Date | May 2015 | Marks available | 9 | Reference code | 15M.1.hl.TZ0.14 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Find | Question number | 14 | Adapted from | N/A |
Question
Sarah is the quality control manager for the Stronger Steel Corporation which makes steel sheets. The steel sheets should have a mean tensile strength of 430 MegaPascals (MPa). If the mean tensile strength drops to 400 MPa, then Sarah must recommend a change in composition. The tensile strength of these steel sheets follows a normal distribution with a standard deviation of 35 MPa. Sarah defines the following hypotheses
\[{H_0}:\mu = 430\]
\[{H_1}:\mu = 400\]
where \(\mu \) denotes the mean tensile strength in MPa. She takes a random sample of \(n\) steel sheets and defines the critical region as \(\bar x \leqslant k\), where \(\bar x\) notes the mean tensile strength of the sample in MPa and \(k\) is a constant.
Given that the \(P{\text{(Type I Error)}} = 0.0851\) and \(P{\text{(Type II Error)}} = 0.115\), both correct to three significant figures, find the value of \(k\) and the value of \(n\).
Markscheme
\(\bar X \sim N\left( {430,{\text{ }}\frac{{{{35}^2}}}{n}} \right)\) (M1)(A1)
Note: The M1 is for considering the distribution of \(\bar X\)
type I error gives \({\text{P}}(\bar X \leqslant k/\mu = 430) = 0.0851\)
\(\frac{{k - 430}}{{\frac{{35}}{{\sqrt n }}}} = - 1.37156 \ldots \) M1A1
type II error gives \({\text{P}}(\bar X > k/\mu = 400) = 0.115\)
\(\frac{{k - 400}}{{\frac{{35}}{{\sqrt n }}}} = 1,20035 \ldots \) M1A1
Note: The two M1 marks above are for attempting to standardize \({\bar X}\) and obtain the corresponding equations with inverse normal values
solving simultaneously (M1)
\(k = 414\) A1
\(n = 9\) A1
Examiners report
This proved to be a difficult question for most candidates with only a minority giving a correct solution. Most candidates either made no attempt at the question or just wrote several lines of irrelevant mathematics.