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, 92 out of 246 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=0.02α=0.02 level of significance.

What is the hypothesized population proportion for this test?
p=p= 
(Report answer as a decimal accurate to 2 decimal places.  Do not report using the percent symbol.)

Based on the statement of this problem, how many tails would this hypothesis test have?

one-tailed test

two-tailed test

Choose the correct pair of hypotheses for this situation:
(A)(B)(C)H0:p=0.33H0:p=0.33
Ha:p<0.33Ha:p<0.33H0:p=0.33H0:p=0.33
Ha:p≠0.33Ha:p≠0.33H0:p=0.33H0:p=0.33
Ha:p>0.33Ha:p>0.33 (D)(E)(F)H0:p=0.374H0:p=0.374
Ha:p<0.374Ha:p<0.374H0:p=0.374H0:p=0.374
Ha:p≠0.374Ha:p≠0.374H0:p=0.374H0:p=0.374
Ha:p>0.374Ha:p>0.374

(A)

(B)

(C)

(D)

(E)

(F)

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=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.)

This P-value (and test statistic) leads to a decision to...

reject the null

accept the null

fail to reject the null

reject the alternative

As such, the final conclusion is that...

There is sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T program.

There is not sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T program.

The sample data support the assertion that there is a different proportion of only children in the G&T program.

There is not sufficient sample evidence to support the assertion that there is a different proportion of only children in the G&T program.

Answer 1

Population proportion is  P = X / n = 92/246 = 0.37

two-tailed test

To Test :-

H0:p=0.33

Ha:p≠0.33

Test Statistic :-
Z = ( P - P0 ) / ( √((P0 * q0)/n)
Z = ( 0.374 - 0.33 ) / ( √(( 0.33 * 0.67) /246))
Z = 1.467

P value = 2 * P ( Z > 1.4671 ) = 0.1423

Looking for the value Z = 1.467 in standard normal table and multiply that value by 2, since it is two tailed test.

Decision based on P value
P value = 2 * P ( Z > 1.4671 )
P value = 0.1423
Reject null hypothesis if P value < α = 0.02
Since P value = 0.1423 > 0.02, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0

fail to reject the null

There is sufficient evidence to warrant rejection of the assertion that there is a different proportion of only children in the G&T program.