nazismes2w7

Answered

2022-11-23

A quick question regarding hypothesis testing (decision variable)

If we have two normally distributed populations ${P}_{1}$ and ${P}_{2}$, then to test the hypothesis ${H}_{0}:{\mu}_{1}={\mu}_{2}$ against an alternative hypothesis, we choose the decision variable according to three cases:

Case 1: If ${\sigma}_{1}$ and ${\sigma}_{2}$ are known, then regardless whether ${n}_{1}$ and ${n}_{2}$ are large or small, we consider the standard normal statistic:

$Z=\frac{{\overline{X}}_{1}-{\overline{X}}_{2}}{\sqrt{\frac{{\sigma}_{1}^{2}}{{n}_{1}}+\frac{{\sigma}_{2}^{2}}{{n}_{2}}}}$

Case 2: If the $\sigma $'s are unknown, but ${n}_{1}$ and ${n}_{2}$ are large, then we consider:

$Z=\frac{{\overline{X}}_{1}-{\overline{X}}_{2}}{\sqrt{\frac{{S}_{{X}_{1}}^{2}}{{n}_{1}}+\frac{{S}_{{X}_{2}}^{2}}{{n}_{2}}}}$

Case 3: If the $\sigma $'s are unknown and ${n}_{1},{n}_{2}$ are small, then we consider the student-t statistic:

$T=\frac{{\overline{X}}_{1}-{\overline{X}}_{2}}{\hat{\sigma}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}$

Where ${\hat{\sigma}}^{2}:=\frac{({n}_{1}-1){S}_{{X}_{1}}^{2}+({n}_{2}-1){S}_{{X}_{2}}^{2}}{{n}_{1}+{n}_{2}-2}$

Is this correct?

If we have two normally distributed populations ${P}_{1}$ and ${P}_{2}$, then to test the hypothesis ${H}_{0}:{\mu}_{1}={\mu}_{2}$ against an alternative hypothesis, we choose the decision variable according to three cases:

Case 1: If ${\sigma}_{1}$ and ${\sigma}_{2}$ are known, then regardless whether ${n}_{1}$ and ${n}_{2}$ are large or small, we consider the standard normal statistic:

$Z=\frac{{\overline{X}}_{1}-{\overline{X}}_{2}}{\sqrt{\frac{{\sigma}_{1}^{2}}{{n}_{1}}+\frac{{\sigma}_{2}^{2}}{{n}_{2}}}}$

Case 2: If the $\sigma $'s are unknown, but ${n}_{1}$ and ${n}_{2}$ are large, then we consider:

$Z=\frac{{\overline{X}}_{1}-{\overline{X}}_{2}}{\sqrt{\frac{{S}_{{X}_{1}}^{2}}{{n}_{1}}+\frac{{S}_{{X}_{2}}^{2}}{{n}_{2}}}}$

Case 3: If the $\sigma $'s are unknown and ${n}_{1},{n}_{2}$ are small, then we consider the student-t statistic:

$T=\frac{{\overline{X}}_{1}-{\overline{X}}_{2}}{\hat{\sigma}\sqrt{\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}}}}$

Where ${\hat{\sigma}}^{2}:=\frac{({n}_{1}-1){S}_{{X}_{1}}^{2}+({n}_{2}-1){S}_{{X}_{2}}^{2}}{{n}_{1}+{n}_{2}-2}$

Is this correct?

Answer & Explanation

Shyanne Meyers

Expert

2022-11-24Added 6 answers

Step 1

You are somewhat correct, but not entirely. When dealing with two population data where the variances are unknown, you must first determine if the population variances are equal or not. This information could be given to you in the question/application or you may need to determine it yourself based on sample data. You can use the F test determine if variances are equal or not. In particular, you are testing

${H}_{0}:\frac{{\sigma}_{1}^{2}}{{\sigma}_{2}^{2}}=1\phantom{\rule{2em}{0ex}}{H}_{1}:\frac{{\sigma}_{1}^{2}}{{\sigma}_{2}^{2}}\{<,\ne ,>\}1$

and the F-statistic is given by

$F=\frac{{s}_{1}^{2}}{{s}_{2}^{2}}$

with degrees of freedom ${v}_{1}={n}_{1}-1$ and ${v}_{2}={n}_{2}-2$. Performing this test lets you know whether the population variances are equal or not. Note that this is only applicable when the populations is normally distributed and independent.

