Bottles of iced tea are supposed to contain 500 ml. A random sample of 8 bottles was selected and the volumes measured (in ml) were as follows:

497.2, 502.0, 501.0, 498.6, 496.3, 499.1, 500.1, 497.7 .

  (i)     Calculate unbiased estimates of the mean and variance.

  (ii)     Test at the \(5\%\) significance level the null hypothesis \({{\rm{H}}_0}:\mu  = 500\) against the alternative hypothesis \({{\rm{H}}_1}:\mu  < 500\) .

A random sample of size four is taken from the distribution N(60, 36) .

Calculate the probability that the sum of the sample values is less than 250.

(i)     497.2, 502.0, 501.0, 498.6, 496.3, 499.1, 500.1, 497.7

using the GDC

\(\overline x  = 499.0\) , \({\sigma ^2} = 3.8(0)\)     A1A1

Note: Accept \(499\).

(ii)     EITHER

\(p\)-value = 0.0950     A1

since \(0.0950 > 0.05\) accept \({H_0}\)     R1A1

OR

\({t_{calc}} = - 1.45\) , \({t_{critical}} = - 1.895\) for \(v = 7\) at 5 % level     A1

since \({t_{calc}} > {t_{critical}}\) accept \({H_0}\)     R1A1

[5 marks]

each \(X \sim {\rm{N}}(60,36)\) so \(\sum\limits_{n = 1}^4 {{X_n} \sim {\rm{N}}(4(60),4(36)) = {\rm{N}}(240,144)} \)     M1A1A1

\({\rm{Pr}}({\rm{Sum}} < 250) = {\rm{Pr}}\left( {z < \frac{{250 - 240}}{{12}} = \frac{5}{6}} \right)\)     (M1)(A1)

\( = 0.798\) (by GDC)     A1

Notes: Accept \(0.797\) (tables).

    Answer only is awarded M0A0A0(M1)(A1)A1.

[6 marks]

(a)(i) Very few mistakes were made in this question, although sometimes variance and standard deviation were confused. Why both variance and standard deviation are needed might be something that teachers could explore.

(ii) Again there were no serious problems although some candidates fail to show all the important parameters such as degrees of freedom.

This was found to be relatively straightforward except for using the correct variance of \(144\). It would be useful here to make clear the distinction between the sum of random variables and a multiple of a random variable.