S3 June 2018 Q4

EdexcelOld spec9 marksIncludes hypothesis testingConfidence Intervals

4. The waiting times, in minutes, of patients at a doctor’s surgery follows a normal distribution with unknown mean \(\mu\) and known standard deviation \(\sigma\)

A random sample of 120 patients was taken.

(a) Find, in the form \(k\sigma\), the width of a 99% confidence interval for \(\mu\) based on this sample. Give the value of \(k\) to 2 decimal places. (3)

A further random sample of 100 patients from the surgery gave a 90% confidence interval for \(\mu\) of (5.14, 6.25)

(b) Use this confidence interval to determine whether or not it provides evidence that \(\mu = 6\)
State the hypotheses being tested here and write down the significance level being used. You do not need to carry out any further calculations. (3)

Mark scheme (a)

Scheme	Marks
\(2 \times 2.5758 \times \dfrac{\sigma}{\sqrt{120}} = 0.47027\ldots\sigma\)	M1B1A1
	(3)

Notes

1st M1 Use of \(2z\dfrac{\sigma}{\sqrt{n}}\) with \(z \gt 2\)

1st B1 2.58 or better

1st A1 awrt 0.47\(\sigma\)

Mark scheme (b)

Scheme	Marks
\(\mathrm{H}_0 : \mu = 6 \qquad \mathrm{H}_1 : \mu \neq 6\)	B1
(Significance level = )10%	B1
(6 is in the interval so not significant, do not reject \(\mathrm{H}_0\)) \(\mu = 6\)	B1
	(3)

Notes

1st B1 Both hypotheses in terms of \(\mu\).

2nd B1 10%

3rd B1 Correct comment leading to accepting \(\mathrm{H}_0\)

Mark scheme (c)

Scheme	Marks
\(1.6449 \times \dfrac{\sigma}{\sqrt{100}} = (6.25 - 5.14) / 2 (= 0.555)\)	M1B1
\(\sigma = 3.374\ldots\)	A1
	(3)
	(9 marks)

Notes

1st M1 for \(z\dfrac{\sigma}{\sqrt{100}} = 0.555\) oe, using \(n = 100\) and where \(|z| \gt 1.5\)

1st B1 for 1.6449 or better in an attempt (could be \(1.6449\sigma = c\) or even \(1.6449\ \sigma^2 = c\))

1st A1 awrt 3.37. Allow awrt 3.38 from use of \(z = 1.64\)