Get documents about "Binomial Distribution Hypothesis Testing Solutions and Mark Scheme
Document Sample
MEI Past Paper Questions S1 Binomial & Hypothesis Testing
Binomial Distribution & Hypothesis Testing
Solutions and Mark Scheme
Question 1 (S1 January 2001)
24
(i) C18 x 0.7518 x 0.256 = 0.185 (3 s.f.) M1 for binomial
coefficient
M1 for product
A1 cao
Sample taken at random E1 for statement 4
(ii) H0: p = 0.75 B1 for H0
H1: p < 0.75 B1 for H1
Interested to see if proportion is less than 75% E1 for explanation
Based on one-tail H1 3
(iii) P(X ≤ 9) = 0.0796 > 0.05 (hence accept H0) B1 for probability
M1 for comparison
There is not enough evidence to reject the hypothesis that A1 for conclusion in
the proportion preferring Fizzicola is 75% words
3
(iv) For n = 6:
Using tables: P(X ≥ 6) = 1 – 0.8220 = 0.178 (3 s.f.) M1 for calculation of
Or P(X ≥ 6) = 0.756 = 0.178 (3 s.f.) probability
which is greater than 0.05 A1 cao
(hence 6 is not a critical value, and so the region is E1
empty) 3
(v) Using tables:
For n = 10: P(X ≥ 10) = 0.0563 > 0.05
[hence 10 is not in the critical region]
For n = 11: P(X ≥ 11) = 0.0422 < 0.05 M1 for either comparison
[hence 11 is in the critical region]
[ or 0.75n < 0.05 ⇒ n > 10.4 ] [ or M1 for inequality ]
Hence smallest value of n for which critical region is not A1 for conclusion
empty is 11 2
15
MEI Past Paper Questions S1 Binomial & Hypothesis Testing
Question 2 (S1 June 2001)
(i) 420 7 B1 for probability as 1
Estimate for p = = = 0.35 fraction, decimal or %
1200 20
(ii) 30 30
(A) C10 x 0.3510 x 0.6520 = 0.15 (2 s.f.) B1 for C10
B1 for 0.3510 x 0.6520
A1 cao 3
(B) P(1 batch produces at least 5 germinating seeds) M1 for use of tables with
= 1 − P(X < 5) correct probabilities
= 1 − 0.3519 = 0.6481 A1 cao
hence
P(each batch produces at least 5 germinating seeds) M1 for their “0.64812”
= 0.64812 = 0.420 (3 s.f.) A1 4
(iii) H0: p = 0.35; H1: p > 0.35 B1 for null hypothesis
B1 for alternative
hypothesis
P(X ≥ 11) = 1 − P(X ≤ 10) = 1 − 0.9788 M1 for sight of 0.0212 or
= 0.0212 < 0.05 (5%) and 0.9788
P(X ≥ 10) = 1 − P(X ≤ 9) = 1 − 0.9403 M1 for sight of 0.0597 or
= 0.0597 > 0.05 (5%) 0.9403
hence Critical Region = {11, 12, 13, 14, 15, 16, 17, 18} B1 for critical region
or {11, …. , 18} or 11 ≤ X ≤ 18 listed or on diagram
B1 for seeing the
acceptance region on the
diagram
Charmaine's seeds must be grown under the same
conditions as Malcolm's E1 for explanation 7
or “seeds selected randomly”
15
MEI Past Paper Questions S1 Binomial & Hypothesis Testing
Question 3 (S1 January 2002)
(i) Let X ~ B(30, 0.4)
B1 for mean value
Mean = np = 30 × 0.4 or 40% of 30 [ = 12 ] [may be seen in part (ii)]
1
30
(ii) P(X = 12) = C12 × 0.412 × 0.618 12
M1 for 0.4 × 0.6
18
30
M1 for C12 × …
= 0.147 (to 3 s.f.) or 0.15 (to 2 s.f.)
