Reverse Hypothesis Testing Questions

Scope Label

Core 9758 support. This branch handles reverse questions for the core $Z$-test path. The ideas are not a new test; they are the usual testing inequalities used backwards.

What Makes a Question “Reverse”

In a normal hypothesis-test question, you are given the sample result and asked for the decision.

In a reverse question, you may be given:

• the decision;
• a required significance level;
• a boundary condition;
• a phrase such as “find the range of values for which $H_0$ is not rejected”.

Then you work backwards.

The core idea is:

A decision about rejection is an inequality for the test statistic.

Caption: In a lower-tailed reverse question, the stated decision becomes an inequality for the observed statistic; upper-tailed and two-tailed questions use the same idea with different critical regions.

Rejection and Non-Rejection Inequalities

For a lower-tailed test:

• reject $H_0$ if $z_{\text{obs}}$ is in the left critical region;
• do not reject $H_0$ if $z_{\text{obs}}$ is outside that region.

For an upper-tailed test:

• reject $H_0$ if $z_{\text{obs}}$ is in the right critical region;
• do not reject $H_0$ if $z_{\text{obs}}$ is outside that region.

For a two-tailed test:

• reject $H_0$ if $z_{\text{obs}}$ is too far from $0$ in either direction;
• do not reject $H_0$ if $z_{\text{obs}}$ lies between the two critical values.

Substituting the Test Statistic

For a $Z$-test of a population mean,

$$z_{\text{obs}}=\frac{\bar{x}-{\mu }_0}{\sigma /\sqrt{n}}.$$

Once you know the inequality for $z_{\text{obs}}$, substitute this expression and solve for the unknown.

The unknown may be:

• $\bar{x}$;
• a sample size;
• a claimed mean ${\mu }_0$;
• a boundary value in the question.

Worked Example 1: Range for Non-Rejection

A population has known standard deviation $6$. A random sample of size $36$ is used to test

$$H_0:\mu =50\text{against}{H}_1:\mu \ne 50$$

at the $10\%$ level.

Find the range of sample means $\bar{x}$ for which $H_0$ is not rejected.

For a two-tailed $10\%$ test, the non-rejection region is approximately

$$-1.645<z_{\text{obs}}<\mathrm{1.645.}$$

Now

$$z_{\text{obs}}=\frac{\bar{x}-50}{6/\sqrt{36}}=\bar{x}-50,$$

because $6/\sqrt{36}=1$.

So

$$-1.645<\bar{x}-50<\mathrm{1.645.}$$

Hence

$$48.355<\bar{x}<\mathrm{51.645.}$$

For sample means in this interval, the test does not reject $H_0$ at the $10\%$ level.

Caption: A two-tailed non-rejection decision gives an interval for the test statistic, which then becomes an interval for $\bar{x}$.

Worked Example 2: Boundary for Rejection

A lower-tailed test is carried out at the $5\%$ level:

$$H_0:\mu =100,H_1:\mu <100.$$

The population standard deviation is $12$ and the sample size is $64$. Find the largest sample mean that would still lead to rejection.

For a lower-tailed $5\%$ test,

$$z_{\text{obs}}<-\mathrm{1.645.}$$

Substitute

$$z_{\text{obs}}=\frac{\bar{x}-100}{12/\sqrt{64}}=\frac{\bar{x}-100}{1.5}.$$

For the boundary,

$$\frac{\bar{x}-100}{1.5}=-\mathrm{1.645.}$$

Thus

$$\bar{x}=100-1.645(1.5)=\mathrm{97.5325.}$$

So rejection occurs for approximately

$$\bar{x}<\mathrm{97.53.}$$

Least Significance Level

If a question asks for the least significance level at which $H_0$ would be rejected, find the p-value.

Reason:

• rejection happens when p-value $<\alpha$;
• the smallest threshold that would just lead to rejection is the observed p-value.

In practice, state the p-value with appropriate rounding and interpret it as the least significance level needed for rejection.

Worked Example 3: Unknown $\bar{x}$ Means No Direct p-Value

A test is carried out at the $5\%$ level:

$$H_0:\mu =3,H_1:\mu <3.$$

The population standard deviation is $0.6$, and a sample of size $15$ is taken. It is given that $H_0$ is rejected. Find the largest possible sample mean.

Because $\bar{x}$ is unknown, we cannot compute a p-value directly. Instead, use the rejection condition.

For a lower-tailed $5\%$ test,

$$z_{\text{obs}}<-\mathrm{1.645.}$$

Now

$$z_{\text{obs}}=\frac{\bar{x}-3}{0.6/\sqrt{15}}.$$

At the boundary,

$$\frac{\bar{x}-3}{0.6/\sqrt{15}}=-\mathrm{1.645.}$$

So

$$\bar{x}=3-1.645\left(\frac{0.6}{\sqrt{15}}\right)\approx \mathrm{2.745.}$$

Therefore, rejection occurs for approximately

$$\bar{x}<\mathrm{2.745.}$$

The largest possible sample mean leading to rejection is about $2.745$.

Common Pitfalls

• Starting with the sample mean instead of the decision condition.
• Using a rejection inequality when the question asks for non-rejection.
• Forgetting to reverse or preserve inequalities correctly when solving.
• Using one-tailed critical values in a two-tailed reverse question.
• Treating “least significance level” as a new formula instead of the p-value idea.

Revision Checklist

• Can you identify whether the question gives a decision and asks for a range?
• Can you write the critical or non-rejection inequality first?
• Can you substitute $z_{\text{obs}}=(\bar{x}-{\mu }_0)/(\sigma /\sqrt{n})$ correctly?
• Can you explain why the least significance level is the p-value?