Standard +0.3 This is a straightforward one-sample hypothesis test with clear data and standard procedure. Students must formulate hypotheses, calculate sample mean and standard deviation, perform a z-test, and compare to critical value at 0.1% level. While it requires multiple steps (5-6 marks typical), each step follows a standard algorithm with no conceptual challenges or novel insights required. Slightly above average difficulty only due to the unusual 0.1% significance level and needing to assume/state that variance is known or use t-distribution.
6 Alex obtained the actual waist measurements, \(w\) inches, of a random sample of 50 pairs of jeans, each of which was labelled as having a 32 -inch waist. The results are summarised by
$$n = 50 , \quad \Sigma w = 1615.0 , \quad \Sigma w ^ { 2 } = 52214.50$$
Test, at the \(0.1 \%\) significance level, whether this sample provides evidence that the mean waist measurement of jeans labelled as having 32 -inch waists is in fact greater than 32 inches. State your hypotheses clearly.
\section*{Jan 2006}
Guidance: One hypothesis correctly stated, not \(x\) or \(\bar{w}\) or \(\bar{w}\). Both completely correct, \(\mu\) used. Sample mean 32.3 seen. Correct formula for \(s^2\) used. Multiply by 50/49 or \(\sqrt{}\)
Insufficient evidence that waists are actually larger
Marks: M1 A1 B1 M1 B1 A1 M1 A1
Guidance: Correct formula for \(z\), can use \(s\), aef, need \(\mu = 32\). \(z = 2.1\) or \(1 - \Phi(z) = 0.0179\), not –2.1. Explicitly compare their 2.1 with 3.09(0) or their 0.0179 with 0.001. CV in range [32.4, 32.5], √on \(k\). \(z = 3.09\) and (later) compare \(\bar{x}\). CV a.r.t. \(32.3\), √on \(k\). Correct conclusion, can be implied, needs essentially correct method including \(\sqrt{n}\), any reasonable \(\sigma\), but not from \(\mu = 32.3\). Interpreted in context
**Answer:** $H_0: \mu = 32; H_1: \mu > 32$, where $\mu$ is population mean waist measurement
$\bar{W} = 32.3$
$s^2 = 52214.50/50 - \bar{W}^2 = [1]$
$\sigma^2 = 50/49 \times s^2 = [50/49 \text{ or } 1.0204]$ | **Marks:** B1 B1 B1 M1 M1 | **Guidance:** One hypothesis correctly stated, not $x$ or $\bar{w}$ or $\bar{w}$. Both completely correct, $\mu$ used. Sample mean 32.3 seen. Correct formula for $s^2$ used. Multiply by 50/49 or $\sqrt{}$
**Answer:** $\alpha: z = \frac{(32.3 - 32) \times \sqrt{49}}{8} = 2.1$
Compare 2.1 with 3.09 or 0.0179 with 0.001
$\beta: \text{CV} = 32 + 3.09 \div \sqrt{49} = 32.44$
Compare CV with 32.3
Do not reject $H_0$
Insufficient evidence that waists are actually larger | **Marks:** M1 A1 B1 M1 B1 A1 M1 A1 | **Guidance:** Correct formula for $z$, can use $s$, aef, need $\mu = 32$. $z = 2.1$ or $1 - \Phi(z) = 0.0179$, not –2.1. Explicitly compare their 2.1 with 3.09(0) or their 0.0179 with 0.001. CV in range [32.4, 32.5], √on $k$. $z = 3.09$ and (later) compare $\bar{x}$. CV a.r.t. $32.3$, √on $k$. Correct conclusion, can be implied, needs essentially correct method including $\sqrt{n}$, any reasonable $\sigma$, but not from $\mu = 32.3$. Interpreted in context
6 Alex obtained the actual waist measurements, $w$ inches, of a random sample of 50 pairs of jeans, each of which was labelled as having a 32 -inch waist. The results are summarised by
$$n = 50 , \quad \Sigma w = 1615.0 , \quad \Sigma w ^ { 2 } = 52214.50$$
Test, at the $0.1 \%$ significance level, whether this sample provides evidence that the mean waist measurement of jeans labelled as having 32 -inch waists is in fact greater than 32 inches. State your hypotheses clearly.
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\hfill \mbox{\textit{OCR S2 Q6 [10]}}