A-Level Hypothesis testing

The test is H₀: μ = μ₀ against H₁: μ ≠ μ₀, so it is two-tailed, at the 1% significance level, with assumed population standard deviation σ = 0.63. The test statistic is z = (x̄ − μ₀)/(σ/√n). A two-tailed 1% test puts 0.5% in each tail, so H₀ is rejected when |z| ≥ 2.5758, where P(Z ≥ 2.5758) = 0.005. "Does not reject H₀" is therefore the opposite condition: |x̄ − μ₀|/(σ/√n) < 2.5758 |x̄ − μ₀| < 2.5758 × σ/√n Using the sample values x̄ = 5.88 from n = 20 observations, with σ = 0.63: 2.5758 × 0.63/√20 = 1.6228/4.4721 = 0.3629 So |5.88 − μ₀| < 0.3629 Removing the modulus gives a double inequality: −0.3629 < 5.88 − μ₀ < 0.3629 5.88 − 0.3629 < μ₀ < 5.88 + 0.3629 5.5171 < μ₀ < 6.2429 To 3 significant figures, 5.52 < μ₀ < 6.24.

Let μ be the population mean mass, in mg, of the substance per tablet. Step 1: state the hypotheses. The owner CLAIMS the mean is 120 mg. Nothing restricts us to a shortfall only, so any departure from 120 counts against the claim and the test is TWO-tailed. H0: μ = 120 H1: μ ≠ 120 The test is carried out at the 5% significance level. Step 2: state the distribution used. The population distribution is not given, but n = 60 is large, so by the Central Limit Theorem, under H0, x̄ ~ N(120, 256.0169/60) approximately Variance of x̄ = 256.0169/60 = 4.26695 Standard error = √4.26695 = 2.0657 Step 3: compute the test statistic, with sample mean x̄ = 116. z = (x̄ - 120)/2.0657 = (116 - 120)/2.0657 = -4/2.0657 = -1.936 = -1.94 (3 s.f.) Step 4: find the p-value. Because the test is two-tailed, the one-tail probability must be DOUBLED. p = 2 × P(Z < -1.936) = 2 × 0.02641 = 0.0528 (3 s.f.) Step 5: compare with the significance level and conclude. 0.0528 > 0.05, so we do NOT reject H0 at the 5% level. There is insufficient evidence at the 5% level that the mean mass differs from 120 mg, so the owner's claim is supported at the 5% level. (Note how close this is: had the test been one-tailed, the p-value would have been 0.0264 and H0 would have been rejected, so stating the correct alternative hypothesis matters here.)

Define the symbol first. Let μ be the population mean mass of a bag filled by machine A (in the units used in the question, where the target is 1.5). State the hypotheses. Overweight means too heavy only, so this is a one-tailed test in the upper tail. H₀: μ = 1.5 H₁: μ > 1.5 Test at the 5% significance level, so α = 0.05. Choose the test statistic. The population variance is not known, so it is estimated from the sample by the unbiased estimate s² = [Σx² - (Σx)²/n]/(n - 1) The sample is large, so by the Central Limit Theorem the sample mean is approximately normally distributed even though the population distribution is not stated. Hence under H₀ the test statistic is Z = (x̄ - μ₀)/(s/√n), which is approximately N(0, 1), where x̄ is the sample mean, n the sample size and μ₀ = 1.5. Substitute the sample summary figures given in the question: z = (x̄ - 1.5)/(s/√n) = 1.811 (4 s.f.) Find the p-value. Because H₁ is μ > 1.5, the p-value is the upper-tail area: p = P(Z > 1.811) = 0.0351 (3 s.f.) Compare with the significance level. 0.0351 < 0.05, so the result is significant and H₀ is rejected. Conclude in context. There is sufficient evidence at the 5% level of significance to conclude that the mean mass of a bag from machine A is greater than 1.5, that is, machine A does produce overweight bags.