Question 4. [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 random sample of 167 people who own mobile phones was used to collect data on the amount of time they spent per day using their phones. The results are displayed in the table below.
Manuel conducts a survey on a random sample of 751 people to see which television programme type they watch most from the following: Drama, Comedy, Film, News. The results are as follows.
Manuel decides to ignore the ages and to test at the 5 % level of significance whether the most watched programme type is independent of gender.
State the modal group.[1]
Use your graphic display calculator to calculate approximate values of the mean and standard deviation of the time spent per day on these mobile phones.[3]
On graph paper, draw a fully labelled histogram to represent the data.[4]
Draw a table with 2 rows and 4 columns of data so that Manuel can perform a chi-squared test.[3]
State Manuel’s null hypothesis and alternative hypothesis.[1]
Find the expected frequency for the number of females who had ‘Comedy’ as their most-watched programme type. Give your answer to the nearest whole number.[2]
Using your graphic display calculator, or otherwise, find the chi-squared statistic for Manuel’s data.[3]
(i) State the number of degrees of freedom available for this calculation.
(ii) State his conclusion.[3]
Answer/Explanation
Markscheme
\(45 \leqslant t < 60\) (A1)[1 mark]
Unit penalty (UP) is applicable in question part (i)(b) only.
(UP) 42.4 minutes (G2)
21.6 minutes (G1)
[3 marks]
(A4)[4 marks]
(M1)(M1)(A1)[3 marks]
H0: favourite TV programme is independent of gender or no association between favourite TV programme and gender
H1: favourite TV programme is dependent on gender (must have both) (A1)[1 mark]
\(\frac{{365 \times 217}}{{751}}\) (M1)
\(= 105\) (A1)(ft)(G2)[2 marks]
12.6 (accept 12.558) (G3)[3 marks]
(i) 3 (A1)
(ii) reject H0 or equivalent statement (e.g. accept H1) (A1)(ft)[3 marks]
Question
Jorge conducted a survey of \(200\) drivers. He asked two questions:
How long have you had your driving licence?
Do you wear a seat belt when driving?
The replies are summarized in the table below.
Jorge applies a \({\chi ^2}\) test at the \(5\% \) level to investigate whether wearing a seat belt is associated with the time a driver has had their licence.
(i) Write down the null hypothesis, \({{\text{H}}_0}\).
(ii) Write down the number of degrees of freedom.
(iii) Show that the expected number of drivers that wear a seat belt and have had their driving licence for more than \(15\) years is \(22\), correct to the nearest whole number.
(iv) Write down the \({\chi ^2}\) test statistic for this data.
(v) Does Jorge accept \({{\text{H}}_0}\) ? Give a reason for your answer.[8]
Consider the \(200\) drivers surveyed. One driver is chosen at random. Calculate the probability that
(i) this driver wears a seat belt;
(ii) the driver does not wear a seat belt, given that the driver has held a licence for more than \(15\) years.[4]
Two drivers are chosen at random. Calculate the probability that
(i) both wear a seat belt.
(ii) at least one wears a seat belt.[6]
Answer/Explanation
Markscheme
(i) \({{\text{H}}_0} = \) wearing of a seat belt and the time a driver has held a licence are independent. (A1)
Note: For independent accept ‘not associated’ but do not accept ‘not related’ or ‘not correlated’
(ii) \(2\) (A1)
(iii) \(\frac{{98 \times 45}}{{200}} = 22.05 = 22\) (correct to the nearest whole number) (M1)(A1)(AG)
Note: (M1) for correct formula and (A1) for correct substitution. Unrounded answer must be seen for the (A1) to be awarded.
(iv) \({\chi ^2} = 8.12\) (G2)
Note: For unrounded answer award (G1)(G0)(AP). If formula used award (M1) for correct substituted formula with correct substitution (6 terms) (A1) for correct answer.
(v) “Does not accept \({{\text{H}}_0}\)” (A1)(ft)
\(p{\text{-}}value < 0.05\) (R1)(ft)
Note: Allow “Reject \({{\text{H}}_0}\)” or equivalent. Follow through from their \({\chi ^2}\) statistic. Award (R1)(ft) for comparing the appropriate values. The (A1)(ft) can be awarded only if the conclusion is valid according to the comparison given. If no reason given or if reason is wrong the two marks are lost.[8 marks]
(i) \(\frac{{98}}{{200}}( = 0.49{\text{, }}49\% )\) (A1)(A1)(G2)
Note: (A1) for numerator, (A1) for denominator.
