Question

A psychologist has developed an aptitude test which consists of a series of mathematical and vocabulary problems. They want to test the hypothesis that the mean test score is 64.6.

A random sample of 40 people have taken the test and their results recorded:

Download the data

69	45	62	71	67	72	68	62	69	64
81	57	95	85	56	60	76	82	61	55
75	48	93	74	63	98	74	52	72	81
44	79	57	54	44	69	55	91	52	88

You may find this Student's t distribution table useful throughout this question.

a)Calculate the test statistic (t) for the hypothesis test. Give your answer to 4 decimal places.

t =

b)A level of significance of α = 0.05 is to be used for the test. The P-value for this test statistic is less, or greater than the level of significance.

c)The result of this test is that the null hypothesis is rejected, or not rejected.

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	69	1
	45	529
	62	36
	71	9
	67	1
	72	16
	68	0
	62	36
	69	1
	64	16
	81	169
	57	121
	95.0	729
	85	289
	56	144
	60	64
	76	64
	82	196
	61	49
	55	169
	75	49
	48	400
	93	625
	74	36
	63	25
	98	900
	74	36
	52	256
	72	16
	81	169
	44	576
	79	121
	57	121
	54	196
	44	576
	69	1
	55	169
	91	529
	52	256
	88	400
Total	2720	8096

Mean X̅ = Σ Xi / n
X̅ = 2720 / 40 = 68
Sample Standard deviation S_X = √ ( (Xi - X̅ )2 / n - 1 )
S_X = √ ( 8096 / 40 -1 ) = 14.408

To Test :-
H0 :- µ = 64.6
H1 :- µ ≠ 64.6

Test Statistic :-
t = ( X̅ - µ ) / ( S / √(n))
t = ( 68 - 64.6 ) / ( 14.408 / √(40) )
t = 1.4925

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, n-1)
t(α/2, n-1) = t(0.05 /2, 40-1) = 2.023
| t | > t(α/2, n-1) = 1.4925 < 2.023
Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 1.4925 ) = 0.1436
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.1436 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis