S3 June 2007 Q6
6. A random sample of the daily sales (in £s) of a small company is taken and, using tables of the normal distribution, a 99% confidence interval for the mean daily sales is found to be
\[(123.5,\ 154.7)\]Find a 95% confidence interval for the mean daily sales of the company. (6)
| Scheme | Marks |
|---|---|
| \(\bar{x} = \tfrac{1}{2}(123.5 + 154.7) = \underline{139.1}\) | B1 |
| 2.5758 | B1 |
| “their 2.5758” \(\dfrac{\sigma}{\sqrt{n}} = 154.7 - 139.1 = 15.6\) | M1 |
| AWRT 1.96 | B1 |
| “their 1.96” \(\dfrac{\sigma}{\sqrt{n}} = \dfrac{15.6 \times 1.96}{2.5758} = (11.87\ldots)\) | M1 |
| So 95% C.I. = \(139.1 \pm 11.87\ldots = (127.22\ldots, 150.97\ldots)\) AWRT (127, 151) | A1 |
| (6) | |
| (6 marks) |
Notes
1st B1 for mean = 139.1 only
1st M1 for UL – mean or mean – LL set equal to \(z\) value times standard error or some equivalent expression for standard error. Follow through their 2.5758 provided a \(z\) value.
May be implied by \(\dfrac{\sigma}{\sqrt{n}} = 6.056\ldots\) [N.B. \(\dfrac{15.6}{2.3263} = 6.705\ldots\)]
Condone poor notation for standard error if it is being used correctly to find CI.
2nd M1 for full method for semi-width (or width) of 95% interval
Follow through their \(z\) values for both M marks
N.B. Use of 2.60 instead of 2.5758 should just lose 2nd B1 since it leads to AWRT (127, 151)