Question

2. A professor was curious as to whether the students in a very large class she was teaching, who turned in their tests first, scored differently from the class mean on the test. The μ on the test was 75 with σ = 10; the scores were approximately normally distributed. The mean score for the first 20 tests was 78. Did the students turning in their tests first score significantly different from the larger mean at the .05 level?

   a) Use the four steps of hypothesis testing b) Illustrate the distributions involved

   c) Calculate the 95% confidence interval (*even if ns result)

Answer 1

Part a)

To Test :-
H0 :- µ = 75
H1 :- µ ≠ 75

Part b)

We will use Z test, because σ is known.

Test Statistic :-
Z = ( X̅ - µ ) / ( σ / √(n))
Z = ( 78 - 75 ) / ( 10 / √( 20 ))
Z = 1.3416

Test Criteria :-
Reject null hypothesis if | Z | > Z( α/2 )
Critical value Z(α/2) = Z( 0.05 /2 ) = 1.96
| Z | > Z( α/2 ) = 1.3416 < 1.96
Result :- Fail to reject null hypothesis

Decision based on P value
Reject null hypothesis if P value < α = 0.05 level of significance
P value = 2 * P ( Z > 1.3416 ) = 2 * 1 - P ( Z < 1.3416 )
P value = 0.1797
Since 0.1797 > 0.05 ,hence we fail to reject null hypothesis
Result :- We fail to reject null hypothesis

There is insufficient evidence to support the claim that students turning in their tests first score significantly different from the larger mean at the .05 level of significance.

Part c)

Confidence Interval :-
X̅ ± Z( α /2) σ / √ ( n )
Z(α/2) = Z (0.05 /2) = 1.96
78 ± Z (0.05/2 ) * 10/√(20)
Lower Limit = 78 - Z(0.05/2) 10/√(20)
Lower Limit = 73.6173
Upper Limit = 78 + Z(0.05/2) 10/√(20)
Upper Limit = 82.3827
95% Confidence interval is ( 73.6173 , 82.3827 )