In a certain school district, it was observed that 25% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 89 out of 269 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 a = 0.02 level of significance. What is the hypothesized population proportion for this test? 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: Но: р — 0.25 | Но:р — 0.25 | Но:? На:р < 0.25 | На:р# 0.25 | На:? (А) | (B) | (C) (A) (B) (C) (D) (E) (F) 30

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06:22
A myopenmath.com
In a certain school district, it was observed that 25% of
the students in the element schools were classified as
only children (no siblings). However, in the special
program for talented and gifted children, 89 out of 269
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
a = 0.02 level of significance.
What is the hypothesized population proportion for this
test?
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:
Но: р — 0.25| Но:р — 0.25 | Но:р
На:р < 0.25 | На:р+0.25 | На:p
(А) | (B) | (С)
(A) (B)
(D) (E) (F)
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.)
This P-value (and test statistic) leads to a decision to...
reject the null
accept the null
fail to reject the null
O 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.
O 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
Transcribed Image Text:06:22 A myopenmath.com In a certain school district, it was observed that 25% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 89 out of 269 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 a = 0.02 level of significance. What is the hypothesized population proportion for this test? 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: Но: р — 0.25| Но:р — 0.25 | Но:р На:р < 0.25 | На:р+0.25 | На:p (А) | (B) | (С) (A) (B) (D) (E) (F) 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.) This P-value (and test statistic) leads to a decision to... reject the null accept the null fail to reject the null O 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. O 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
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