Answer:
We conclude that the proportion of only children in the special program is significantly different from the proportion for the school district.
Explanation:
We are given that in a certain school district, it was observed that 24% of the students in the element schools were classified as only children
However, in the special program for talented and gifted children, 119 out of 415 students are only children.
Let p = proportion of only children in the special program.
So, Null Hypothesis,
: p = 24% {means that the proportion of only children in the special program is equal to the proportion for the school district}
Alternate Hypothesis,
: p
24% {means that the proportion of only children in the special program is significantly different from the proportion for the school district}
The test statistics that would be used here One-sample z-test for proportions;
T.S. =
~ N(0,1)
where,
= sample proportion of only children in the special program =
= 0.29
n = sample of students = 415
The hypothesized population proportion for this test is 0.24.
So, the test statistics =
= 2.385
The value of z test statistic is 2.385.
Now, at 0.05 significance level the z table gives critical value of -1.96 and 1.96 for two-tailed test.
Since our test statistic doesn't lie within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that the proportion of only children in the special program is significantly different from the proportion for the school district.