Week 8 Tutorial

How to Solve Hypothesis Testing Problems

A step-by-step walkthrough with worked examples

Clearview Accounting Group

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What This Tutorial Covers

This tutorial walks you through exactly how to solve hypothesis testing problems, step by step. We will work through two complete examples together:

Example 1

Testing a mean (t-test)

"Are audit hours exceeding the benchmark?"

Example 2

Testing a proportion (Z-test)

"Is the late-filing rate above the target?"

Before you start any problem, always ask yourself three questions:

Am I testing a mean (an average) or a proportion (a percentage)?
What direction does the question point? (less than, greater than, or different from?)
Do I know the population standard deviation, or only the sample standard deviation?

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The 5-Step Checklist

Every hypothesis test follows the same five steps. Think of it like a recipe — follow the steps in order and you'll get the right answer.

Step	What To Do	How To Do It
1	State the hypotheses	Read the question carefully. Write $H_0$ (the claim / status quo) and $H_1$ (what you suspect). The direction of $H_1$ determines the tail.
2	Choose $\alpha$	Usually given in the question (0.05 is the default). Write it down.
3	Calculate the test statistic	Plug numbers into the correct formula (t-test for means, Z-test for proportions). This gives one number.
4	Make your decision	Compare your test statistic to the critical value from the table, or compare the p-value to $\alpha$.
5	Write a conclusion	Translate back into plain English. Always mention the significance level and the business context.

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Which Formula Do I Use?

In almost every real-world problem, you will use the t-test (because the population standard deviation $\sigma$ is almost never known). The Z-test for proportions always uses Z because proportions have a known formula for the standard error.

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How to Identify the Tails

Read the question carefully and look for direction words:

If the question says...	Then $H_1$ is...	Test type
"less than", "below", "decreased", "under", "lower"	$\mu < \text{value}$ or $p < \text{value}$	Left-tailed
"more than", "above", "increased", "exceeds", "higher"	$\mu > \text{value}$ or $p > \text{value}$	Right-tailed
"different from", "changed", "not equal to"	$\mu \neq \text{value}$ or $p \neq \text{value}$	Two-tailed

Tip: The words "at least" ($\geq$) and "no more than" ($\leq$) go in $H_0$. The suspicion or concern always goes in $H_1$.

Quick practice:

"A manager suspects training costs have fallen below \$2,400."

$H_0: \mu \geq 2{,}400$ $H_1: \mu < 2{,}400$ → Left-tailed

Worked Example 1

Testing a Mean (t-test) — Audit Completion Hours

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Example 1 — Read the Problem

Scenario

Clearview's quality manual says a standard financial audit should take 40 hours. The managing partner, Sarah, suspects that audits are now taking longer than this benchmark. She randomly selects 36 recent audits and records the completion times.

The sample results are: $\bar{X} = 43.2$ hours, $s = 7.8$ hours, $n = 36$

Test at $\alpha = 0.05$.

Before we start calculating, let's extract the key information:

What I'm given

Claimed value: $\mu_0 = 40$ | Sample mean: $\bar{X} = 43.2$ | Sample SD: $s = 7.8$ | Sample size: $n = 36$ | $\alpha = 0.05$

Key clues from the wording

"Suspects audits are taking longer" → this means "greater than" → right-tailed test

We have $s$ (sample SD), not $\sigma$ (population SD) → t-test

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Example 1 — Steps 1 and 2

Step 1 — State the Hypotheses

The status quo (what is claimed): audits take 40 hours → this goes in $H_0$

The suspicion (what we want to test): audits take longer → this goes in $H_1$

$H_0: \mu \leq 40$ $H_1: \mu > 40$ (right-tailed)

Step 2 — Choose the Significance Level

The question tells us: $\alpha = 0.05$

This means we are willing to accept a 5% risk of wrongly concluding audits take longer when they actually don't.

