Question

A high school principal estimates that the dropout rate for seniors at high schools in Maryland is 14%. Last year in a random sample of 350 Maryland seniors, 34 withdrew from school. At α = 0.05, is there enough evidence to reject the principal’s claim?

Answer 1

Here, we have to use one sample z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

H0: p = 0.14 versus Ha: p ≠ 0.14

This is a two tailed test.

We are given

Level of significance = α = 0.05

Test statistic formula for this test is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

x = number of items of interest = 34

n = sample size = 350

p̂ = x/n = 34/350 = 0.097142857

p = 0.14

q = 1 - p = 0.86

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.097142857 - 0.14)/sqrt(0.14*0.86/350)

Z = -2.3107

Test statistic = -2.3107

P-value = 0.0208

(by using z-table)

P-value < α = 0.05

So, we reject the null hypothesis

There is enough evidence to reject the principal’s claim.