S3 January 2006 Q3
3. The drying times of paint can be assumed to be normally distributed. A paint manufacturer paints 10 test areas with a new paint. The following drying times, to the nearest minute, were recorded.
82, 98, 140, 110, 90, 125, 150, 130, 70, 110.
(a) Calculate unbiased estimates for the mean and the variance of the population of drying times of this paint. (5)
Given that the population standard deviation is 25,
(b) find a 95% confidence interval for the mean drying time of this paint. (5)
Fifteen similar sets of tests are done and the 95% confidence interval is determined for each set.
(c) Estimate the expected number of these 15 intervals that will enclose the true value of the population mean \(\mu\). (2)
| Scheme | Marks |
|---|---|
| \(\hat{\mu} = \dfrac{82 + 98 + 140 + 110 + 90 + 125 + 150 + 130 + 70 + 110}{10}\) | M1 |
| \(= 110.5\) | A1 |
| \(\hat{\sigma}^2 = \dfrac{1}{9}\left(128153 - 10 \times 110.5^2\right)\) 128153 | B1 M1 |
| \(= 672.28\) (AWRT 672) | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| 95% confidence limits are (condone use of 5 instead of 25) (for 1.96) \(110.5 \pm 1.96 \times \dfrac{25}{\sqrt{10}}\) | M1 B1 A1ft |
| 95% conf. lim. = AWRT(95, 126) | A1 A1 |
| (5) |
| Scheme | Marks |
|---|---|
| Number of intervals \(= \dfrac{95}{100} \times 15\) | M1 |
| \(= 14.25\) (Allow 14 or 14.3 if method is clear) | A1 |
| (2) | |
| (12 marks) |