Step 2

In both cases, you are performing the T-test.

If the variances are equal

$t=\frac{({X}_{1}-{X}_{2})-({\mu}_{1}-{\mu}_{2})}{\sqrt{{s}_{p}^{2}(\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}})}}$

where ${s}_{p}$ is the pooled variance given by

${s}_{p}^{2}=\frac{({n}_{1}-1){s}_{1}^{2}+({n}_{2}-1){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}$

Here, the degrees of freedom is given by $v={n}_{1}+{n}_{2}-2$.

If population variances are unequal, the test statistic is given by

$t=\frac{({X}_{1}-{X}_{2})-({\mu}_{1}-{\mu}_{2})}{\sqrt{\frac{{s}_{1}^{2}}{{n}_{1}}+\frac{{s}_{2}^{2}}{{n}_{2}}}}$

with degrees of freedom given by

$v=\frac{{(\frac{{s}_{1}^{2}}{{n}_{1}}+\frac{{s}_{2}^{2}}{{n}_{2}})}^{2}}{\frac{({s}_{1}^{2}/{n}_{1}{)}^{2}}{{n}_{1}-1}+\frac{({s}_{2}^{2}/{n}_{2}{)}^{2}}{{n}_{2}-1}}$

Here, most likely ${H}_{0}:{\mu}_{1}-{\mu}_{2}=0$

You are somewhat correct, but not entirely. When dealing with two population data where the variances are unknown, you must first determine if the population variances are equal or not. This information could be given to you in the question/application or you may need to determine it yourself based on sample data. You can use the F test determine if variances are equal or not. In particular, you are testing

${H}_{0}:\frac{{\sigma}_{1}^{2}}{{\sigma}_{2}^{2}}=1\phantom{\rule{2em}{0ex}}{H}_{1}:\frac{{\sigma}_{1}^{2}}{{\sigma}_{2}^{2}}\{<,\ne ,>\}1$

and the F-statistic is given by

$F=\frac{{s}_{1}^{2}}{{s}_{2}^{2}}$

with degrees of freedom ${v}_{1}={n}_{1}-1$ and ${v}_{2}={n}_{2}-2$. Performing this test lets you know whether the population variances are equal or not. Note that this is only applicable when the populations is normally distributed and independent.

Step 2

In both cases, you are performing the T-test.

If the variances are equal

$t=\frac{({X}_{1}-{X}_{2})-({\mu}_{1}-{\mu}_{2})}{\sqrt{{s}_{p}^{2}(\frac{1}{{n}_{1}}+\frac{1}{{n}_{2}})}}$

where ${s}_{p}$ is the pooled variance given by

${s}_{p}^{2}=\frac{({n}_{1}-1){s}_{1}^{2}+({n}_{2}-1){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}$

Here, the degrees of freedom is given by $v={n}_{1}+{n}_{2}-2$.

If population variances are unequal, the test statistic is given by

$t=\frac{({X}_{1}-{X}_{2})-({\mu}_{1}-{\mu}_{2})}{\sqrt{\frac{{s}_{1}^{2}}{{n}_{1}}+\frac{{s}_{2}^{2}}{{n}_{2}}}}$

with degrees of freedom given by

$v=\frac{{(\frac{{s}_{1}^{2}}{{n}_{1}}+\frac{{s}_{2}^{2}}{{n}_{2}})}^{2}}{\frac{({s}_{1}^{2}/{n}_{1}{)}^{2}}{{n}_{1}-1}+\frac{({s}_{2}^{2}/{n}_{2}{)}^{2}}{{n}_{2}-1}}$

Here, most likely ${H}_{0}:{\mu}_{1}-{\mu}_{2}=0$

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