Question

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.)*

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H_{0} : p = 0.31

H_{a} : p
0.31

= x / n = 77/204 = 0.3775

P_{0} = 0.31

1 - P_{0} = 1-0.31 =0.69

Test statistic = z

=
- P_{0} / [P_{0
*} (1 - P_{0} ) / 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 p_{0}, at the α = 0.05 significance
level.

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, 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 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 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...

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 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 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 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...

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...

You wish to test the following claim (HaHa) at a significance
level of α=0.05α=0.05.
Ho:p1=p2Ho:p1=p2
Ha:p1<p2Ha:p1<p2
You obtain 92.6% successes in a sample of size n1=759n1=759 from
the first population. You obtain 96.3% successes in a sample of
size n2=269n2=269 from the second population. For this test, you
should NOT use the continuity correction, and you should use the
normal distribution as an approximation for the binomial
distribution.
What is the test statistic for this sample? (Report answer accurate
to...

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 3 minutes ago

asked 6 minutes ago

asked 17 minutes ago

asked 50 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 3 hours ago

asked 3 hours ago