A1 cao
Compare P(X = 12) with P(X = 11) and P(X = 13) E1 for convincing
explanation 4
or equivalent
(iii) [Let p represent the probability that a young person chosen
at random fails the fitness test and Let X ~ B(20, p) ]
H0: p = 0.4 B1 for H0
H1: p ≠ 0.4 (i.e. two-tail test) B1 for H1 2
P(X ≥ 12) = 1 – P(X < 12) B1 for P(X ≥ 12)
= 1 – 0.9435 = 0.0565 B1 for 0.0565 cao
Since 0.0565 > 0.025, there is not enough evidence to M1 for comparison
reject H0, i.e. accept the hypothesis that the proportion of A1 for conclusion in
young people who fail the fitness test is 40% words in context 4
The critical region is the set of x-values for which we reject
H0 :
Using tables:
Lower tail: P(X ≤ 3) = 0.0160 < 0.025 B1 for 0.0160 & 0.0510
P(X ≤ 4) = 0.0510 > 0.025
Upper tail: P(X ≥ 12) = 0.0565 > 0.025 B1 for 0.0565 & 0.0210
P(X ≥ 13) = 0.0210 < 0.025
B1 for lower tail
Critical region:
{0, 1, 2, 3, 13, 14, 15, 16, 17, 18, 19, 20} B1 for upper tail 4
15
MEI Past Paper Questions S1 Binomial & Hypothesis Testing
Question 4 (S1 January 2003)
(i) p = 1 – 0.8 = 0.2 B1 for probability 1
(ii) P(exactly 6 callers get through immediately) M1 for 0.26 × 0.819
= 25
C6 × 0.26 × 0.819 M1 for 25C6 ×
= 0.16 (to 2 s.f.) = 0.163 (to 3 s.f.) A1 cao
P(at least 2 callers get through immediately)
M1 for
= 1 – P(0 or 1 callers get through immediately) 25
C1 × 0.2 × 0.824
= 1 – [0.825 + 25
C1 × 0.2 × 0.824]
M1 for 1 – […. + ….]
= 1 – [0.00378 + 0.02361] 6
A1 cao
= 0.973 (to 3 s.f.) or 0.97 (to 2 s.f.)
(iii) Either
H0: p = 0.2 and H1: p > 0.2 B1 for H0 , B1 for H1
Consider X ~ B(20, 0.2):
M1 for probability
P(X ≥ 10) = 1 – P(X < 10) = 1 – 0.9974 = 0.0026 A1
Since 0.0026 < 0.05, do not accept H0 M1 for comparison
There is enough evidence to accept the hypothesis that A1 for conclusion in
the proportion of callers getting through first time has words (dependent)
increased.
B1 for 8 (at start of upper
tail)
The critical region for the test is {8, 9, …, 19, 20}. B1 for {9, 10, .., 19, 20}
or
H0: p = 0.8 and H1: p < 0.8 B1 for H0 , B1 for H1
Consider X ~ B(20, 0.8): M1 for probability
P(X ≤ 10) = 0.0026 A1
Since 0.0026 < 0.05, do not accept H0 M1 for comparison
There is enough evidence to accept the hypothesis that A1 for conclusion in
the proportion of callers getting through first time has words (dependant)
increased.
B1 for 12 (at end of lower
tail)
The critical region for the test is {0, 1, …, 11, 12} . B1 for {0, 1, .., 10, 11} 8
If critical region found first, all marks are possible, with
probability marks implied. Explicit conclusion to test still
required, using position in critical region.
15
MEI Past Paper Questions S1 Binomial & Hypothesis Testing
Question 5 (S1 November 2003)
(i) [ Let X ~ B(12, 0.17) ] ⎛12 ⎞
M1 for “ ⎜ ⎟ × ”
⎝2⎠
⎛12 ⎞ M1 for “0.244 × 0.7562”
(A) P(X = 2) = ⎜ ⎟ × 0.172 × 0.8310 = 0.296 (3 s.f.)
⎝2⎠ A1
(B) P(X < 2) M1 for 1st term
= 12 × 0.17 × 0.8311 + 0.8312 M1 for sum of 2 terms
= 0.37 (2 s.f.) or 0.370 (3 s.f.) A1
6
(ii) [ Let X ~ B(18, 0.2) ]
(A) P(X ≥ 4) = 1 – P(X ≤ 3) M1 for use of tables
= 1 – 0.5010 = 0.50 (2 s.f.) or 0.499 (3 s.f.) A1
(B) P(X ≤ k) < 0.05 for k = 0 B1 for k = 0
B1 for k = 8
P(X ≥ k) < 0.05 for k = 8, 9, 10, …, 18 5
B1 for rest of values
(iii) H0: p = 0.2 B1 for H0
H1: p ≠ 0.2 B1 for H1
CR Critical Region G1 for lower tail
G1 for upper tail
0 2 4 6 8 10 12 14 16 18
4
15
MEI Past Paper Questions S1 Binomial & Hypothesis Testing
Question 6 (S1 January 2004)
(i) X ~ B(21, 0.85) B1 for distribution
or ‘binomial’ if parameters stated or used in part (ii)
Departure times are independent of each other
or E1 for assumption
Probability of “success” (p = 0.85) is the same for each
trial 2
(ii) (A) P(all departures leave on time) M1 for probability
21
= 0.85 = 0.0329 (to 3 s.f.) = 0.033 (to 2 s.f.) A1 2
(B) P(exactly 3 departures are late leaving) M1 for “0.8518 × 0.153”
⎛ 21⎞
⎛ 21⎞ M1 for “ ⎜ ⎟ × …”
= ⎜ ⎟ × 0.8518 0.153 = 0.241 (to 3 s.f.) ⎝ 18 ⎠
⎝ 18 ⎠
A1
= 0.24 (to 2 s.f.)
3
(C) P(during a week ≤ 1 departure late leaving)
= 0.7166 [ from tables using B(7, 0.15) ] B1 for 0.7166
⇒ P(during each week ≤ 1 departure late leaving)
= (0.7166)3 = 0.368 (to 3 s.f.) M1 for probability
= 0.37 (to 2 s.f.) A1 cao
3
(iii) H0: p = 0.85 B1 for H0
H1: p < 0.85 B1 for H1
[ Assuming X ~ B(15, 0.85) ]
P(X ≤ 10) = 0.0617 (from tables) B1 for probability
Since 0.0617 > 0.05, [ accept H0 ] M1 (dep) for comparison
There is not enough evidence to accept the hypothesis that A1 for conclusion in
the proportion of sailings leaving on time has decreased. words
Allow approach via critical regions: Third and fourth marks:
Diagram with division between 9 and 10 B1
“10 is not in the critical region” or equivalent M1 (dependent) 5
15
MEI Past Paper Questions S1 Binomial & Hypothesis Testing
Question 7 (S1 June 2004)
(i) Expected number of packets with vouchers
B1 cao
= 30 × 0.2 = 6
M1 for “0.26 × 0.824”
P(6 packets contain vouchers)
⎛ 30 ⎞
⎛ 30 ⎞ M1 for “ ⎜ ⎟ × p6 × q24”
= ⎜ ⎟ × 0.2 × 0.8 = 0.179 (to 3 s.f.)
6 24 ⎝6⎠
⎝6⎠ A1 cao
4
= 0.18 (to 2 s.f.)
