n a certain school district, it was observed that 27% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 63 out of 184 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= Correct (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 Correct Choose the correct pair of hypotheses for this situation: (A) (B) (C) H0:p=0.27H0:p=0.27 Ha:p<0.27Ha:p<0.27 H0:p=0.27H0:p=0.27 Ha:p≠0.27Ha:p≠0.27 H0:p=0.27H0:p=0.27 Ha:p>0.27Ha:p>0.27 (D) (E) (F) H0:p=0.342H0:p=0.342 Ha:p<0.342Ha:p<0.342 H0:p=0.342H0:p=0.342 Ha:p≠0.342Ha:p≠0.342 H0:p=0.342H0:p=0.342 Ha:p>0.342Ha:p>0.342 (A) (B) (C) (D) (E) (F) Correct 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

n a certain school district, it was observed that 27% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 63 out of 184 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= Correct (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 Correct Choose the correct pair of hypotheses for this situation: (A) (B) (C) H0:p=0.27H0:p=0.27 Ha:p<0.27Ha:p<0.27 H0:p=0.27H0:p=0.27 Ha:p≠0.27Ha:p≠0.27 H0:p=0.27H0:p=0.27 Ha:p>0.27Ha:p>0.27 (D) (E) (F) H0:p=0.342H0:p=0.342 Ha:p<0.342Ha:p<0.342 H0:p=0.342H0:p=0.342 Ha:p≠0.342Ha:p≠0.342 H0:p=0.342H0:p=0.342 Ha:p>0.342Ha:p>0.342 (A) (B) (C) (D) (E) (F) Correct 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

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Question

In a certain school district, it was observed that 27% of the students in the element schools were classified as only children (no siblings). However, in the special program for talented and gifted children, 63 out of 184 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= Correct
(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

Correct

Choose the correct pair of hypotheses for this situation:

(A)	(B)	(C)	H0:p=0.27H0:p=0.27 Ha:p<0.27Ha:p<0.27	H0:p=0.27H0:p=0.27 Ha:p≠0.27Ha:p≠0.27	H0:p=0.27H0:p=0.27 Ha:p>0.27Ha:p>0.27		(D)	(E)	(F)	H0:p=0.342H0:p=0.342 Ha:p<0.342Ha:p<0.342	H0:p=0.342H0:p=0.342 Ha:p≠0.342Ha:p≠0.342	H0:p=0.342H0:p=0.342 Ha:p>0.342Ha:p>0.342

(A)

(B)

(C)

(D)

(E)

(F)

Correct

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

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