Question

Consider the data labelled “problem 2”. It presents a random sample of students scores (in percentage) obtained by students in a standardized national test.

(i) Construct a 94% confidence interval on the fraction of students who scored above 70% on the test. (10)






(ii) Test the claim that more than 35% of the students score higher than 78% on the test. Write the hypotheses, identify test-statistics and compute its value.

84
69.2
31.9
85.5
47.8
88.8
52.1
78.2
78.3
79.5
69.5
42.4
59.8
48.4
53.7
46.7
76.7
34.4
69.5
56.8
75.9
80
61.4
78.6
62.7
45.7
61.9
73.3
75.6
83.5
74.5
38.2
38.4
52.2
96.7
59.2
69.4
32.9
98
66.3
99
94.6
90.1
28.2
72.8

Answer 1

Let x be the number of scores above 70%

So there are 20 scores in the given data set are above 70

Given : x = 20 , n = 45 , Therefore = x/n = 0.4444

i) 94% confidence interval on the fraction of students who scored above 70% on the test.

Confidence level (c) = 0.94

Therefore α = 1 - 0.94 = 0.06 , 1 - (α/2) = 0.9700

So we have to find z score corresponding to area 0.9700 on z score table

So z = 1.88

Confidence interval =

=

=

= 0.3051 and 0.5837

94% confidence interval on the fraction of students who scored above 70% on the test is ( 0.3051 , 0.5837 )

ii )   Claim : more than 35% of the students score higher than 70% on the test.

P = 0.35 vs P > 0.35

Given : x = 20, n = 45,   p = 0.35 , q = 1- p = 0.65 , = x/n = 0.4444

Test statistic:

Z = ;  

Z =

Z = 0.0944 / 0.0711

Z = 1.33

P-value :

As Ha contain > sign , this is right tail test,

P( z > 1.33 ) = 1- P( z < 1.33 ) = 1 - 0.9082 ----( from z score table )

P value = 0.0918

Assume Level of significance α = 0.05

Decision: As P-value is greater than α= 0.05, we fail to reject the null hypothesis H0

Conclusion: There is insufficient evidence that the more than 35% of the students score higher than 70% on the test.