Hypothesis Testing

Board	Paper	Notes
AQA	Paper 1, 2	Binomial tests in P1; normal tests in P2
Edexcel	P1, P2	Similar
OCR (A)	Paper 1, 2	Includes critical regions
CIE (9709)	P1, P6	Basic hypothesis testing in P6

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Definition.

• The null hypothesis $H_0$ is the default assumption ( “no effect” or “no change”).
• The alternative hypothesis $H_1$ is what we are trying to find evidence for.

• One-tailed: $H_1: p \gt p_0$ (right-tailed) or $H_1: p \lt p_0$ (left-tailed).
• Two-tailed: $H_1: p \neq p_0$.

The choice depends on the research question. Use a one-tailed test only when you have a specific Directional prediction before seeing the data.

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Definition. The significance level $\alpha$ is the maximum probability of incorrectly Rejecting $H_0$ when it is true. Common values: 1%, 5%, 10%.

The critical value is the boundary between the acceptance and rejection regions.

The critical region (rejection region) is the set of values of the test statistic that lead to Rejection of $H_0$.

For discrete distributions, the actual significance level may differ from the nominal level $\alpha$ Because we cannot achieve exactly $\alpha$.

Example. For $X \sim B(15, 0.5)$A right-tailed test at $\alpha = 0.05$:

$P(X \geq 12) = 1 - P(X \leq 11)$. We find the smallest $c$ such that $P(X \geq c) \leq 0.05$.

$P(X \geq 12) \approx 0.0176$, $P(X \geq 11) \approx 0.0592$.

Critical region: $X \geq 12$. Actual significance level: 1.76%.

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Definition.

• Type I error: Rejecting $H_0$ when $H_0$ is true (false positive). $P(\mathrm{Type I}) = \alpha \mathrm{ (the significance level)}$

• Type II error: Failing to reject $H_0$ when $H_0$ is false (false negative). $P(\mathrm{Type II}) = \beta$

• The power of a test is $1 - \beta = P(\mathrm{reject } H_0 \mid H_0 \mathrm{ is false})$.

Decreasing $\alpha$ (making the test stricter) generally increases $\beta$ (more false Negatives). There is always a trade-off between Type I and Type II errors.

Intuition. Think of a courtroom: a Type I error is convicting an innocent person; a Type II Error is acquitting a guilty person. Making the standard of proof higher (beyond reasonable doubt) Reduces Type I errors but increases Type II errors. You cannot eliminate both simultaneously.

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1. Define the random variable and its distribution under $H_0$.
2. State $H_0$ and $H_1$.
3. State the significance level $\alpha$.
4. Calculate the critical region (or critical value).
5. Determine the test statistic from the data.
6. Compare the test statistic to the critical value.
7. Conclude in context.

Alternatively: 1–3. Same as above. 4. Calculate the $p$-value: the probability of obtaining a Result at least as extreme as the observed value, assuming $H_0$ is true. 5. If $p$-value $\lt \alpha$Reject $H_0$. Otherwise, do not reject $H_0$. 6. Conclude in context.

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Example. A coin is tossed 20 times and lands heads 15 times. Test at the 5% significance level Whether the coin is biased towards heads.

$X \sim B(20, p)$.

$H_0: p = 0.5$, $H_1: p \gt 0.5$. One-tailed, $\alpha = 0.05$.

Under $H_0$: $X \sim B(20, 0.5)$.

Find $c$ such that $P(X \geq c) \leq 0.05$.

$P(X \geq 14) = 1 - P(X \leq 13) \approx 0.0577 \gt 0.05$. $P(X \geq 15) = 1 - P(X \leq 14) \approx 0.0207 \lt 0.05$.

Critical region: $X \geq 15$. Since $X = 15$ is in the critical region, we reject $H_0$.

There is sufficient evidence at the 5% level that the coin is biased towards heads.