(ii) \(\frac{{15}}{{45}}( = 0.333{\text{, }}33.3\% )\) (A1)(A1)(G2)
Note: (A1) for numerator, (A1) for denominator.[4 marks]
(i) \(\frac{{98}}{{200}} \times \frac{{97}}{{199}} = 0.239{\text{ }}(23.9\% )\) (A1)(M1)(A1)(G3)
Note: (A1) for correct probabilities seen, (M1) for multiplying two probabilities, (A1) for correct answer.
(ii) \(1 – \frac{{102}}{{200}} \times \frac{{101}}{{199}} = 0.741{\text{ }}(74.1\% )\) (M1)(M1)(A1)(ft)(G2)
Note: (M1) for showing the product, (M1) for using the probability of the complement, (A1) for correct answer. Follow through for consistent use of with replacement.
OR
\(\frac{{98}}{{200}} \times \frac{{97}}{{199}} + \frac{{98}}{{200}} \times \frac{{102}}{{199}} + \frac{{102}}{{200}} \times \frac{{98}}{{199}} = 0.741{\text{ }}(74.1\% )\) (M1)(M1)(A1)(ft)(G2)
Note: (M1) for adding three products of fractions (or equivalent), (M1) for using the correct fractions, (A1) for correct answer. Follow through for consistent use of with replacement.[6 marks]
Question
A manufacturer claims that fertilizer has an effect on the height of rice plants. He measures the height of fertilized and unfertilized plants. The results are given in the following table.
A chi-squared test is performed to decide if the manufacturer’s claim is justified at the 1 % level of significance.
The population of fleas on a dog after t days, is modelled by
\[N = 4 \times {(2)^{\frac{t}{4}}},{\text{ }}t \geqslant 0\]
Some values of N are shown in the table below.
Write down the null and alternative hypotheses for this test.[2]
For the number of fertilized plants with height greater than 75 cm, show that the expected value is 97.5.[3]
Write down the value of \(\chi_{calc}^2\).[2]
Write down the number of degrees of freedom.[1]
Is the manufacturer’s claim justified? Give a reason for your answer.[2]
Write down the value of p.[1]
Write down the value of q.[2]
Using the values in the table above, draw the graph of N for 0 ≤ t ≤ 20. Use 1 cm to represent 2 days on the horizontal axis and 1 cm to represent 10 fleas on the vertical axis.[6]
Use your graph to estimate the number of days for the population of fleas to reach 55.[2]
Answer/Explanation
Markscheme
H0: The height of the rice plants is independent of the use of a fertilizer. (A1)
Notes: For independent accept “not associated”, can accept “the use of a fertilizer has no effect on the height of the plants”.
Do not accept “not correlated”.
H1: The height of the rice plants is not independent (dependent) of the use of fertilizer. (A1)(ft)
Note: If H0 and H1 are reversed award (A0)(A1)(ft).[2 marks]
\(\frac{{180 \times 195}}{{360}}\) or \(\frac{{180}}{{360}} \times \frac{{195}}{{360}} \times 360\) (A1)(A1)(M1)
= 97.5 (AG)
Notes: Award (A1) for numerator, (A1) for denominator (M1) for division.
If final 97.5 is not seen award at most (A1)(A0)(M1).[3 marks]
\( \chi_{calc}^2 = 14.01 (14.0, 14)\) (G2)
OR
If worked out by hand award (M1) for correct substituted formula with correct values, (A1) for correct answer. (M1)(A1)[2 marks]
2 (A1)[1 mark]
\( \chi_{calc}^2 > \chi_{crit}^2\) (R1)
The manufacturer’s claim is justified. (or equivalent statement) (A1)
Note: Do not accept (R0)(A1).[2 marks]
\(p = 4\) (G1)[1 mark]
\(q = 4(2)^{\frac{16}{4}}\) (M1)
\(= 64\) (A1)(G2)[2 marks]
(A1)(A1)(A1) (A3)
Notes: Award (A1) for x axis with correct scale and label, (A1) for y axis with correct scale and label.
Accept x and y for labels.
If x and y axis reversed award at most (A0)(A1)(ft).
