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 80.

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

Download the data

85 73 73 65 63 68 65 79 92 54
65 40 62 60 54 65 85 56 58 58
61 69 76 95 96 78 41 87 66 74
70 85 91 54 68 85 70 43 51 44
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)At a level of significance of ? = 0.05, the result of this test is that the null hypothesis is rejectednot rejected.

Answer 1

(a) Test statistic (t) for the hypothesis test is -5.0498

(b) Null hypothesis is rejected .

Explanation:

We are given that 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 80, i.e.;

Null Hypothesis,
`H_0` :
`\mu` = 80

Alternate Hypothesis,
`H_1` :
`\mu\\eq` 80

We are given the sample data of 40 people, i.e. n = 40 ;

Sample mean, X bar =
`(\sum X)/(n)` =
`(2724)/(40)` = 68.1

Sample standard deviation, s =
`\sqrt{(\sum (X-Xbar)^(2) )/(n-1) }` = 14.904

The test statistics we will use here is;

T.S. =
`(Xbar-\mu)/((s)/(√(n) ) )` ~
`t_n_-_1`

=
`(68.1-80)/((14.904)/(√(40) ) )` ~
`t_3_9`

= -5.0498

So, test statistic (t) for the hypothesis test is -5.0498 .

(b) Now, at level of significance of 0.05, t table gives critical value of between [-2.0225 , 2.0225] . Since our test statistics does not lie in the range of critical value means it lies in the rejection region so we have sufficient evidence to reject null Hypothesis.

Therefore, null hypothesis is rejected .