Question

In a sample of 1000 students, it was found that they consumed on average 23 grams of sugar with a standard deviation of 4 grams and that 230 planned to transfer.

a) Can we conclude that with 99% confidence that students consume on average over 20 grams of sugar? Use the rejection region approach.

b) Can we conclude that with 99% confidence that under 25% of students plan to transfer? Use the P-value approch.

Answer 1

Part a)

To Test :-

H0 :- µ = 20
H1 :- µ > 20

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 23 - 20 ) / ( 4 / √(1000) )
t = 23.7171

Test Criteria :-
Reject null hypothesis if t > t(α, n-1)
Critical value t(α, n-1) = t(0.01 , 1000-1) = 2.33
t > t(α, n-1) = 23.7171 > 2.33
Result :- Reject null hypothesis

There is sufficient evidence to support the claim that students consume on average over 20 grams of sugar.

Part b)

To Test :-

H0 :- P = 0.25
H1 :- P < 0.25

P = X / n = 230/1000 = 0.23

Test Statistic :-
Z = ( P - P0) / √(P0 * q0 / n)
Z = ( 0.23 - 0.25 ) / √(( 0.25 * 0.75) /1000)
Z = -1.461

Decision based on P value
P value = P ( Z < -1.4606 ) = 0.0721
Reject null hypothesis if P value < α = 0.01
Since P value = 0.0721 > 0.01, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

There is insufficient evidence to support the claim that under 25% of students plan to transfer.