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Date May Example question Marks available 6 Reference code EXM.1.SL.TZ0.8
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Test Question number 8 Adapted from N/A

Question

In an effort to study the level of intelligence of students entering college, a psychologist collected data from 4000 students who were given a standard test. The predictive norms for this particular test were computed from a very large population of scores having a normal distribution with mean 100 and standard deviation of 10. The psychologist wishes to determine whether the 4000 test scores he obtained also came from a normal distribution with mean 100 and standard deviation 10. He prepared the following table (expected frequencies are rounded to the nearest integer):

 

Copy and complete the table, showing how you arrived at your answers.

[5]
a.

Test the hypothesis at the 5% level of significance.

[6]
b.

Markscheme

To calculate expected frequencies, we multiply 4000 by the probability of each cell:

    p ( 80.5 X 90.5 ) = p ( 80.5 100 10 Z 90.5 100 10 )        (M1)

          = p ( 19.5 Z 0.95 )

           = 0.1711 0.0256

           = 0.1455

Therefore, the expected frequency  = 4000 × 0.1455        (M1)

           582        (A1)

Similarly:  p ( 90.5 X 100.5 ) = 0.5199 0.1711

           = 0.3488

    Frequency  = 4000 × 0.3488

           1396        (A1)

And  p ( 100.5 X 110.5 ) = 0.8531 0.5199

           = 0.3332

    Frequency  = 4000 × 0.3332

           1333        (A1)

[5 marks]

a.

To test the goodness of fit of the normal distribution, we use the χ 2  distribution. Since the last cell has an expected frequency less than 5, it is combined with the cell preceding it. There are therefore 7 – 1 = 6 degrees of freedom.          (C1)  

χ 2 = ( 20 6 ) 2 6 + ( 90 96 ) 2 96 + ( 575 582 ) 2 582 + ( 1282 1396 ) 2 1396 + ( 1450 1333 ) 2 1333 + ( 499 507 ) 2 507 + ( 84 80 ) 80 2             (M1)

= 53.03          (A1)          

H0: Distribution is Normal with μ = 100  and σ = 10 .

H1: Distribution is not Normal with μ = 100  and  σ = 10 .            (M1)

χ ( 0.95 , 6 ) 2 = 14.07

Since  χ 2 = 53.0 > χ critical 2 = 14.07 , we reject H0          (A1) 

Or use of p-value

Therefore, we have enough evidence to suggest that the normal distribution with mean 100 and standard deviation 10 does not fit the data well.          (R1) 

Note: If a candidate has not combined the last 2 cells, award (C0)(M1)(A0)(M1)(A1)(R1) (or as appropriate).

[6 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 » SL 4.11—Expected, observed, hypotheses, chi squared, gof, t-test
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