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, 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 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:

(A) | (B) | (C) | H0:p=0.31H0:p=0.31 Ha:p<0.31Ha:p<0.31 |
H0:p=0.31H0:p=0.31 Ha:p≠0.31Ha:p≠0.31 |
H0:p=0.31H0:p=0.31 Ha:p>0.31Ha:p>0.31 |
(D) | (E) | (F) | H0:p=0.363H0:p=0.363 Ha:p<0.363Ha:p<0.363 |
H0:p=0.363H0:p=0.363 Ha:p≠0.363Ha:p≠0.363 |
H0:p=0.363H0:p=0.363 Ha:p>0.363Ha:p>0.363 |
---|

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
- 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.

**PLEASE SHOW WORK! THANK YOU!!**

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

n = 215

x = 78

= x / n = 78 /215 = 0.36

P_{0} = 0.31

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

Test statistic = z

=
- P_{0} / [P_{0
*} (1 - P_{0} ) / n]

= 0.36 - 0.31/ [0.31 *0.69 / 215 ]

= 1.674

Test statistic = z = 1.674

P-value = 0.0942

= 0.01

P-value ≥

0.0942 ≥ 0.01

Fail to reject the null hypothesis .

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

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 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 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...

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...

You wish to test the following claim (HaHa) at a significance
level of ?=0.002.
Ho:p=0.18
Ha:p?0.18
You obtain a sample of size n=435 in which there are 73 successful
observations. 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 three decimal places.)
test statistic =
What is the p-value for this sample? (Report...

You wish to test the following claim (HaHa) at a significance
level of α=0.02α0.02.
Ho:p=0.22Hop0.22
Ha:p<0.22Hap0.22
You obtain a sample of size n=167n167 in which there are 22
successful observations. 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 three decimal places.)
test statistic =
What is the p-value for this sample? (Report...

you wish to test the following claim (HaHa) at a significance
level of α=0.01α=0.01.
Ho:p=0.14Ho:p=0.14
Ha:p<0.14Ha:p<0.14
You obtain a sample of size n=709n=709 in which there are 94
successful observations. 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 three decimal places.)
test statistic =
What is the p-value for this sample? (Report...

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