IBDP MAI : Topic 4 Statistics and probability - AHL 4.14 Linear transformation AI HL Paper 3
Question
Mr Sailor owns a fish farm and he claims that the weights of the fish in one of his lakes have a mean of 550 grams and standard deviation of 8 grams.
Assume that the weights of the fish are normally distributed and that Mr Sailor’s claim is true.
Kathy is suspicious of Mr Sailor’s claim about the mean and standard deviation of the weights of the fish. She collects a random sample of fish from this lake whose weights are shown in the following table.
Using these data, test at the $5 \%$ significance level the null hypothesis $H_0: \mu=550$ against the alternative hypothesis $H_1: \mu<550$, where $\mu$ grams is the population mean weight.
Kathy decides to use the same fish sample to test at the $5 \%$ significance level whether or not there is a positive association between the weights and the lengths of the fish in the lake. The following table shows the lengths of the fish in the sample. The lengths of the fish can be assumed to be normally distributed.
a.i. Find the probability that a fish from this lake will have a weight of more than 560 grams.
a.ii.The maximum weight a hand net can hold is $6 \mathrm{~kg}$. Find the probability that a catch of 11 fish can be carried in the hand net.
b.i.State the distribution of your test statistic, including the parameter.
b.iiFind the $p$-value for the test.
b.iiiState the conclusion of the test, justifying your answer.
c.i. State suitable hypotheses for the test.
c.ii.Find the product-moment correlation coefficient $r$.
c.iiiState the $p$-value and interpret it in this context.
d. Use an appropriate regression line to estimate the weight of a fish with length $360 \mathrm{~mm}$.
▶️Answer/Explanation
a.i. Note: Accept all answers that round to the correct 2 sf answer in (a), (b) and (c) but not in (d).
$$
\begin{aligned}
& X \sim N\left(550,8^2\right) \quad \text { (M1) } \\
& \mathrm{P}(X>560)-0.10564 \ldots=0.106
\end{aligned}
$$
A1
[2 marks]
a.ii.Note: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
$$
\begin{aligned}
& X_i \sim N\left(550,8^2\right), i=1, \ldots, 11 \\
& \text { let } Y=\sum_{i=1}^{11} X_i \\
& \mathrm{E}(\mathrm{Y})=11 \times 550(6050) \quad \text { A1 } \\
& \operatorname{Var}(\mathrm{Y})=11 \times 8^2 \quad(704) \quad \text { (M1)A1 } \\
& \mathrm{P}(Y \leqslant 6000)=0.02975 \ldots=0.0298
\end{aligned}
$$
A1
[4 marks]
b.i. Note: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
$t$ distribution with 7 degrees of freedom
A1A1
[2 marks]
b.ii.Note: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
$$
p=0.25779 \ldots=0.258 \quad \text { A2 }
$$
[2 marks]
b.iiiNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
$$
p>0.05
$$
R1
therefore we conclude that there is no evidence to reject $H_0$
A1
Note: $\boldsymbol{F T}$ their $p$-value.
Note: Only award $\mathbf{A 1}$ if $\boldsymbol{R 1}$ awarded.
[2 marks]
c.i. Note: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
$$
H_0: \rho=0, H_1: \rho>0
$$
A1
Note: Do not accept $r$ in place of $\rho$.
[1 mark]
c.ii.Note: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
$$
r=0.782 \quad \text { A2 }
$$
[2 marks]
c.iiiNote: Accept all answers that round to the correct 2sf answer in (a), (b) and (c) but not in (d).
0.01095… $=0.0110 \quad$ A1
since $0.0110<0.05 \quad \boldsymbol{R 1}$
there is positive association between weight and length
A1
Note: $\boldsymbol{F T}$ their p-value.
Note: Only award $\boldsymbol{A 1}$ if $\boldsymbol{R 1}$ awarded.
Note: Conclusion must be in context.
[3 marks]
d. Note: Accept all answers that round to the correct 2 sf answer in (a), (b) and (c) but not in (d).
regression line of $y$ (weight) on $x$ (length) is $\quad$ (M1)
$$
y=0.8267 \ldots x+255.96 \ldots
$$
$x=360$ gives $y=554$
A1
Note: Award M1AOAO for the wrong regression line, that is $y=0.7393 \ldots x-51.62 \ldots$.
[3 marks]