binomial distribution

Question

In a certain school district, it was observed that 27% of the students in the element schools were classified as only children (no siblings).  However, in the special program for talented and gifted children, 79 out of 226 students are only children.  The school district administrators want to know if the proportion of only children in the special program is significantly different from the proportion for the school district. Test at the  level of significance.

What is the hypothesized population proportion for this test?

 

(Report answer as a decimal accurate to 2 decimal places.  Do not report using the percent symbol.)

Using the normal approximation for the binomial distribution (without the continuity correction), what is the test statistic for this sample based on the sample proportion?

 

(Report answer as a decimal accurate to 3 decimal places.)

You are now ready to calculate the P-value for this sample.

P-value = 

(Report answer as a decimal accurate to 4 decimal places.)

Answer 1

Hypothesized proportion = 0.27

The sample proportion is 79/226 = 0.35

The z-statistic is (0.35 - 0.27)/(sqrt(0.27*0.73/226) = 2.709

This is a two-sided test so p-value = P(Z > 2.709) + P(Z < -2.709). Since the normal distribution is symmetric, this is equal to 2 * P(Z < 2.709) = 2 * 0.0034 = 0.0068

Since the p-value is less than 0.02, we would reject the null hypothesis and there is sufficient evidence that the proportion of only children differs from 0.27.