A2 June 2019 Q6
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
Stating your hypotheses clearly, test, at the 5% level of significance, whether or not Stan’s belief is supported. (9)
| Scheme | Marks | AO |
|---|---|---|
| 99% confidence interval for Var uses \(\chi^2\) values of 1.735 or 23.589 | B1 | 3.3 |
| \(\dfrac{9s^2}{1.735} = 0.2328 \quad \text{or} \quad \dfrac{9s^2}{23.589} = 0.01712\) | M1 | 2.1 |
| \(s^2 = \dfrac{0.2328 \times \text{“}1.735\text{”}}{9} \text{ or } \dfrac{0.01712 \times \text{“}23.589\text{”}}{9} \quad [= 0.04487\ldots]\) | dM1 | 1.1b |
| \(\bar{x} = 4.84\) | B1 | 1.1b |
| \(\mathrm{H}_0 : \mu = 5 \quad \mathrm{H}_1 : \mu \lt 5\) | B1 | 2.5 |
| CV \(t_9 = -1.833\) | B1 | 1.1b |
| \(t = \pm\dfrac{\text{“}4.84\text{”} - 5}{\sqrt{\text{“}0.0449\text{”}/10}}\) | M1 | 1.1b |
| = awrt \(-2.39\) | A1 | 1.1b |
| Stan’s belief is supported or there is evidence that the mean diameter of the bolts is less than 5mm | A1ft | 2.2b |
| (9) | ||
| (9 marks) |
Notes
B1: For realising a \(\chi^2\) distribution must be used as a model and finding a correct value
M1: For realising the need to set \(\dfrac{9s^2}{\text{“}\text{smallest } \chi^2\text{”}} = 0.2328\) or \(\dfrac{9s^2}{\text{“}\text{largest } \chi^2\text{”}} = 0.01712\)
dM1: correct method used to solve equation to find \(s^2\)
B1: awrt 4.84
B1: Both hypotheses correct using the notation \(\mu\)
B1: \(\pm\) 1.833
M1: For us of correct formula ie \(\pm\dfrac{\text{“}\text{their } 4.84\text{”} - 5}{\sqrt{\text{“}\text{their } 0.0449\text{”}/10}}\) If “4.84” not shown it must be correct here
A1: \(-2.39\)
A1ft: Drawing a correct inference following through their CV and test statistic (must have matching signs)
NB if chi squared values not shown
\(s^2 = 0.045\) or 0.0449 award B0 M1M1 for awrt 0.04487 award B1 M1 A1
Use of \(2(2.5758)\dfrac{\sigma}{\sqrt{10}} = 0.21568\) gives \(\sigma = \sqrt{0.0175}\) could get B0M0M0B1B1B1M0A0A0
Unless continue to get \(s^2 = \dfrac{10}{9}0.0175 = 0.0194\ldots\)
Use of \(2(1.833)\dfrac{s}{\sqrt{10}} = 0.21568\) gives \(s = 0.1860\) could get B0M0M0B1B1B1M1A0A1