An educator estimates that the dropout rate for seniors at high schools in Istanbul is 15%. Last year, 38 seniors from a random sample of 200 Istanbul seniors withdrew. At α=0.05, is there enough evidence to reject the educator’s claim?

1. Formulate the hypotheses:

H_0 : p = 0.15

H_a : p \neq 0.15

2. Calculate the sample proportion:

\hat{p} = \frac{38}{200} = 0.19

3. Determine the sample size and significance level:

n = 200

\alpha = 0.05

4. Compute the test statistic:

z=\frac{\hat{p} - p_0}{\sqrt{\frac{p_0 \times(1 - p_0)}{n}}}=\frac{0.19 - 0.15}{\sqrt{\frac{0.15 \times(1 - 0.15)}{200}}}\approx1.58

5. Determine the p-value:

P(z)=2\times P(Z>1.58)\approx2\times0.05705=0.1141

Since the p-value is greater than the significance level (0.1141 > 0.05), we do not reject the null hypothesis.

Answer: There is not enough evidence to reject the educator’s claim.