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.

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, 114 out of 386 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=0.05
level of significance.
What is the hypothesized...

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, 78 out of 215 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=0.01α=0.01 level of significance.
What is the hypothesized...

In a certain school district, it was observed that 26% of the
students in the element schools were classified as only children
(no siblings). However, in the special program for talented and
gifted children, 138 out of 425 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=0.01α=0.01 level of significance.
What is the hypothesized...

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, 106 out of 289 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.
What is the hypothesized...

In a certain school district, it was observed that 29% of the
students in the element schools were classified as only children
(no siblings). However, in the special program for talented and
gifted children, 91 out of 257 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...

In a certain school district, it was observed that 26% of the
students in the element schools were classified as only children
(no siblings). However, in the special program for talented and
gifted children, 93 out of 313 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 = 0.05
level of significance.
What is...

in a certain school district, it was observed that 28% of the
students in the element schools were classified as only children
(no siblings). However, in the special program for talented and
gifted children, 98 out of 272 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 = 0.02
level of significance. What is...

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...

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...

7.
You wish to test the following at a significance level of
α=0.05α=0.05.
H0:p=0.85H0:p=0.85
H1:p>0.85H1:p>0.85
You obtain a sample of size n=250n=250 in which there are 225
successful observations.
For this test, we use the normal distribution as an approximation
for the binomial distribution.
For this sample...
The test statistic (zz) for the data = (Please show
your answer to three decimal places.)
The p-value for the sample = (Please show your
answer to four decimal places.)
The p-value is...
greater than...

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