S4 June 2012 Q5
5. Boxes of chocolates manufactured by Philippe have a mean weight of \(\mu\) grams and a standard deviation of \(\sigma\) grams. A random sample of 25 of these boxes are weighed. Using this sample, the unbiased estimate of \(\mu\) is 455 and the unbiased estimate of \(\sigma^2\) is 55.
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \sigma^2 = 36;\ \mathrm{H}_1 : \sigma^2 \gt 36\) | B1 |
| \(\upsilon = 24,\ \chi^2_{24}(0.05) = 36.415\) | B1 |
| \(\dfrac{(n-1)S^2}{\sigma^2} = \dfrac{24 \times 55}{36} = 36.67\) | M1 A1 |
| Since \(36.67 \gt 36.415\) there is sufficient evidence to reject \(\mathrm{H}_0\). | A1 ft |
| There is evidence to suggest that the variance is greater than 36. | A1 ft |
| (6) |
Notes
B1 both correct. Also allow \(\mathrm{H}_0 : \sigma = 6;\ \mathrm{H}_1 : \sigma \gt 6\)
B1 36.415
M1 use of \(\dfrac{(n-1)S^2}{\sigma^2}\)
A1 awrt 36.7
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : \mu = 450 \qquad \mathrm{H}_1 : \mu \gt 450\) | B1 |
| \(t_{24} = 1.711\) | B1 |
| \(t = \pm\dfrac{455 - 450}{\sqrt{\dfrac{55}{25}}} = \pm 3.37\ldots\) | M1 A1 |
| Significant; The mean weight of chocolates is greater than 450, Or \(\mu\) is more than 450 | A1ft; A1ft |
| (6) |
Notes
M1 \(\pm\dfrac{455 - 450}{\sqrt{\dfrac{55}{25}}}\)
A1 awrt 3.4
A1ft any statement – no conflicting
A1ft contextual statement must include “weight of chocolate” and is “greater than 450” (corrected from the printed mark scheme: printed as “greater than 50”) CHECK
| Scheme | Marks |
|---|---|
| The weights are normally distributed | B1 |
| (1) | |
| (13 marks) |