Practise formulating hypotheses, selecting one- or two-tailed tests, calculating critical regions or p-values and stating contextual conclusions with Type I or II errors.
A researcher read a magazine article which stated that boys aged 1 to 3 prefer green to orange. It claimed that, when offered a green cube and an orange cube to play with, a boy is more likely to choose the green one.
The researcher disagrees with this claim. She believes that boys of this age are equally likely to choose either colour. In order to test her belief, the researcher carried out a hypothesis test at the 5\% significance level. She offered a green cube and an orange cube to each of 10 randomly chosen boys aged 1 to 3, and recorded the number, X, of boys who chose the green cube.
Out of the 10 boys, 8 boys chose the green cube.
Assuming that the researcher's belief that either colour cube is equally likely to be chosen is valid, a student correctly calculates that P(X=8)=0.0439, correct to 3 significant figures. He says that, because this value is less than 0.05, the null hypothesis should be rejected.
Explain why this statement is incorrect.
Need to find P(X >=slant 8)
oe (e.g. invalid because it should be a tail probability
compared with 0.05).
Carry out the test on the researcher's claim that either colour cube is equally likely to be chosen.
H_0: P( green )=0.5
H_1: P( green )>0.5
Allow p=0.5.
Allow p>0.5.
P(X >=slant 8)=0.0439+ ^10 C_9 x (0.5) x (0.5)^9+0.5^10
Attempt 0.0439+P(X=9)+P(X=10).
Must see Binomial expressions B(10,0.5).
=0.0547 or 0.0546(3 sf)
SC B1 0.0547 or 0.0546 with no working.
0.0547 > 0.05
Valid comparison of tail probability with 0.05.
[Do not reject H_0 ] 'There is insufficient evidence [at the 5\% level] to accept
the hypothesis that boys prefer green.’
Or 'There is sufficient evidence to support the researcher's claim.'
A1FT
In context, not definite. No contradictions.
Allow 'There is insufficient evidence to reject the
hypothesis that boys like green and orange equally’.
Not definite, e.g. not 'They don't prefer green" or 'Researchers claim true' 'Magazine's claim untrue'.
Any mention of 'claim' must be clear which claim it is.
Another researcher claims that a Type I error was made in carrying out the test.
Explain why this cannot be true.
A similar test, at the 5\% significance level, was carried out later using 10 other randomly chosen boys aged 1 to 3.
H_0 was not rejected
Mark independently.
Find the probability of a Type I error.
^10 C_9 x (0.5) x (0.5)^9+0.5^10 or " 0.0547 " -0.0439
Finding P(9,10) using B(10,0.5).
Could be seen in (a)(ii).
P( Type I error )=11 / 1024 or 0.0107(3 sf) or 0.0108
In this question you should not use an approximating distribution.
At an election in Menham last year, 24 % of voters supported the Today Party. A student wishes to test whether support for the Today Party has decreased since last year. He chooses a random sample of 25 voters in Menham and finds that exactly 2 of them say that they support the Today Party.
Test at the 5\% significance level whether support for the Today Party has decreased.
H0:p=0.24H1:p<0.24P(X≤2)=0.7625+25×0.7624×0.24+25C2×0.7623×0.242 or 0.0010479+0.0082732+0.0313513
Expression must be seen. No end errors.
= 0.0407
SC B1 for unsupported 0.0407.
0.0407 < 0.05
For valid comparison.
[Evidence to reject H0.] There is sufficient evidence to suggest that the support for the Today Party has decreased.
A1FT
FT their probability. In context, not definite, no contradictions. SC: if H1:p=0.24 and compare with 0.025 ;
max B0 M1 A1 M1 A0.
5
The numbers of green sweets in 200 randomly chosen packets of Frutos are summarised in the table.
| Number of green sweets | 0 | 1 | 2 | 3 | >3 |
|---|---|---|---|---|---|
| Number of packets | 32 | 50 | 97 | 21 | 0 |
The manufacturers of Frutos claim that the mean number of green sweets in a packet is 1.65 .
Anji believes that the true value of the mean, μ, is less than 1.65 . She uses the results from the 200 randomly chosen packets to test the manufacturers' claim.
State suitable null and alternative hypotheses for the test.
H0:μ=1.65H1:μ<1.65
Marking guidance:
Accept 'population mean' but not just 'mean'.
Show that the result of Anji's test is significant at the 5 % level but not at the 1 % level.
0.783÷200′1.535′−1.65
Standardising with their mean.
=-1.838 or -1.84
A1*
Φ(0.05) and Φ(0.01) attempted
Or P(z<−ˋ1.838ˋ) attempted.
SC: Condone Φ(0.025)=2.807 and Φ(0.005)=3.291 following two-tailed test in (b).
-1.645>-1.838>-2.326
[Hence significant at 5\% but not 1\% level]
AG
=0.033 and 0.05>0.033>0.01
SC: use of 1.54 or 1.53 for the mean leading to -1.645>-1.758>- 2.326
or -1.645>-1.918>-2.326 or 0.95 < 0.9606 or 0.9724 < 0.99
scores M1 M1 A1.
Marking guidance:
Accept use of critical value method
1.535 < 1.547
or accept 1.65>1.638.
It is given that Anji made a Type I error.
Explain how this shows that the significance level that Anji used in her test was not 1 %.
At the 1 % level H0 is not rejected Or a Type I error can only occur if H0 is rejected.
OE