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 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=0.29H0:p=0.29 Ha:p<0.29Ha:p<0.29 |
H0:p=0.29H0:p=0.29 Ha:p≠0.29Ha:p≠0.29 |
H0:p=0.29H0:p=0.29 Ha:p>0.29Ha:p>0.29 |
(D) | (E) | (F) |
H0:p=0.354H0:p=0.354 Ha:p<0.354Ha:p<0.354 |
H0:p=0.354H0:p=0.354 Ha:p≠0.354Ha:p≠0.354 |
H0:p=0.354H0:p=0.354 Ha:p>0.354Ha:p>0.354 |
(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.
Solution :
This is the two tailed test .
One sample proportion test
Two tailed test .
The null and alternative hypothesis is ,
H0 : p = 0.29
Ha : p 0.29
n = 257
x = 91
= x / n = 91 / 257 = 0.3541
P0 = 0.29
1 - P0 = 1 - 0.29 = 0.71
z = - P0 / [P0 * (1 - P0 ) / n]
= 0.3541 - 0.29 / [(0.29 * 0.71) / 257]
= 2.264
P(z > 2.264) = 1 - P(z < 2.264) = 1 - 0.9882 = 0.0118
P-value = 0.0236
= 0.02
P-value >
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
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