Standard +0.3 This is a straightforward one-tail z-test requiring calculation of sample mean, standard deviation, test statistic, and comparison with critical value. All steps are routine S2 procedures with no conceptual challenges beyond standard hypothesis testing mechanics, making it slightly easier than average.
5 The mean solubility rating of widgets inserted into beer cans is thought to be 84.0, in appropriate units. A random sample of 50 widgets is taken. The solubility ratings, \(x\), are summarised by
$$n = 50 , \quad \Sigma x = 4070 , \quad \Sigma x ^ { 2 } = 336100$$
Test, at the \(5 \%\) significance level, whether the mean solubility rating is less than 84.0 .
\(\frac{c - 84.0}{\sqrt{(s^2/50)}} = -1.645 \Rightarrow c = 81.697\)
M1
For critical value calculation, inc use of \(\sqrt{50}\)
A1
For correct value 81.697
\(\bar{x}\) is in the critical region since \(81.4 < 81.697\)
M1
For comparing sample mean to critical region
Hence \(H_0\) is rejected
A1
For stating or implying rejection of \(H_0\)
There is sufficient evidence to conclude that the mean solubility rating is less than 84.0
A1
For stating the outcome in context
Total: 10 marks
$\bar{x} = \frac{4070}{50} = 81.4$ | B1 | For correct value of sample mean
$s^2 = \frac{336100}{49} - \frac{4070^2}{49 \times 50} = 98$ | M1 | For calculation of unbiased or biased estimate
| A1 | For correct value of unbiased estimate
$H_0: \mu = 84.0$; $H_1: \mu < 84.0$ | B1 | For correct statement of null hypothesis
| B1 | For correct statement of alt hypothesis
**Either:**
$z = \frac{\bar{x} - 84.0}{\sqrt{(s^2/50)}} = -1.857$ | M1 | For standardising, including use of $\sqrt{50}$
| A1 | For correct value -1.857
This is significant, since $-1.857 < -1.645$ | M1 | For comparing $z$ value to -1.645 or equiv
**Or:**
$\frac{c - 84.0}{\sqrt{(s^2/50)}} = -1.645 \Rightarrow c = 81.697$ | M1 | For critical value calculation, inc use of $\sqrt{50}$
| A1 | For correct value 81.697
$\bar{x}$ is in the critical region since $81.4 < 81.697$ | M1 | For comparing sample mean to critical region
Hence $H_0$ is rejected | A1 | For stating or implying rejection of $H_0$
There is sufficient evidence to conclude that the mean solubility rating is less than 84.0 | A1 | For stating the outcome in context
**Total: 10 marks**
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5 The mean solubility rating of widgets inserted into beer cans is thought to be 84.0, in appropriate units. A random sample of 50 widgets is taken. The solubility ratings, $x$, are summarised by
$$n = 50 , \quad \Sigma x = 4070 , \quad \Sigma x ^ { 2 } = 336100$$
Test, at the $5 \%$ significance level, whether the mean solubility rating is less than 84.0 .
\hfill \mbox{\textit{OCR S2 Q5 [10]}}