H2 Math Hypothesis Testing for a Population Mean

JC2 · H2 Mathematics · Study guide

Carry out and interpret one- and two-sided mean tests, estimate variance, find critical regions and solve reverse-test inequalities.

Concepts · worked methods · graphs10 min read · practice linked below

Choose the alternative from the claim

Let $\mu$ denote the population mean of the quantity being investigated, including its units. The question’s claim decides the alternative before the sample result is used. “Decreased” gives $H_1:\mu\lt \mu_0$ , “increased” gives $H_1:\mu\gt \mu_0$ , and “changed” gives $H_1:\mu\ne\mu_0$ . In the usual JC test write the null at the boundary $H_0:\mu=\mu_0$ .

Test a directional claim

Has a new method reduced the filling time?

Filling times have population standard deviation 12 s. A random sample of 36 independent fills using a new method has mean 46.4 s. Previously the population mean was 50 s. Test at the 5% level whether the population mean time has decreased.

Let $\mu$ be the population mean filling time with the new method, in seconds.

\begin{aligned}H_0&:\mu=50,\\H_1&:\mu<50.\end{aligned}

Because n = 36 is sufficiently large for a CLT approximation in this setting, under the null hypothesis the sample mean is approximately normal:

\begin{aligned}\overline X&\sim N\left(50,\frac{12^2}{36}\right),\\z&=\frac{46.4-50}{12/\sqrt{36}}\\&=-1.8,\\p&=P(Z\leq-1.8)\\&\approx0.03593.\end{aligned}

Since $0.03593\lt 0.05$ , reject $H_0$ . There is sufficient evidence at the 5% significance level that the population mean filling time has decreased.

For a small sample, the same known-variance z-test would require a normally distributed population. A large sample supports a CLT approximation; it does not establish independence or repair biased selection.

A two-sided test counts equally extreme results in both directions

Has the mean changed?

Use the same sample, but now investigate whether the population mean has changed from 50 s. The null remains $\mu=50$ ; the alternative is $\mu\ne50$ . The observed z is still −1.8.

\begin{aligned}p&=P(Z\leq-1.8)+P(Z\geq1.8)\\&=2P(Z\leq-1.8)\\&\approx0.07186.\end{aligned}

Since $0.07186\gt 0.05$ , do not reject $H_0$ . There is insufficient evidence at the 5% level of a change in the population mean. This does not prove the population mean is 50 s.

In original units, the upper equally extreme sample mean is 53.6 s, reflected from 46.4 about 50. The p-value is the probability of a sample mean at most 46.4 or at least 53.6, assuming the population mean really is 50 s.

The same sample, two different pre-specified questions

Alternative being tested

Changed: two-sided p = 0.0719. At 5%, do not reject H₀.

The tail must follow the question asked before seeing the data.

Explain p-value and significance level in context

For the lower-tail test above, the p-value 0.03593 is the probability of obtaining a sample mean of at most 46.4 s in 36 fills when the population mean is 50 s, under the model assumptions. It is not the probability that the null hypothesis is true.

A 5% significance level means the procedure has a 0.05 probability of concluding that the population mean filling time has decreased when the population mean is actually 50 s. It describes the risk of a false rejection under the null, not the percentage of individual fills below 50 s.

Use the decision rule reject when p ≤ the significance level. The boundary is retained consistently in the reverse inequalities below. Calculate with unrounded values before comparing.

Unknown variance: estimate it before testing

Coded sample summaries

A new independent random sample of 40 filling times gives $\sum(x-50)=-80$ and $\sum(x-50)^2=1600$ . The population variance is unknown. Use a large-sample z-test at 5% to investigate a reduction from 50 s.

\begin{aligned}\bar x&=50+\frac{-80}{40}=48,\\s^2&=\frac1{39}\left[1600-\frac{(-80)^2}{40}\right]\\&=\frac{1440}{39}\\&\approx36.9231\text{ s}^2.\end{aligned}

Let $\mu$ again be the population mean filling time. Test $H_0:\mu=50$ against $H_1:\mu\lt 50$ . With a sufficiently large sample, the estimated standard error gives the approximate z statistic:

\begin{aligned}z&=\frac{48-50}{\sqrt{(1440/39)/40}}\\&\approx-2.08167,\\p&\approx P(Z\leq-2.08167)\\&\approx0.01869.\end{aligned}

Reject $H_0$ , since the p-value is below 0.05. There is sufficient evidence of a decrease in the population mean at the 5% level.

If a question gives the sample variance using denominator n, multiply it by $n/(n-1)$ first. If it explicitly gives an unbiased variance estimate $s^2$ , use that directly. A small unknown-variance sample does not justify this large-sample approximation by itself.

