Question

Suppose that we want to check whether the average performance at a school is more than 60. The sample based information about the test scores is provided below. X-bar = 76, σ = 7, n = 11 In order to check whether the overall average score is over 60, (a). State the null hypothesis H0 and the alternative hypothesis H1 . (3 points) (b). What is the observed value for the test-statistic ? (4 points) (c ). What is the critical value at 5 % significance level ? (5 points) (d ). What will be your conclusion at 5 % significance level ? and why ? .

Answer 1

H0 :- µ = 60
H1 :- µ > 60

Test Statistic :-
Z = ( X - µ ) / ( σ / √(n))
Z = ( 76 - 60 ) / ( 7 / √( 11 ))
Z = 7.5809

Test Criteria :-
Reject null hypothesis if Z > Z(α)
Critical Value Z(α) = Z(0.05) = 1.64
Z > Z(α) = 7.5809 > 1.64
Result :- Reject null hypothesis

Decision based on P value
Reject null hypothesis if P value < α = 0.05 level of significance
P value = P ( Z < 7.5809 )
P value = 0
Since 0 < 0.05 ,hence we reject null hypothesis
Result :- Reject null hypothesis

There is sufficient evidence to support the claim that the average performance at a school is more than 60.