The shopkeeper claims that when one of the coins is tossed, the probability of obtaining a head is \(0.6\). Bill wishes to test this claim by tossing the coin \(250\) times and counting the number of heads obtained.

  (i)     State suitable hypotheses for this test.

  (ii)     He obtains \(140\) heads. Find the \(p\)-value of this result and determine whether or not it supports the shopkeeper’s claim at the \(5\%\) level of significance.

Bill tosses the other coin a large number of times and counts the number of heads obtained. He correctly calculates a \(95\%\) confidence interval for the probability that when this coin is tossed, a head is obtained. This is calculated as [\(0.35199\), \(0.44801\)] where the end-points are correct to five significant figures.

Determine

  (i)     the number of times the coin was tossed;

  (ii)     the number of heads obtained.

(i)     \({{\rm{H}}_0}:p = 0.6\) ; \({{\rm{H}}_1}:p \ne 0.6\)     A1A1

(ii)     EITHER

using a normal approximation, \(p\)-value \( = 0.197\)     A2

Note: Award A1 for \(0.0984\).

the shopkeeper’s claim is supported     A1

because \(0.197 > 0.05\)     R1

OR

using binomial distribution, \(p\)-value \( = 0.221\)     A2

Note: Award A1 for \(0.110\).

the shopkeeper’s claim is supported     A1

because \(0.221 > 0.05\)     R1

Note: Follow through the candidate’s \(p\)-value for A1R1.

Note: Accept \(p\)-values correct to two significant figures.

[6 marks]

(i)     \(\hat p = \frac{{0.35199 + 0.44801}}{2} = 0.4\)     A1

width of CI \( = 3.92\sqrt {\frac{{0.4 \times 0.6}}{n}} \)    M1

\(3.92\sqrt {\frac{{0.4 \times 0.6}}{n}} = 0.44801 - 0.35199 = 0.096(02)\)     A1

solving,

\(n = {\left( {\frac{{3.92}}{{0.096(02)}}} \right)^2} \times 0.24\)     (M1)

\( = 400\)     A1

(ii)     \(x = n\widehat p = 400 \times 0.4 = 160\)     M1A1

[7 marks]

Part (a) was well answered in general, using the calculator either to carry out a significance test on proportions or to find the \(p\)-value directly using the binomial distribution. Some candidates gave their conclusion in the form "Accept H₀", this was not accepted since the question asked whether or not the shopkeeper’s claim was supported and a direct answer to this question was required.

Part (b) caused problems for some candidates who were unsure how to proceed. Some candidates used a trial and error method which involved showing that \(\hat p = 0.4\) and then using their calculator to find the confidence interval for appropriate pairs of values for \(n\) and \(p\) until reaching \(400,160\). This was accepted as a valid method although it is not recommended as a general method since its success was based upon the value of \(n\) being one that would probably be tested.