(ii) P(at least 1 packet contains a voucher)
= 1 – P(0 packets contain a voucher)
M1 for their attempt at
= 1 – 0.830 “1 – P(X = 0)”
= 1 – 0.00124
= 0.99876 (5 s.f.) or 0.9988 (4 s.f.) or 0.999 (3 s.f.) A1 cao
which is very nearly 1 or equivalent E1 for comment 3
(iii) H0: p = 0.2 B1 for H0
2
H1: p ≠ 0.2 B1 for H1
(iv) P(X ≤ 0) = 0.0687 M1 for probability
P(X ≥ 5) = 0.0726 or P(X ≥ 6) = 0.0194 M1 for probability
At least one comparison with 0.025 (or 0 not in CR) M1 for comparison
There is not enough evidence to reject the hypothesis that A1 for conclusion in
the proportion of packets of crisps with vouchers is 0.2. words
dep on1st and 3rd M1
The critical region for the test is {6, 7, 8, 9, 10, 11, 12} A1 for region 5
or {6 ≤ x ≤ 12} dep 3rd M1
(v) From tables the first value of n for which P(X ≤ 0) < 0.025 B1 for value of n = 17
seen
is n = 17 1
15
MEI Past Paper Questions S1 Binomial & Hypothesis Testing
Question 8 (S1 November 2004)
(i) [Let X ~ B(10, 0.18)]
M1 for 0.183 × 0.827
P(3 people left-handed) ⎛ 30 ⎞
M1 for “ ⎜ ⎟ × p3 × q7”
⎛10 ⎞ ⎝6⎠
⎜ ⎟ × 0.18 × 0.82 = 0.174 (to 3 s.f.)
3 7
= A1 cao
⎝3⎠
3
(ii) P(at least 2 people left-handed)
= 1 – P(0 or 1 people left-handed) M1 for
1 – ‘some binomial terms’
= 1 – [ P(X = 0) + P(X = 1) ]
M1 for ‘– P(X = 0)’
= 1 – [ 0.8210 – 10 × 0.18 × 0.829 ]
M1 for ‘– P(X = 1)’
= 0.561 (to 3 s.f.) 3
A1
(iii) [Let p represent the probability that a Welsh person chosen
at random is left-handed.]
H0: p = 0.18 B1 for H0
H1: p < 0.18 B1 for H1 2
(iv) [Let X ~ B(15, 0.18)]
P(X ≤ 0) = 0.8215 = 0.051 < 10% M1 for 0.051
P(X ≤ 1) = 0.8215 + 15 × 0.18 × 0.8214 M1 for tail comparison
with 10% - for either
= 0.219 > 10% P(X = 0) or P(X ≤ 1)
M1 for 0.219
The critical region for the test is {0} A1 dep. on previous M1 4
being correct
(v) There would be no critical region, E1
because 0.8210 = 0.137 > 10% E1 2
15
MEI Past Paper Questions S1 Binomial & Hypothesis Testing
Question 9 (S1 January 2005)
(i) [ Let X ~ B(18, 0.25)] M1 for P(X ≤ 6)
P(packet contains no more than 6 red Scruffies)
A1
= P(X ≤ 6) = 0.861 (to 3 s.f.) using tables 2
(ii) P(packet contains exactly 4 red Scruffies) = ⎛18 ⎞
M1 for ⎜ ⎟
⎜4⎟
= P(X = 4) ⎝ ⎠
⎛18 ⎞ M1 for 0.254 × 0.7514
= ⎜ ⎟ × 0.254 × 0.7514 = 0.213 (to 3 s.f.)
⎝4⎠ A1
or
or 0.5187 – 0.3057 using tables
M1 for correct table
= 0.213 (to 3 s.f.) M1 for correct subtraction 3
A1
(iii) [ Let X ~ B(18, 0.25) and Y ~ B(19, 0.25) ]
P(packet contains exactly 4 red Scruffies) =
M1 for 0.9 and 0.1
= 0.9 × P( X = 4) + 0.1 × P( Y = 4)
M1 for 0.213
= 0.9 × 0.213 + 0.1 × 0.2023
M1, A1 for 0.2023
= 0.212 (to 3 s.f.)
0.4654 – 0.2631 A1
using tables 5
(iv) [Let p represent the probability of selecting a red Scruffie.]
H0: p = 0.25 B1 for H0
H1: p > 0.25 B1 for H1
P(X ≥ 8) = 1 – P(X ≤ 7) = 1 – 0.9341 M1 for 1 – P(X ≤ 7)
= 0.0569 A1 for 0.0569
> 5% M1 for comparison 6
[Cannot reject H0]
There is not enough evidence to accept the hypothesis that E1 for context
the proportion of red Scruffies has increased.
16
Related docs
