}}{n_{1}}}+{\frac {\sigma _{2}^{2}}{n_{2}}}}}}=2\cdot t$$

(d)

False, because according to the formula on page 18 in lesson 10, the square root of $SEM$ will be smaller and therefore the confidence-interval will be smaller and not bigger.

(e)

False, the chance of the null hypothesis being true is not determined based on the signifcance level.

(f)

True, because the significance level defines the percentage of the above chance.

--Simplex 18:24, 2. Feb. 2023 (CET)

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