The length of time, T, in months, that a football manager stays in his job before he is removed can be approximately modelled by a normal distribution with population mean $\mu$ and population variance ${\sigma ^2}$. An independent sample of five values of T is given below.

6.5, 12.4, 18.2, 3.7, 5.4

(a)     Given that ${\sigma ^2} = 9$,

(i)     use the above sample to find the 95 % confidence interval for $\mu$, giving the bounds of the interval to two decimal places;

(ii)     find the smallest number of values of T that would be required in a sample for the total width of the 90 % confidence interval for $\mu$ to be less than 2 months.

(b)     If the value of ${\sigma ^2}$ is unknown, use the above sample to find the 95 % confidence interval for $\mu$, giving the bounds of the interval to two decimal places.

(a)     (i)     as ${\sigma ^2}$ is known ${\bar x}$ is ${\text{N}}\left( {\mu ,\frac{{{\sigma ^2}}}{n}} \right)$     (M1)

CI is $\bar x - {z^ * }\frac{\sigma }{{\sqrt n }} < \mu  < \bar x + {z^ * }\frac{\sigma }{{\sqrt n }}$     (M1)

$\bar x = 9.24,{\text{ }}{z^ * } = 1.960$ for 95 % CI     (A1)

CI is $6.61 < \mu < 11.87$ by GDC     A1A1

(ii)     CI is $\bar x - {z^ * }\frac{\sigma }{{\sqrt n }} < \mu < \bar x + {z^ * }\frac{\sigma }{{\sqrt n }}$

require $2 \times 1.645\frac{3}{{\sqrt n }} < 2$     R1A1

$4.935 < \sqrt n$     (A1)

$24.35 < n$     A1

so smallest value for n = 25     A1

Note: Accept use of table.

[10 marks]

(b)     as ${\sigma ^2}$ is not known ${\bar x}$ has the t distribution with v = 4     (M1)(A1)

CI is $\bar x - {t^ * }\frac{{{s_{n - 1}}}}{{\sqrt n }} < \mu < \bar x + {t^ * }\frac{{{s_{n - 1}}}}{{\sqrt n }}$

$\bar x = 9.24,{\text{ }}{s_{n - 1}} = 5.984,{\text{ }}{t^ * } = 2.776$ for 95 % CI     (A1)

CI is $1.81 < \mu < 16.67$ by GDC     A1A1

[5 marks]

Total [15 marks]

The 2 confidence intervals were generally done well by using a calculator. Some marks were dropped by not giving the answers to 2 decimal places as required. Weak candidates did not realise that (b) was a t interval. Part (a) (ii) was not as well answered and often it was the first step that was the problem.