In a certain school district, it was observed that 32% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 106 out of 278 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.)
In a certain school district, it was observed that 32% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 106 out of 278 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=0.32H0:p=0.32 Ha:p<0.32Ha:p<0.32 |
H0:p=0.32H0:p=0.32 Ha:p≠0.32Ha:p≠0.32 |
H0:p=0.32H0:p=0.32 Ha:p>0.32Ha:p>0.32 |
(D) | (E) | (F) | H0:p=0.381H0:p=0.381 Ha:p<0.381Ha:p<0.381 |
H0:p=0.381H0:p=0.381 Ha:p≠0.381Ha:p≠0.381 |
H0:p=0.381H0:p=0.381 Ha:p>0.381Ha:p>0.381 |
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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=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.
Given data
number of success , x= 106
sample size, n =278
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