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Example. A machine fills bags with mean weight 500g. A sample of 30 bags gives $\bar{x} = 497$G. Test at the 5% level whether the mean weight has decreased, given $\sigma = 6$G.

$H_0: \mu = 500$, $H_1: \mu \lt 500$. $\alpha = 0.05$.

Under $H_0$: $\bar{X} \sim N(500, 6^2/30) = N(500, 1.2)$.

$z = \dfrac{497 - 500}{\sqrt{1.2}} = \dfrac{-3}{1.0954} = -2.739$.

Critical value: $P(Z \lt -1.645) = 0.05$.

Since $-2.739 \lt -1.645$We reject $H_0$.

There is sufficient evidence that the mean weight has decreased.

For large $n$: $\hat{p} \sim N\!\left(p, \dfrac{p(1-p)}{n}\right)$ approximately.

Test statistic: $z = \dfrac{\hat{p} - p_0}{\sqrt{p_0(1-p_0)/n}}$.

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Use a one-tailed test when:

• The research question has a specific directional prediction established before data collection.
• Only one direction of deviation is practically meaningful.
• The consequence of missing an effect in the unexpected direction is negligible.

Use a two-tailed test when:

• You are interested in any difference from $H_0$Regardless of direction.
• You want a more conservative test that is harder to reach significance with.
• There is no strong prior reason to expect the effect in one specific direction.

Example. Testing whether a new teaching method changes exam scores:

• One-tailed ($H_1: \mu \gt \mu_0$): justified only if prior research strongly suggests the method improves scores, and you would not act on a decrease.
• Two-tailed ($H_1: \mu \neq \mu_0$): appropriate if the method is new and could either help or harm, and either outcome matters.

For a test at significance level $\alpha$The allocation of the significance level differs:

• One-tailed: The entire $\alpha$ goes into one tail. The critical value is at the $1 - \alpha$ quantile (right-tailed) or $\alpha$ quantile (left-tailed).
• Two-tailed: $\alpha/2$ goes into each tail. The critical values are at the $\alpha/2$ and $1 - \alpha/2$ quantiles.

This means the two-tailed test has a higher bar for each individual tail.

Example. Standard normal test at $\alpha = 0.05$:

• One-tailed ($H_1: \mu \gt \mu_0$): reject if $z \gt 1.645$.
• Two-tailed ($H_1: \mu \neq \mu_0$): reject if $z \gt 1.960$ or $z \lt -1.960$.

An observed $z = 1.80$ is significant for the one-tailed test ($1.80 \gt 1.645$) but not for the Two-tailed test ($|1.80| \lt 1.960$).

### 8.3 Effect on power

For the same $\alpha$A one-tailed test has greater power than a two-tailed test against an Alternative in the predicted direction, because the critical value is closer to the null value. However, a one-tailed test has zero power to detect an effect in the opposite direction.

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When $n$ is sufficiently large, the binomial distribution $B(n, p)$ can be approximated by a normal Distribution. The standard conditions are:

$np \gt 5 \quad \mathrm{and} \quad n(1 - p) \gt 5$

Under these conditions:

$X \approx N(np, np(1-p))$

Equivalently, for the sample proportion $\hat{p} = X/n$:

$\hat{p} \approx N\!\left(p, \dfrac{p(1-p)}{n}\right)$

Caution: Warning $H_0$), not the observed sample proportion $\hat{p}$.

Since the binomial distribution is discrete and the normal distribution is continuous, a continuity correction improves the accuracy of the approximation:

• For $P(X \leq k)$Use $P\!\left(Z \leq \dfrac{k + 0.5 - np}{\sqrt{np(1-p)}}\right)$.
• For $P(X \geq k)$Use $P\!\left(Z \geq \dfrac{k - 0.5 - np}{\sqrt{np(1-p)}}\right)$.
• For $P(X = k)$Use $P(k - 0.5 \lt X \lt k + 0.5)$ in the normal.