Checkpoint: Before moving on, double-check: does your $H_1$ match the suspicion in the question? If Sarah suspects audits are longer, then $H_1$ must say $\mu > 40$, not $\mu < 40$.

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Example 1 — Step 3: Calculate the Test Statistic

Step 3 — Compute the Test Statistic

Since we have $s$ (not $\sigma$), we use the t-test formula:

$$t = \frac{\bar{X} - \mu_0}{s / \sqrt{n}}$$

Let's plug in our numbers one piece at a time:

Numerator (top)

$\bar{X} - \mu_0 = 43.2 - 40 = 3.2$

This tells us: the sample mean is 3.2 hours above the claimed value.

Denominator (bottom) — the "standard error"

$s / \sqrt{n} = 7.8 / \sqrt{36} = 7.8 / 6 = 1.3$

This tells us: given normal sampling variability, sample means would typically differ from the true mean by about 1.3 hours.

Put it together

$t = 3.2 / 1.3 = \mathbf{2.462}$

This tells us: our sample mean is about 2.5 standard errors above 40. That's quite far!

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Example 1 — Step 4: Make the Decision

Step 4 — Compare to the Critical Value

We need the critical value from the t-table. To look it up, we need two things:

Degrees of freedom

$df = n - 1 = 36 - 1 = 35$

Significance level and tail

Right-tailed test, $\alpha = 0.05$ → look up $t_{0.05, 35}$

From the t-table: $t_{\text{critical}} = 1.690$

The decision rule

If $t_{\text{stat}} > t_{\text{critical}}$, reject $H_0$.

$2.462 > 1.690$? Yes! → Reject $H_0$

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Example 1 — Step 5: Write the Conclusion

Step 5 — State Your Conclusion in Plain English

Now translate the statistical result back into the language of the original problem. A good conclusion has three parts:

1. State the significance level

"At the 5% significance level..."

2. State the evidence

"...there is sufficient evidence to conclude that..."

3. Answer the business question

"...the mean audit completion time at Clearview exceeds the 40-hour benchmark."

Complete conclusion

"At the 5% significance level, there is sufficient evidence to conclude that the mean audit completion time at Clearview exceeds the 40-hour benchmark. Sarah's concern is supported by the data, and the firm should review its audit resourcing and scheduling."

Never say: "We accept $H_0$" or "The null hypothesis is true." Instead say "We do not reject $H_0$" or "There is insufficient evidence to reject $H_0$."

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Example 1 — All Five Steps Together

Here is the complete solution laid out as you would write it in an exam or assignment:

Step	Working
Step 1	$H_0: \mu \leq 40$ $H_1: \mu > 40$ (right-tailed)
Step 2	$\alpha = 0.05$
Step 3	$t = \dfrac{43.2 - 40}{7.8 / \sqrt{36}} = \dfrac{3.2}{1.3} = 2.462$
Step 4	Critical value: $t_{0.05, 35} = 1.690$. Since $2.462 > 1.690$, reject $H_0$.
Step 5	At the 5% significance level, there is sufficient evidence to conclude that the mean audit completion time exceeds 40 hours.

This is the format you should follow for every problem. Write it out exactly like this — five clear steps, each labelled.

Worked Example 2

Testing a Proportion (Z-test) — Late Tax Filings

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Example 2 — Read the Problem

Scenario

Clearview's tax division aims for at least 90% of business activity statements (BAS) to be lodged on time. The division head suspects the on-time rate has dropped below this target. A random sample of 150 BAS lodgements finds that 126 were on time.

Test at $\alpha = 0.05$.

