S2 January 2013 Q6
6.
Mrs George claims that 45% of voters would vote for her.
In an opinion poll of 20 randomly selected voters it was found that 5 would vote for her.
In a second opinion poll of \(n\) randomly selected people it was found that no one would vote for Mrs George.
| Scheme | Marks |
|---|---|
| A statement concerning a population parameter | B1 |
Notes
It must be a statement including the words population parameter.
| Scheme | Marks |
|---|---|
| A critical region is the range / set of values / answers or a test statistic or region/area or values (where the test is significant) | B1 |
| that would lead to the rejection of H0 / acceptance of \(\mathrm{H}_1\) | B1 |
| (3) |
Notes
The scheme prints (3) here as the total for parts (a) and (b).
| Scheme | Marks |
|---|---|
| \(\mathrm{H}_0 : p = 0.45 \qquad \mathrm{H}_1 : p \lt 0.45\) (or \(p \ne 0.45\)) | |
| \(X \sim \mathrm{B}(20, 0.45)\) | M1 |
| \(\mathrm{P}(X \leqslant 5) = 0.0553\) CR \(X \leqslant 4\) | A1 |
| Accept \(\mathrm{H}_0\). Not significant. 5 does not lie in the Critical region. | M1d |
| There is no evidence that the proportion who voted for Mrs George is not 45% or there is evidence to support Mrs George’s claim | A1cso |
| (4) |
Notes
1st M1 using B(20, 0.45) and finding \(\mathrm{P}(X \leqslant 5)\) or \(\mathrm{P}(X \geqslant 6)\) Using the normal approximation to the binomial is M0
A1 0.0553 (allow 0.9447) if not using CR or CR \(X \leqslant 4\) or \(X \lt 5\)
2nd M1 dependent on previous M being awarded. A correct statement (do not allow if there are contradicting non contextual statements nor award if 2 probabilities are given which would result in different conclusions)
A1cso Conclusion must contain the words Mrs George. There must be no incorrect working seen. If there are no hypotheses you cannot award this mark.
NB A correct contextual statement on it’s own will score M1 A1.
| Scheme | Marks |
|---|---|
| B(8, 0.45): P(0) = 0.0084 | M1 |
| B(7, 0.45): P(0) = 0.0152 | A1 |
| Hence smallest value of \(n\) is 8 | B1 |
| (3) | |
| (10 marks) |
Notes
Alternative
| Scheme | Marks |
|---|---|
| \((0.55)^n \lt 0.01\) | M1 |
| \(n\log 0.55 \lt \log 0.01\) | |
| \(n \gt 7.7\ldots\) | A1 |
| Hence smallest value of \(n\) is 8 | B1cso |
M1 Attempt to find P(0) from B(\(n\), 0.45) or \((0.55)^n \lt 0.01\) or \((0.55)^n = 0.01\) or \((0.55)^n \gt 0.01\)
A1 P(0) = 0.0084 and P(0) = 0.0152 or getting 7.7 May be implied by correct answer.
B1 cso. \(n = 8\) should not come from incorrect working.
NB An answer of 8 on its own with no working gains M1A1B1