Question

Note: you can copy any of the following symbols from here: p o o2 S It has been reported that the probability that an individual will develop schizophrenia over their lifetime is 0.004. Medical researches selected a random sample of 3000 individuals and determined that 17 of them developed schizophrenia. Is there evidence to support the claim that the true proportion of people who will develop schizophrenia is different from 0.004 at the a = 0.05 level of significance? (Source: Saha S, Chant D, Welham J, McGrath J. A systematic review of the prevalence of schizophrenia. PLoS Med 2(5): e141.) a. Determine the null and alternative hypotheses: Ho: H1: 5. What type of test is being conducted in this problem? c. What parameter is being tested? d. Explain what it would mean to make a Type I error. e. Explain what it would mean to make a Type II error. Suppose the null hypothesis is rejected. State the conclusion based on the results of the test.

Answer 1

1. a) Null and Alternative hypothesis:

Ho : p = 0.004

H1 : p ≠ 0.004

b) we will perform a one sample z proportion test.

c) Parameter being tested

n = 3000, x = 17

p̄ = x/n = 0.0057

d) Type I error is rejecting the null hypothesis when it is true.

In this case, Type I error would be to reject the null that proportion of people who develop schizophrenia is 0.04, when in reality it is 0.004.

e) Type II error is failing to reject the null hypothesis when it is not true.

In this case, Type II error would be to fail to reject the null that proportion of people who develop schizophrenia is 0.04, when in reality it is different from 0.004.

f) if null hypothesis is reject then conclusion will be:

There is enough evidence to conclude that proportion of people who develop schizophrenia is different from 0.004 at 0.05 significance level.

g) Test statistic:

z = (p̄ -p)/√(p*(1-p)/n) = (0.0057 - 0.004)/√(0.004 * 0.996/3000) = 1.4463

h) p-value using excel:

p-value = 2*(1-NORM.S.DIST(ABS(1.4463), 1)) = 0.1481

i) conclusion:

p-value > α, Do not reject the null hypothesis.

There is not enough evidence to conclude that proportion of people who develop schizophrenia is different from 0.004 at 0.05 significance level.

j) 95% Confidence interval :

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = p̄ - z_c*√( p̄ *(1- p̄ )/n) = 0.0057 - 1.96 *√(0.0057*0.9943/3000) = 0.0030

Upper Bound = p̄ + z_c*√( p̄ *(1- p̄ )/n) = 0.0057 + 1.96 *√(0.0057*0.9943/3000) = 0.0084

Yes, 0.004 is within the interval.