Question

In a certain school district, it was observed that 31% of the students in the element...

In a certain school district, it was observed that 31% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 77 out of 204 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 α=0.05α=0.05 level of significance.

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?
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.)

Homework Answers

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H0 : p = 0.31

Ha : p 0.31

= x / n = 77/204 = 0.3775

P0 = 0.31

1 - P0 = 1-0.31 =0.69

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.3775- 0.31/ [0.31*(0.69) /204 ]

= 2.083

P(z >2.083 ) = 1 - P(z < 2.083) = 0.0372

P-value = 0.0372

= 0.05    

p= 0.0372 < 0.05, it is concluded that the null hypothesis is rejected.

Reject the null hypothesis .

There is enough evidence to claim that the population proportion p is different than p0​, at the α = 0.05 significance level.

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