S3 June 2016 Q7

EdexcelOld spec11 marksConfidence Intervals

7. A restaurant states that its hamburgers contain 20% fat. Paul claims that the mean fat content of their hamburgers is less than 20%. Paul takes a random sample of 50 hamburgers from the restaurant and finds that they contain a mean fat content of 19.5% with a standard deviation of 1.5%

You may assume that the fat content of hamburgers is normally distributed.

(a) Find the 90% confidence interval for the mean fat content of hamburgers from the restaurant. (4)

(b) State, with a reason, what action Paul should recommend the restaurant takes over the stated fat content of their hamburgers. (2)

The restaurant changes the mean fat content of their hamburgers to \(\mu\)% and adjusts the standard deviation to 2%. Paul takes a sample of size \(n\) from this new batch of hamburgers. He uses the sample mean \(\bar{X}\) as an estimator of \(\mu\).

Mark scheme (a)

Scheme	Marks
\(19.5 \pm 1.6449 \times \dfrac{1.5}{\sqrt{50}}\)	M1B1
\(= (19.151\ldots,\ 19.848\ldots)\) awrt 19.2, awrt 19.8	A1A1
	(4)

Notes

M1 correct with their \(z\) i.e. \(19.5 \pm (z \text{ value}) \times \dfrac{1.5}{\sqrt{50}}\)

B1 for 1.6449

A1 awrt 19.2, A1 awrt 19.8(5)

Mark scheme (b)

Scheme	Marks
CI does not contain 20 oe	M1
Fast Food restaurant statement is too high; they should reduce the stated value.	A1
	(2)

Notes

M1 Require 20 compared to their interval

A1 Accept statement that relates to 20 being above the interval.

Mark scheme (c)

Scheme	Marks
\(\mathrm{P}(\|\bar{X} - \mu\| \lt 0.5) = 0.9\) \(\dfrac{0.5}{\frac{2}{\sqrt{n}}} = 1.6449\)	M1A1
\(n = \left(2 \times \dfrac{1.6449}{0.5}\right)^2 = 43.29\ldots\)	dM1A1
Sample size required is 44	A1
	(5)
	(11 marks)

Notes

M1 \(\dfrac{0.5}{\frac{2}{\sqrt{n}}} = z\) value or equivalent expression

A1 All correct

dM1 Attempt to solve \(\dfrac{0.5}{\frac{2}{\sqrt{n}}} =\) their \(z\) value

A1 awrt 43.3

A1 44 cao