Standard +0.3 This is a straightforward one-sample t-test with clearly stated hypotheses (H₀: μ = 5 vs H₁: μ < 5). Students must calculate the test statistic from given summary statistics and compare to critical value. While it requires understanding of hypothesis testing procedure, it's a standard textbook application with no conceptual complications or novel elements—slightly easier than average for Further Maths.
6 A company manufactures bolts. The diameter of the bolts follows a normal distribution with a mean diameter of 5 mm .
Stan believes that the mean diameter of the bolts is less than 5 mm . He takes a random sample of 10 bolts and measures their diameters. He calculates some statistics but spills ink on his work before completing them. The only information he has left is as follows
\includegraphics[max width=\textwidth, alt={}, center]{67df73d4-6ce4-45f7-8a69-aa94292ea814-16_394_1150_527_456}
Stating your hypotheses clearly, test, at the \(5 \%\) level of significance, whether or not Stan's belief is supported.
Correct formula; if "4.84" not shown must be correct here
\(= \text{awrt} -2.39\)
A1
Stan's belief is supported / evidence that mean diameter of bolts is less than 5mm
A1ft
Correct inference following through their CV and test statistic (must have matching signs); NB if \(\chi^2\) values not shown: \(s^2=0.045\) or \(0.0449\) award B0 M1M1
# Question 6:
| Answer/Working | Mark | Guidance |
|---|---|---|
| 99% CI for variance uses $\chi^2$ values of 1.735 or 23.589 | B1 | For realising $\chi^2$ distribution must be used; finding a correct value |
| $\frac{9s^2}{1.735} = 0.2328$ or $\frac{9s^2}{23.589} = 0.01712$ | M1 | Setting $\frac{9s^2}{\text{"smallest }\chi^2\text{"}} = 0.2328$ or $\frac{9s^2}{\text{"largest }\chi^2\text{"}} = 0.01712$ |
| $s^2 = \frac{0.2328 \times \text{"1.735"}}{9}$ or $\frac{0.01712 \times \text{"23.589"}}{9}$ $[= 0.04487...]$ | dM1 | Correct method to solve equation to find $s^2$ |
| $\bar{x} = 4.84$ | B1 | awrt 4.84 |
| $H_0: \mu = 5$, $H_1: \mu < 5$ | B1 | Both hypotheses correct using $\mu$ |
| CV $t_9 = -1.833$ | B1 | $\pm 1.833$ |
| $t = \pm\frac{\text{"4.84"}-5}{\sqrt{\frac{\text{"0.0449"}}{10}}}$ | M1 | Correct formula; if "4.84" not shown must be correct here |
| $= \text{awrt} -2.39$ | A1 | |
| Stan's belief is supported / evidence that **mean diameter** of bolts is less than **5mm** | A1ft | Correct inference following through their CV and test statistic (must have matching signs); NB if $\chi^2$ values not shown: $s^2=0.045$ or $0.0449$ award B0 M1M1 |
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6 A company manufactures bolts. The diameter of the bolts follows a normal distribution with a mean diameter of 5 mm .
Stan believes that the mean diameter of the bolts is less than 5 mm . He takes a random sample of 10 bolts and measures their diameters. He calculates some statistics but spills ink on his work before completing them. The only information he has left is as follows\\
\includegraphics[max width=\textwidth, alt={}, center]{67df73d4-6ce4-45f7-8a69-aa94292ea814-16_394_1150_527_456}
Stating your hypotheses clearly, test, at the $5 \%$ level of significance, whether or not Stan's belief is supported.
\hfill \mbox{\textit{Edexcel FS2 2019 Q6 [9]}}