Example. Historically, 40% of students at a school take the bus. In a survey of 120 students, 58 Take the bus. Test at the 5% level whether the proportion has changed.

$X \sim B(120, p)$. $H_0: p = 0.4$, $H_1: p \neq 0.4$. Two-tailed, $\alpha = 0.05$.

Check conditions using $p_0 = 0.4$: $np_0 = 120 \times 0.4 = 48 \gt 5$ and $n(1 - p_0) = 120 \times 0.6 = 72 \gt 5$. Conditions satisfied.

Under $H_0$: $X \approx N(48, 28.8)$So $\sigma = \sqrt{28.8} \approx 5.367$.

Using continuity correction:

$z = \dfrac{58 - 0.5 - 48}{5.367} = \dfrac{9.5}{5.367} = 1.770$

Two-tailed critical values: $\pm 1.96$. Since $|1.770| \lt 1.96$do not reject $H_0$.

There is insufficient evidence at the 5% level that the proportion of bus users has changed.

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A confidence interval gives a range of plausible values for a population parameter, together With a specified level of confidence.

Definition. A $100(1 - \alpha)\%$ confidence interval for a parameter $\theta$ is an interval $(L, U)$ constructed from sample data such that, in repeated sampling, $100(1 - \alpha)\%$ of such Intervals would contain the true value of $\theta$.

### 10.2 95% confidence interval for a population proportion

For large $n$ where $n\hat{p} \gt 5$ and $n(1 - \hat{p}) \gt 5$The sample proportion $\hat{p}$ Is approximately normal. The $100(1-\alpha)\%$ confidence interval for $p$ is:

$\hat{p} \pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}$

For a 95% confidence interval, $z_{\alpha/2} = 1.96$:

$\hat{p} \pm 1.96\sqrt{\dfrac{\hat{p}(1 - \hat{p})}{n}}$

The margin of error is $1.96\sqrt{\hat{p}(1-\hat{p})/n}$Which decreases as $n$ increases.

There is a direct and important link between confidence intervals and two-tailed hypothesis tests:

• A $100(1-\alpha)\%$ confidence interval contains exactly those values of $p_0$ that would not be rejected by a two-tailed test of $H_0: p = p_0$ at level $\alpha$.
• If $p_0$ falls outside the confidence interval, then $H_0$ is rejected at level $\alpha$.
• If $p_0$ falls inside the confidence interval, then $H_0$ is not rejected at level $\alpha$.

Example. Using the bus survey data: $\hat{p} = 58/120 \approx 0.483$, $n = 120$.

$95\%\mathrm{ CI} = 0.483 \pm 1.96\sqrt{\dfrac{0.483 \times 0.517}{120}} = 0.483 \pm 1.96 \times 0.0456$

$95\%\mathrm{ CI} = 0.483 \pm 0.0894 = (0.394, 0.573)$

Since $p_0 = 0.4$ lies inside $(0.394, 0.573)$We do not reject $H_0: p = 0.4$ at the 5% level. This is consistent with the hypothesis test result in Section 9.3.

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Definition. The $p$-value is the probability of obtaining a test statistic at least as Extreme as the observed value, assuming $H_0$ is true.

$p\mathrm{-value} = P(\mathrm{test statistic} \geq \mathrm{observed} \mid H_0 \mathrm{ true})$

For a two-tailed test, “at least as extreme” means at least as far from the null value in either Direction, so the $p$-value is doubled.

• If $p\mathrm{-value} \lt \alpha$: reject $H_0$. The result is statistically significant.
• If $p\mathrm{-value} \geq \alpha$: do not reject $H_0$. The result is not statistically significant.