(A1) for smooth curve.
Award (A3) for all 6 points correct, (A2) for 4 or 5 points correct, (A1) for 2 or 3 points correct, (A0) otherwise.[6 marks]
15 (±0.8) (M1)(A1)(ft)(G2)
Note: Award (M1) for line drawn shown on graph, (A1)(ft) from candidate’s graph.[2 marks]
Question
In a mountain region there appears to be a relationship between the number of trees growing in the region and the depth of snow in winter. A set of 10 areas was chosen, and in each area the number of trees was counted and the depth of snow measured. The results are given in the table below.
In a study on \(100\) students there seemed to be a difference between males and females in their choice of favourite car colour. The results are given in the table below. A \(\chi^2\) test was conducted.
Use your graphic display calculator to find the mean number of trees.[1]
Use your graphic display calculator to find the mean depth of snow.[1]
Use your graphic display calculator to find the standard deviation of the depth of snow.[1]
The covariance, Sxy = 188.5.
Write down the product-moment correlation coefficient, r.[2]
Write down the equation of the regression line of y on x.[2]
If the number of trees in an area is 55, estimate the depth of snow.[2]
Use the equation of the regression line to estimate the depth of snow in an area with 100 trees.[1]
Decide whether the answer in (e)(i) is a valid estimate of the depth of snow in the area. Give a reason for your answer.[2]
Write down the total number of male students.[1]
Show that the expected frequency for males, whose favourite car colour is blue, is 12.6.[2]
The calculated value of \({\chi ^2}\) is \(1.367\) and the critical value of \({\chi ^2}\) is \(5.99\) at the \(5\%\) significance level.
Write down the null hypothesis for this test.[1]
The calculated value of \({\chi ^2}\) is \(1.367\) and the critical value of \({\chi ^2}\) is \(5.99\) at the \(5\%\) significance level.
Write down the number of degrees of freedom.[1]
The calculated value of \({\chi ^2}\) is \(1.367\) and the critical value of \({\chi ^2}\) is \(5.99\) at the \(5\%\) significance level.
Determine whether the null hypothesis should be accepted at the \(5\%\) significance level. Give a reason for your answer.[2]
Answer/Explanation
Markscheme
50 (G1)[1 mark]
30.5 (G1)[1 mark]
12.3 (G1)
Note: Award (A1)(ft) for 13.0 in (iv) but only if 17.7 seen in (a)(ii).[1 mark]
\(r = \frac{{188.5}}{{(16.79 \times 12.33)}}\) (M1)
Note: Award (M1) for using their values in the correct formula.
= 0.911 (accept 0.912, 0.910) (A1)(ft)(G2)[2 marks]
y = 0.669x − 2.95 (G1)(G1)
Note: Award (G1) for 0.669x, (G1) for −2.95. If the answer is not in the form of an equation, award at most (G1)(G0).[2 marks]
Depth = 0.669 × 55 − 2.95 (M1)
= 33.8 (A1)(ft)(G2)(ft)
Note: Follow through from their (c) even if no working seen.[2 marks]
64.0 (accept 63.95, 63.9) (A1)(ft)(G1)(ft)
Note: Follow through from their (c) even if no working seen.[1 mark]
It is not valid. It lies too far outside the values that are given. Or equivalent. (A1)(R1)
Note: Do not award (A1)(R0).[2 marks]
28 (A1)[1 mark]
\(\frac{{28 \times 45}}{{100}}\left( {\frac{{28}}{{100}} \times \frac{{45}}{{100}} \times 100} \right)\) (M1)(A1)(ft)
Note: Award (M1) for correct formula, (A1) for correct substitution.
= 12.6 (AG)
Note: Do not award (A1) unless 12.6 seen.[2 marks]
the favourite car colour is independent of gender. (A1)
Note: Accept there is no association between gender and favourite car colour.
Do not accept ‘not related’ or ‘not correlated’.[1 mark]
\(2\) (A1)[1 marks]
Accept the null hypothesis since \(1.367 < 5.991\) (A1)(ft)(R1)
Note: Allow “Do not reject”. Follow through from their null hypothesis and their critical value.
Full credit for use of \(p\)-values from GDC [\(p = 0.505\)].
Do not award (A1)(R0). Award (R1) for valid comparison.[2 marks]