Find a critical region in sample-mean units

Which sample means reject a claimed minimum?

A maker claims mean content is at least 100 g. Contents are normal with known standard deviation 4 g. Ten independent items are sampled. Find the critical region for a 5% test of the claim.

To challenge “at least 100”, use $H_0:\mu=100$ and $H_1:\mu\lt 100$ , where $\mu$ is the population mean content in grams. Under $H_0$ ,

\begin{aligned}\overline X&\sim N\left(100,\frac{16}{10}\right).\end{aligned}

Reject when the lower-tail probability is at most 0.05:

\begin{aligned}\frac{\bar x-100}{4/\sqrt{10}}&\leq-1.64485363,\\\bar x&\leq100-1.64485363\frac4{\sqrt{10}}\\&\approx97.9194\text{ g}.\end{aligned}

The critical region is $\bar x\leq97.9194\ldots$ g, with the unrounded boundary understood. A claim of “at most 100” would be challenged using an upper-tail alternative instead.

Work backwards from rejection or non-rejection

Find the possible population standard deviations

A random independent sample of 80 visits has mean 16 min. The population standard deviation is $\sigma\gt 0$ . A large-sample test of $H_0:\mu=15$ against $H_1:\mu\gt 15$ rejects at 5%. Find the range of $\sigma$ .

\begin{aligned}z&=\frac{16-15}{\sigma/\sqrt{80}}\\&=\frac{\sqrt{80}}{\sigma}.\end{aligned}

Rejection requires z to be at least the upper 5% critical value. Since $\sigma$ is positive,

\begin{aligned}\frac{\sqrt{80}}{\sigma}&\geq1.64485363,\\\sigma&\leq\frac{\sqrt{80}}{1.64485363},\\0<\sigma&\leq5.437731\ldots\text{ min}.\end{aligned}

A smaller standard deviation makes the same one-minute difference more significant. The endpoint follows the stated p ≤ 0.05 decision rule.

Find null means that are not rejected

A normal population has known standard deviation 0.6 min. A random independent sample of 25 has mean 5.9 min. Find the values of $\mu_0$ for which $H_0:\mu=\mu_0$ is not rejected against $H_1:\mu\ne\mu_0$ at 1%.

Non-rejection places the statistic strictly between the two critical values:

\begin{aligned}\left|\frac{5.9-\mu_0}{0.12}\right|&<2.5758293,\\|5.9-\mu_0|&<0.3090995,\\\mu_0&>5.5909005,\\\mu_0&<6.2090995.\end{aligned}

The bounds are approximate; the exact values use $5.9\pm0.12\Phi^{-1}(0.995)$ . Keep both sides of the inequality when solving for the unknown null mean.

Variation: least sample size to reject

With known population standard deviation 12 s and observed sample mean 46.4 s, find the least n giving rejection of $H_0:\mu=50$ against $H_1:\mu\lt 50$ at 5%. Assume normal observations so the test is valid for every n.

\begin{aligned}\frac{-3.6}{12/\sqrt n}&\leq-1.64485363,\\0.3\sqrt n&\geq1.64485363,\\n&\geq30.06159\ldots.\end{aligned}

The least n is 31. At n = 30, z ≈ −1.6432 does not reach the critical value; at n = 31, z ≈ −1.6703 does. This holds the observed mean and known variance fixed; it is not a statement that collecting more data will leave the observed mean unchanged.

Recover the significance level from a rejection rule

A fixed cutoff and a different null mean

A sample mean has distribution $N(50,2^2)$ under the null. A test rejects when $\overline X\leq46$ . Find its significance level. What happens to that level if the null mean becomes 51, with the same standard error and cutoff?

\begin{aligned}\alpha&=P(\overline X\leq46\mid\mu=50)\\&=\Phi\left(\frac{46-50}{2}\right)\\&=\Phi(-2)\\&\approx0.02275.\end{aligned}

The significance level is about 2.28%. With null mean 51,

\begin{aligned}\alpha_{\text{new}}&=\Phi\left(\frac{46-51}{2}\right)\\&=\Phi(-2.5)\\&\approx0.00621.\end{aligned}

The cutoff lies farther below the null mean in standard-error units, so the false-rejection probability is smaller. The significance level is a probability under the null distribution; it is not calculated using the observed mean as the distribution’s centre.

Before you finish

Define the population mean, choose the tail from the claim, justify the sampling model, show the standard error and p-value, compare with the stated level, then conclude in context. “Do not reject” means insufficient evidence for the alternative.