The smaller the $p$-value, the stronger the evidence against $H_0$:

$p$-value range	Strength of evidence against $H_0$
$p \geq 0.10$	Little to no evidence
$0.05 \leq p \lt 0.10$	Weak evidence
$0.01 \leq p \lt 0.05$	Moderate evidence
$0.001 \leq p \lt 0.01$	Strong evidence
$p \lt 0.001$	Very strong evidence

### 11.5 Worked example

Example. A factory produces components with mean length 50 mm. A sample of 40 components gives $\bar{x} = 50.8$ mm. Given $\sigma = 3$ mm, find the $p$-value for testing $H_0: \mu = 50$ vs $H_1: \mu \neq 50$.

Under $H_0$: $\bar{X} \sim N(50, 3^2/40) = N(50, 0.225)$.

$z = \dfrac{50.8 - 50}{\sqrt{0.225}} = \dfrac{0.8}{0.4743} = 1.687$

$p\mathrm{-value} = 2 \times P(Z \gt 1.687) = 2 \times (1 - 0.9542) = 0.0916$

Since $0.0916 \gt 0.05$We do not reject $H_0$ at the 5% level.

Interpretation: If the true mean were 50 mm, there would be approximately a 9.2% chance of Observing a sample mean at least as far from 50 mm as 50.8 mm. This is not unusual enough to provide Convincing evidence against $H_0$.

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Problem 1

A die is rolled 60 times and a 6 appears 16 times. Test at the 5% level whether the die is biased.

Solution 1

$X \sim B(60, p)$. $H_0: p = 1/6$, $H_1: p \neq 1/6$. Two-tailed, $\alpha = 0.05$.

Under $H_0$: $X \sim B(60, 1/6)$. $\mu = 10$ $\sigma = \sqrt{60 \times \frac{1}{6} \times \frac{5}{6}} = \sqrt{50/6} \approx 2.887$.

Using normal approximation: $z = \dfrac{16 - 10}{2.887} = 2.078$.

Two-tailed: critical values $\pm 1.96$. $|2.078| \gt 1.96$So reject $H_0$.

There is evidence at the 5% level that the die is biased.

If you get this wrong, revise: Binomial Hypothesis Tests — Section 5.

Problem 2

A manufacturer claims that 90% of their products pass quality control. In a sample of 200, 170 pass. Test the claim at the 5% significance level.

Solution 2

$X \sim B(200, p)$. $H_0: p = 0.9$, $H_1: p \lt 0.9$. Left-tailed, $\alpha = 0.05$.

$\hat{p} = 170/200 = 0.85$.

Under $H_0$: $\hat{p} \sim N(0.9, \frac{0.9 \times 0.1}{200}) = N(0.9, 0.00045)$.

$z = \dfrac{0.85 - 0.9}{\sqrt{0.00045}} = \dfrac{-0.05}{0.0212} = -2.358$.

Critical value: $-1.645$. Since $-2.358 \lt -1.645$reject $H_0$.

There is evidence that the proportion passing quality control is less than 90%.

If you get this wrong, revise: Normal Hypothesis Tests — Section 6.

Problem 3

Explain the difference between a Type I error and a Type II error in the context of medical testing.

Solution 3

**Type I error:** The test says a healthy person is sick (false positive). This leads to unnecessary treatment and anxiety.

Type II error: The test says a sick person is healthy (false negative). This means the person Goes untreated, potentially with serious consequences.

If you get this wrong, revise: Type I and Type II Errors — Section 3.

Problem 4

Find the critical region for a test of $H_0: p = 0.3$ vs $H_1: p \gt 0.3$ using $X \sim B(10, p)$ at the 5% level.

Solution 4

Under $H_0$: $X \sim B(10, 0.3)$.

$P(X \geq 6) = 1 - P(X \leq 5) = 1 - 0.9527 = 0.0473 \lt 0.05$. $P(X \geq 5) = 1 - P(X \leq 4) = 1 - 0.8497 = 0.1503 \gt 0.05$.

Critical region: $X \geq 6$. Actual significance level: 4.73%.

If you get this wrong, revise: Critical Region — Section 2.3.