What I'm given

Claimed proportion: $p_0 = 0.90$ | On-time count: 126 out of 150 | $n = 150$ | $\alpha = 0.05$

Key clues from the wording

"Dropped below" → $H_1$ is "less than" → left-tailed test

We are testing a percentage (proportion), not an average → Z-test for proportions

First calculation: the sample proportion

$\hat{p} = 126 / 150 = 0.84$

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Example 2 — Steps 1 and 2

Step 1 — State the Hypotheses

Claim: at least 90% on time → $H_0: p \geq 0.90$

Suspicion: it has dropped below 90% → $H_1: p < 0.90$

$H_0: p \geq 0.90$ $H_1: p < 0.90$ (left-tailed)

Step 2 — Significance Level

$\alpha = 0.05$

Condition check (do this before Step 3): For the proportion Z-test to be valid, we need:

$np_0 = 150 \times 0.90 = 135 \geq 5$ ✓

$n(1 - p_0) = 150 \times 0.10 = 15 \geq 5$ ✓

Both conditions are met, so we can proceed.

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Example 2 — Step 3: Calculate the Test Statistic

Step 3 — Compute the Z Statistic

The formula for testing a proportion is:

$$Z = \frac{\hat{p} - p_0}{\sqrt{\dfrac{p_0(1 - p_0)}{n}}}$$

Let's plug in the numbers step by step:

Numerator (top)

$\hat{p} - p_0 = 0.84 - 0.90 = -0.06$

The sample proportion is 6 percentage points below the target.

Denominator (bottom) — the standard error

$\sqrt{\dfrac{0.90 \times 0.10}{150}} = \sqrt{\dfrac{0.09}{150}} = \sqrt{0.0006} = 0.02449$

Put it together

$Z = \dfrac{-0.06}{0.02449} = \mathbf{-2.449}$

The sample proportion is about 2.45 standard errors below the target. That's quite far from what we'd expect if the target were being met.

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Example 2 — Steps 4 and 5

Step 4 — Make the Decision

Critical value

Left-tailed, $\alpha = 0.05$ → From the Z-table: $Z_{\text{critical}} = -1.645$

Decision rule

If $Z_{\text{stat}} < -Z_{\text{critical}}$, reject $H_0$.

$-2.449 < -1.645$? Yes! → Reject $H_0$

Step 5 — Conclusion

Complete conclusion

"At the 5% significance level, there is sufficient evidence to conclude that the proportion of BAS lodgements filed on time has fallen below the 90% target. The tax division head's concern is supported — the on-time rate appears to be approximately 84%, and management should investigate the cause of the delays."

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Example 2 — All Five Steps Together

Step	Working
Step 1	$H_0: p \geq 0.90$ $H_1: p < 0.90$ (left-tailed)
Check	$np_0 = 135 \geq 5$ ✓ $n(1-p_0) = 15 \geq 5$ ✓
Step 2	$\alpha = 0.05$
Step 3	$\hat{p} = 126/150 = 0.84$. $Z = \dfrac{0.84 - 0.90}{\sqrt{0.90 \times 0.10 / 150}} = \dfrac{-0.06}{0.02449} = -2.449$
Step 4	Critical value: $Z_{0.05} = -1.645$. Since $-2.449 < -1.645$, reject $H_0$.
Step 5	At the 5% significance level, there is sufficient evidence that the on-time BAS lodgement rate has fallen below 90%.

Common Mistakes

And how to avoid them

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Common Mistakes to Avoid

Mistake	Why it's wrong	What to do instead
Writing "Accept $H_0$"	We can never prove the null is true	Write "Do not reject $H_0$"
Putting the suspicion in $H_0$	$H_0$ is always the status quo / claim	The research question goes in $H_1$
Using Z-test when $\sigma$ is unknown	You're underestimating uncertainty	Use t-test when you only have $s$
Forgetting to check conditions for proportion test	The normal approximation may not hold	Always verify $np_0 \geq 5$ and $n(1-p_0) \geq 5$
Writing a conclusion without business context	A statistical answer alone is useless	Always link back to the scenario
Using $\hat{p}$ in the denominator of the proportion test	The standard error must be calculated under $H_0$	Always use $p_0$ (the hypothesised value)

