Question

If the sample proportion of first-year students who suffer from depression is 328/450

1- Find and interpret the 95% condence interval for the proportion of students who are suffering from depression. Is there evidence to suggest that more than half

of the population of first-year students suffer from depression? Explain.

2- Give an explanation for your conclusion in part 1.

Please show steps. No steps, no rate.

Answer 1

Part a)

p̂ = X / n = 328/450 = 0.7289
p̂ ± Z(α/2) √( (p * q) / n)
0.7289 ± Z(0.05/2) √( (0.7289 * 0.2711) / 450)
Z(α/2) = Z(0.05/2) = 1.96
Lower Limit = 0.7289 - Z(0.05) √( (0.7289 * 0.2711) / 450) = 0.6878
upper Limit = 0.7289 + Z(0.05) √( (0.7289 * 0.2711) / 450) = 0.77
95% Confidence interval is ( 0.6878 , 0.77 )
( 0.6878 < P < 0.77 )

We are 95% confident that the true population proportion of students who are suffering from depression lies within the interval ( 0.6878 , 0.77 ).

Since the values in the confidence interval is positive and greater than 0.5, we can conclude that more than half of the population of first-year students suffer from depression

Part 2)

To Test :-
H0 :- P = 0.5
H1 :- P > 0.5

P = X / n = 328/450 = 0.7289
Test Statistic :-
Z = ( P - P0 ) / √ ((P0 * q0)/n))
Z = ( 0.728889 - 0.5 ) / √(( 0.5 * 0.5) /450))
Z = 9.71

Test Criteria :-
Reject null hypothesis if Z > Z(α)
Z(α) = Z(0.05) = 1.645
Z > Z(α) = 9.71 > 1.645, hence we reject the null hypothesis
Conclusion :- We Reject H0

There is sufficient evidence to support the claim that more than half of the population of first-year students suffer from depression.