Problem 5

The mean lifetime of a bulb is claimed to be 1000 hours. A sample of 50 bulbs gives $\bar{x} = 985$ hours with $s = 40$ hours. Test at the 1% level whether the mean lifetime is less than 1000 hours.

Solution 5

$H_0: \mu = 1000$, $H_1: \mu \lt 1000$. $\alpha = 0.01$.

$\bar{X} \sim N(1000, 40^2/50) = N(1000, 32)$ approximately.

$z = \dfrac{985 - 1000}{\sqrt{32}} = \dfrac{-15}{5.657} = -2.652$.

Critical value at 1%: $-2.326$. Since $-2.652 \lt -2.326$reject $H_0$.

There is evidence at the 1% level that the mean lifetime is less than 1000 hours.

If you get this wrong, revise: Normal Hypothesis Tests — Section 6.

Problem 6

For $X \sim B(20, 0.5)$Find the critical region for a two-tailed test at the 10% significance level.

Solution 6

Under $H_0$: $X \sim B(20, 0.5)$.

For each tail, we need $P(X \leq c_L) \leq 0.05$ and $P(X \geq c_U) \leq 0.05$.

Lower: $P(X \leq 5) \approx 0.0207 \leq 0.05$, $P(X \leq 6) \approx 0.0577 \gt 0.05$. So $c_L = 5$. Upper: $P(X \geq 15) \approx 0.0207 \leq 0.05$, $P(X \geq 14) \approx 0.0577 \gt 0.05$. So $c_U = 15$.

Critical region: $X \leq 5$ or $X \geq 15$. Actual significance level: $2 \times 0.0207 = 0.0414$.

If you get this wrong, revise: Critical Values and Significance Levels — Section 2.

Problem 7

A teacher claims that the average score on a test is 70%. In a class of 25, the mean score is 66% with standard deviation 12%. Test at the 5% level.

Solution 7

$H_0: \mu = 70$, $H_1: \mu \neq 70$. Two-tailed, $\alpha = 0.05$.

$\bar{X} \sim N(70, 12^2/25) = N(70, 5.76)$ approximately.

$z = \dfrac{66 - 70}{\sqrt{5.76}} = \dfrac{-4}{2.4} = -1.667$.

Two-tailed critical values: $\pm 1.96$. $|-1.667| \lt 1.96$So do not reject $H_0$.

There is insufficient evidence at the 5% level that the mean score differs from 70%.

If you get this wrong, revise: Normal Hypothesis Tests — Section 6.

Problem 8

A drug is effective for 60% of patients. After a new treatment, 18 out of 25 patients are cured. Test whether the new treatment is more effective at the 5% level.

Solution 8

$X \sim B(25, p)$. $H_0: p = 0.6$, $H_1: p \gt 0.6$. Right-tailed, $\alpha = 0.05$.

Under $H_0$: $X \sim B(25, 0.6)$.

$P(X \geq 19) = 1 - P(X \leq 18) \approx 1 - 0.9264 = 0.0736 \gt 0.05$. $P(X \geq 20) = 1 - P(X \leq 19) \approx 1 - 0.9773 = 0.0227 \lt 0.05$.

Critical region: $X \geq 20$. Since $X = 18 \lt 20$do not reject $H_0$.

Insufficient evidence that the new treatment is more effective.

If you get this wrong, revise: Binomial Hypothesis Tests — Section 5.

Problem 9

Explain why failing to reject $H_0$ does not mean $H_0$ is true.

Solution 9

Failing to reject $H_0$ means the data is consistent with $H_0$ but does not prove it. The test may lack sufficient power to detect a real effect. For example, if a drug has a small but real benefit, a small sample may not detect it, leading us to fail to reject $H_0$ even though the drug is effective. The absence of evidence is not evidence of absence.

If you get this wrong, revise: Interpreting Results — Section 7.

Problem 10

For a test of $H_0: \mu = 50$ vs $H_1: \mu \gt 50$ at the 5% level with $\sigma = 4$ and $n = 16$Find the probability of a Type II error if the true mean is $\mu = 52$.