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Quick Reference Card

t-test for a Mean

$$t = \frac{\bar{X} - \mu_0}{s / \sqrt{n}}$$

Degrees of freedom: $df = n - 1$

Use when: testing an average, $\sigma$ unknown

Z-test for a Proportion

$$Z = \frac{\hat{p} - p_0}{\sqrt{p_0(1 - p_0)/n}}$$

Conditions: $np_0 \geq 5$ and $n(1-p_0) \geq 5$

Use when: testing a percentage

Decision rules:
Left-tailed: Reject $H_0$ if test stat $< -$critical value (or if p-value $< \alpha$)
Right-tailed: Reject $H_0$ if test stat $>$ critical value (or if p-value $< \alpha$)
Two-tailed: Reject $H_0$ if $|$test stat$| >$ critical value (or if p-value $< \alpha$)

You're now ready to attempt the tutorial quiz!

Mistake	Why it's wrong	What to do instead
Writing "Accept \(H_0\)"	We can never prove the null is true	Write "Do not reject \(H_0\)"
Putting the suspicion in \(H_0\)	\(H_0\) is always the status quo / claim	The research question goes in \(H_1\)
Using Z-test when \(\sigma\) is unknown	You're underestimating uncertainty	Use t-test when you only have \(s\)
Forgetting to check conditions for proportion test	The normal approximation may not hold	Always verify \(np_0 \geq 5\) and \(n(1-p_0) \geq 5\)
Writing a conclusion without business context	A statistical answer alone is useless	Always link back to the scenario
Using \(\hat{p}\) in the denominator of the proportion test	The standard error must be calculated under \(H_0\)	Always use \(p_0\) (the hypothesised value)

Step	What To Do	How To Do It
1	State the hypotheses	Read the question carefully. Write \(H_0\) (the claim / status quo) and \(H_1\) (what you suspect). The direction of \(H_1\) determines the tail.
2	Choose \(\alpha\)	Usually given in the question (0.05 is the default). Write it down.
3	Calculate the test statistic	Plug numbers into the correct formula (t-test for means, Z-test for proportions). This gives one number.
4	Make your decision	Compare your test statistic to the critical value from the table, or compare the p-value to \(\alpha\).
5	Write a conclusion	Translate back into plain English. Always mention the significance level and the business context.

If the question says...	Then \(H_1\) is...	Test type
"less than", "below", "decreased", "under", "lower"	\(\mu < \text{value}\) or \(p < \text{value}\)	Left-tailed
"more than", "above", "increased", "exceeds", "higher"	\(\mu > \text{value}\) or \(p > \text{value}\)	Right-tailed
"different from", "changed", "not equal to"	\(\mu \neq \text{value}\) or \(p \neq \text{value}\)	Two-tailed

Step	Working
Step 1	\(H_0: \mu \leq 40\) \(H_1: \mu > 40\) (right-tailed)
Step 2	\(\alpha = 0.05\)
Step 3	\(t = \dfrac{43.2 - 40}{7.8 / \sqrt{36}} = \dfrac{3.2}{1.3} = 2.462\)
Step 4	Critical value: \(t_{0.05, 35} = 1.690\). Since \(2.462 > 1.690\), reject \(H_0\).
Step 5	At the 5% significance level, there is sufficient evidence to conclude that the mean audit completion time exceeds 40 hours.

Step	Working
Step 1	\(H_0: p \geq 0.90\) \(H_1: p < 0.90\) (left-tailed)
Check	\(np_0 = 135 \geq 5\) ✓ \(n(1-p_0) = 15 \geq 5\) ✓
Step 2	\(\alpha = 0.05\)
Step 3	\(\hat{p} = 126/150 = 0.84\). \(Z = \dfrac{0.84 - 0.90}{\sqrt{0.90 \times 0.10 / 150}} = \dfrac{-0.06}{0.02449} = -2.449\)
Step 4	Critical value: \(Z_{0.05} = -1.645\). Since \(-2.449 < -1.645\), reject \(H_0\).
Step 5	At the 5% significance level, there is sufficient evidence that the on-time BAS lodgement rate has fallen below 90%.