IBDP Maths AI: Topic: SL 4.11: Formulation of null and alternative hypotheses : IB style Questions HL Paper 2

Question 3. [Maximum mark: 13]

The stopping distances for bicycles travelling at 20 km h-1 are assumed to follow a normal distribution with mean 6.76 m and standard deviation 0.12 m.

a. Under this assumption, find, correct to four decimal places, the probability that a bicycle chosen at random travelling at 20 km h-1 manages to stop

(i) in less than 6.5 m.

(ii) in more than 7 m. [3]

1000 randomly selected bicycles are tested and their stopping distances when travelling at 20 km h-1 are measured.

b. Find, correct to four significant figures, the expected number of bicycles tested that stop between

(ii)6.5 m and 6.75 m.

(ii) 6.75 m and 7 m. [3]

The measured stopping distances of the 1000 bicycles are given in the table.

It is decided to perform a χ 2 goodness of fit test at the 5 % level of significance to decide whether the stopping distances of bicycles travelling at 20 km h-1 can be modelled by a normal distribution with mean 6.76 m and standard deviation 0.12 m.

a. State the null and alternative hypotheses. [2]

b. Find the p-value for the test. [3]

c. State the conclusion of the test. Give a reason for your answer. [2]

▶️Answer/Explanation

(a) (i)  \( P(X<6.5) 0.0151\)

(ii) 0.0228

(b)(i) multiplying their probability by 1000

451.7

(ii) 510.5

(c)H0: stopping distances can be modelled by N(6.76,0.122

H1: stopping distances can not be modelled by N(6.76,0.122)

(d)  15.1 or 22.8 seen 0.0727 (0.0756542…,7.27%)

(e) \(0.05< 0.0727\) there is insufficient evidence to reject H0 (or accept H0)

Question

A Principal would like to compare the students in his school with a national standard.
He decides to give a test to eight students made up of four boys and four girls. One of
the teachers offers to find the volunteers from his class.
(a) Name the type of sampling that best describes the method used by the Principal.

The marks out of 40, for the students who took the test, are:
25, 29, 38, 37, 12, 18, 27, 31.

(b) For the eight students find
(i) the mean mark.
(ii) the standard deviation of the marks.
The national standard mark is 25.2 out of 40.
(c) Perform an appropriate test at the 5% significance level to see if the mean marks
achieved by the students in the school are higher than the national standard. It can
be assumed that the marks come from a normal population.
(d) State one reason why the test might not be valid.
Two additional students take the test at a later date and the mean mark for all ten students
is 28.1 and the standard deviation is 8.4.
For further analysis, a standardized score out of 100 for the ten students is obtained by
multiplying the scores by 2 and adding 20.
(e) For the ten students, find
(i) their mean standardized score.
(ii) the standard deviation of their standardized score.

▶️Answer/Explanation

Ans:

(a) quote
(b)(i) \(27.125 \approx 27.1\)
(ii) \(8.29815… \approx 8.30\)
(c) (let \(\mu be the national mean)
\(H_0: \mu = 25.2\)
\(H_1: \mu > 25.2\)
recognizing t-test
p-value = 0.279391…
0.279391…> 0.05

insufficient evidence to reject the null hypothesis (that the mean for the school is 25.2)
(d) EITHER
the sampling process is not random
For example:
the school asked for volunteers
the students were selected from a single class
OR
the quota might not be representative of the student population
For example:
the school may have only 4 boys and 400 girls.
(e) (i) \((28.1 \times 2 + 20 = ) 76.2
(ii) \(8.4 \times 2\)
= 16.8

Question

A shop sells reddish and broccoli. The weights of reddish can be modelled by a normal distribution with variance 25 grams2 and the weights of broccoli can be modelled by a normal distribution with variance 80 grams2. The shopkeeper claims that the mean weight of reddish is 130 grams and the mean weight of broccoli is 400 grams.

    1. Assuming that the shopkeeper’s claim is correct, find the probability that the weight of six randomly chosen reddish is more than two times the weight of one randomly chosen broccoli. [6]

      Aayush decides to investigate the shopkeeper’s claim that the mean weight of reddish is 130 grams. He plans to take a random sample of n reddish in order to calculate a 98 % confidence interval for the population mean weight.

    2. Find the least value of n required to ensure that the width of the confidence interval is less than 2 grams. [3]

      Maya thinks the mean weight, μ grams, of the broccoli is less than 400 grams. She decides to perform a hypothesis test, using a random sample of size 8. Her hypotheses are

      H0 : μ = 400 ; H1 : μ < 400.

      She decides to reject H0 if the sample mean is less than 395 grams.

    3. Find the significance level for this test. [3]

    4. Given that the weights of the broccoli actually follow a normal distribution with mean 392 grams and variance 80 grams2, find the probability of Maya making a Type II error. [3]

▶️Answer/Explanation

Ans: 

(a)

Let X = \(\sum_{i=1}^{6}C_{i} -2B\)

\(E(X)= 6\times 130-2\times 400=-20\)

\(Var(X)= 6\times 25+4\times 80=470\)

\(P(X>0)\)=0.178

(b)

z = 2.326…

\(\frac{2z\sigma }{\sqrt{n}}<2\)

\(\sqrt{n}>11.6…\)

\(n>135.2….n\)=136 

(c)

variance = \(\frac{80}{8}=10\)

under \(H_{0}\bar{B}\sim N(400,10)\)

signuficance level= p(\bar{B}<395)

= 0.0569 or 5.69% 

(d)

Type II error probability = = \(P(Accept H_{0}|H_{1}true)\)

= \(p (\bar{B}>395|\bar{B}\approx N(392,10))\)

= 0.171

Question

An automatic machine is used to fill bottles of water. The amount delivered, \(Y\) ml , may be assumed to be normally distributed with mean \(\mu \) ml and standard deviation \(8\) ml . Initially, the machine is adjusted so that the value of \(\mu \) is \(500\). In order to check that the value of \(\mu \) remains equal to \(500\), a random sample of 10 bottles is selected at regular intervals, and the mean amount of water, \(\overline y \) , in these bottles is calculated. The following hypotheses are set up.

\({{\rm{H}}_0}:\mu  = 500\) ; \({{\rm{H}}_1}:\mu  \ne 500\)

The critical region is defined to be \(\left( {\overline y  < 495} \right) \cup \left( {\overline y  > 505} \right)\) .

(i)     Find the significance level of this procedure.

(ii)     Some time later, the actual value of \(\mu \) is \(503\). Find the probability of a Type II error.

▶️Answer/Explanation

Markscheme

(i)     Under \({{\rm{H}}_0}\) , the distribution of \({\overline y }\) is N(500, 6.4) .     (A1)

Significance level \( = {\rm{P}}\overline y  < 495\) or \( > 505|{{\rm{H}}_0}\)     M2

\( = 2 \times 0.02405\)     (A1)

\( = 0.0481\)     A1 N5

Note: Using tables, answer is \(0.0478\).

(ii)     The distribution of \(\overline y \) is now N(\(503\), \(6.4\)) .     (A1)

P(Type ΙΙ error) \( = {\rm{P}}(495 < \overline y  < 505)\)     (M1)

\( = 0.785\)     A1 N3

Note: Using tables, answer is \(0.784\). 

[8 marks]

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