Solution 10

Under $H_0$: $\bar{X} \sim N(50, 16/16) = N(50, 1)$.

Critical value: $P(Z \gt c) = 0.05 \implies c = 50 + 1.645(1) = 51.645$.

Type II error = failing to reject $H_0$ when $\mu = 52$.

$\bar{X} \sim N(52, 1)$ under the true distribution.

$P(\bar{X} \lt 51.645 \mid \mu = 52) = P\!\left(Z \lt \dfrac{51.645 - 52}{1}\right) = P(Z \lt -0.355) \approx 0.3613$.

So $\beta \approx 0.361$ and the power is $1 - \beta \approx 0.639$.

If you get this wrong, revise: Type I and Type II Errors — Section 3.

Problem 11

A researcher tests whether a new drug changes recovery time. She uses a two-tailed test of $H_0: \mu = \mu_0$ vs $H_1: \mu \neq \mu_0$ at $\alpha = 0.05$ and obtains $z = 1.85$. (a) What is her conclusion? (b) If she had instead used a right-tailed test $H_1: \mu \gt \mu_0$ at the same level, would her conclusion change? Explain.

Solution 11

(a) Two-tailed test: critical values $\pm 1.96$. $|1.85| = 1.85 \lt 1.96$So **do not reject** $H_0$. There is insufficient evidence that recovery time has changed.

(b) One-tailed test: critical value $1.645$. Since $1.85 \gt 1.645$We reject $H_0$. There is Sufficient evidence that recovery time has increased.

The conclusion changes because a one-tailed test allocates the entire 5% significance level to one Tail, making the critical value less extreme. This illustrates why the choice between one-tailed and Two-tailed must be made before seeing the data.

If you get this wrong, revise: One-Tailed vs Two-Tailed Tests in Depth — Section 8.

Problem 12

A survey of 200 households in a town finds that 45 regularly recycle. The national recycling rate is 20%. Test at the 5% level whether the recycling rate in this town differs from the national rate, using a normal approximation with continuity correction.

Solution 12

$X \sim B(200, p)$. $H_0: p = 0.20$, $H_1: p \neq 0.20$. Two-tailed, $\alpha = 0.05$.

Check conditions using $p_0 = 0.20$: $np_0 = 200 \times 0.20 = 40 \gt 5$ and $n(1 - p_0) = 160 \gt 5$. Conditions satisfied.

Under $H_0$: $X \approx N(40, 200 \times 0.20 \times 0.80) = N(40, 32)$ $\sigma = \sqrt{32} \approx 5.657$.

Using continuity correction:

$z = \dfrac{45 - 0.5 - 40}{5.657} = \dfrac{4.5}{5.657} = 0.796$

Two-tailed critical values: $\pm 1.96$. $|0.796| \lt 1.96$So do not reject $H_0$.

There is insufficient evidence at the 5% level that the recycling rate differs from 20%.

If you get this wrong, revise: Binomial Tests with Normal Approximation — Section 9.

Problem 13

In a random sample of 150 voters, 87 support a new policy. (a) Construct a 95% confidence interval for the true proportion of support. (b) Since the interval does not contain 0.5, a politician claims "a majority of voters support the policy." Is this claim justified?

Solution 13

(a) $\hat{p} = 87/150 = 0.58$.

Check: $n\hat{p} = 150 \times 0.58 = 87 \gt 5$ and $n(1 - \hat{p}) = 150 \times 0.42 = 63 \gt 5$.

$95\%\mathrm{ CI} = 0.58 \pm 1.96\sqrt{\dfrac{0.58 \times 0.42}{150}} = 0.58 \pm 1.96 \times 0.0403$

$95\%\mathrm{ CI} = 0.58 \pm 0.0790 = (0.501, 0.659)$

(b) The 95% CI is $(0.501, 0.659)$. Since the entire interval lies above 0.5, we can reject $H_0: p = 0.5$ at the 5% level. However, the lower bound is only 0.501, so the evidence for a Majority is borderline. The claim is technically supported by the test, but the narrow margin should Be communicated carefully.

