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Date May Example question Marks available 10 Reference code EXM.2.AHL.TZ0.26
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Test Question number 26 Adapted from N/A

Question

Scientists have developed a type of corn whose protein quality may help chickens gain weight faster than the present type used. To test this new type, 20 one-day-old chicks were fed a ration that contained the new corn while another control group of 20 chicks was fed the ordinary corn. The data below gives the weight gains in grams, for each group after three weeks.

The scientists wish to investigate the claim that Group B gain weight faster than Group A. Test this claim at the 5% level of significance, noting which hypothesis test you are using. You may assume that the weight gain for each group is normally distributed, with the same variance, and independent from each other.

[6]
a.

The data from the two samples above are combined to form a single set of data. The following frequency table gives the observed frequencies for the combined sample. The data has been divided into five intervals.

Test, at the 5% level, whether the combined data can be considered to be a sample from a normal population with a mean of 380.

[10]
b.

Markscheme

This is a t-test of the difference of two means. Our assumptions are that the two populations are approximately normal, samples are random, and they are independent from each other.          (R1)

H0: μ1μ2 = 0

H1: μ1 − μ2 < 0          (A1)                

t = −2.460,          (A1)

degrees of freedom = 38          (A1)

Since the value of critical t = −1.686 we reject H0.          (A1)

Hence group B grows faster.          (R1)

[6 marks]

a.

This is a χ 2  goodness-of-fit test.

To finish the table, the frequencies of the respective cells have to be calculated. Since the standard deviation is not given, it has to be estimated using the data itself. s = 49.59, eg the third expected frequency is 40 × 0.308 = 12.32, since P(350.5 < W < 390.5) = 0.3078...

The table of observed and expected frequencies is:

      (M1)(A2)

Since the first expected frequency is 3.22, we combine the two cells, so that the first two rows become one row, that is,

      (M1)

Number of degrees of freedom is 4 – 1 – 1 = 2           (C1)          

H0: The distribution is normal with mean 380

H1: The distribution is not normal with mean 380         (A1)

The test statistic is

χ c a l c 2 = ( f e f 0 ) 2 f e = ( 11 11.04 ) 2 11.04 + ( 8 12.32 ) 2 12.32 + ( 15 10.48 ) 2 10.48 + ( 6 6.17 ) 2 6.17

= 3.469          (A1)

With 2 degrees of freedom, the critical number is χ 2  = 5.99           (A2)

So, we do not have enough evidence to reject the null hypothesis. Therefore, there is no evidence to say that the distribution is not normal with mean 380.           (R1)

[10 marks]

b.

Examiners report

[N/A]
a.
[N/A]
b.

Syllabus sections

Topic 4—Statistics and probability » SL 4.9—Normal distribution and calculations
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Topic 4—Statistics and probability

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