Question

In a certain school district, it was observed that 33% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 107 out of 287 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 ? = 0.05 level of significance.

H0:p=0.33
Ha:p?0.33

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

z =_______(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

Null Hypothesis H0: p= 0.33

Alternative Hypothesis Ha: p ? 0.33

The proportion of only child ( no siblings) is

Under H0, the test statistic using the normal approximation for the binomial distribution is

The P-Value is 0.1228

Since p value is greater than significance level, Fail to Reject H0.

Hence, at 5% level of significance, we have enough evidence to conclude that 33% of the students in the element schools were classified as only children (no siblings).