If you get this wrong, revise: Confidence Intervals — Section 10.

Problem 14

A 95% confidence interval for a population mean is $(48.2, 53.8)$. State whether $H_0$ would be rejected or not rejected at the 5% level for each of the following null values: (a) $\mu_0 = 50$(b) $\mu_0 = 47$(c) $\mu_0 = 54$. Justify using the connection between confidence intervals and hypothesis tests.

Solution 14

A 95% confidence interval contains exactly those values of $\mu_0$ that would **not** be rejected By a two-tailed test at the 5% level.

(a) $\mu_0 = 50$: $50 \in (48.2, 53.8)$So do not reject $H_0$. (b) $\mu_0 = 47$: $47 \notin (48.2, 53.8)$So reject $H_0$. (c) $\mu_0 = 54$: $54 \notin (48.2, 53.8)$So reject $H_0$.

If you get this wrong, revise: Confidence Intervals — Section 10.

Problem 15

A sample of 35 students has mean score 62.4 with known population standard deviation $\sigma = 8$. (a) Find the $p$-value for testing $H_0: \mu = 60$ vs $H_1: \mu \gt 60$. (b) State your conclusion at the 5% significance level and interpret the $p$-value.

Solution 15

(a) Under $H_0$: $\bar{X} \sim N(60, 8^2/35) = N(60, 1.829)$.

$z = \dfrac{62.4 - 60}{\sqrt{1.829}} = \dfrac{2.4}{1.352} = 1.775$

$p\mathrm{-value} = P(Z \gt 1.775) = 1 - 0.9620 = 0.0380$

(b) Since $0.038 \lt 0.05$reject $H_0$ at the 5% level. There is sufficient evidence that The true mean score exceeds 60. The $p$-value of 0.038 means that if the true mean were 60, there Would be a 3.8% chance of observing a sample mean of 62.4 or higher. This provides moderate evidence Against $H_0$.

If you get this wrong, revise: Interpreting p-Values — Section 11.

Problem 16

For a test of $H_0: \mu = 100$ vs $H_1: \mu \gt 100$ with $\sigma = 15$, $n = 25$And $\alpha = 0.05$: (a) Find the critical value in terms of $\bar{x}$. (b) Find the probability of a Type II error and the power of the test if the true mean is $\mu = 108$. (c) How would the power change if $\alpha$ were increased to 0.10?

Solution 16

(a) Under $H_0$: $\bar{X} \sim N(100, 15^2/25) = N(100, 9)$So $\sigma_{\bar{X}} = 3$.

Critical value: $c = 100 + 1.645 \times 3 = 104.935$. Reject $H_0$ if $\bar{X} \gt 104.935$.

(b) Type II error when $\mu = 108$: $P(\bar{X} \leq 104.935 \mid \mu = 108)$.

$\bar{X} \sim N(108, 9)$ under the true distribution.

$P\!\left(Z \leq \dfrac{104.935 - 108}{3}\right) = P(Z \leq -1.022) \approx 0.153$

So $\beta \approx 0.153$ and power $= 1 - 0.153 = 0.847$.

(c) If $\alpha = 0.10$The critical value becomes $c = 100 + 1.282 \times 3 = 103.846$.

$\beta_{\mathrm{new}} = P\!\left(Z \leq \dfrac{103.846 - 108}{3}\right) = P(Z \leq -1.385) \approx 0.083$

Power $= 1 - 0.083 = 0.917$. Increasing $\alpha$ from 0.05 to 0.10 increases the power (from 0.847 to 0.917) but also increases the probability of a Type I error. This illustrates the trade-off Between Type I and Type II errors.

If you get this wrong, revise: Type I and Type II Errors — Section 3.

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