P=z=p-value = In a certain school district, it was observed that 25% of the students in the element schools were classified asonly children (no siblings). However, in the special program for talented and gifted children, 128 out of 407students are only children. The school district administrators want to know if the proportion of only childrenin the special program is significantly different from the proportion for the school district. Test at the:0.01 level of significance.αWhat is the hypothesized population proportion for this test?(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 testtwo-tailed testChoose the correct pair of hypotheses for this situation:Но:р %3D 0.25 | Но:р 3 0.25 | Но :р %3D 0.25На :р < 0.25||На:р+ 0.25 |На:р > 0.25Но:р 3 0.314 |Но:р — 0.314 | Но:р — 0.314Ha:p < 0.314| Ha:p + 0.314 Ha:p > 0.314(A) (B) (C)(D) (E) (F)(A) (B) (C) (D) (E) (F)Using the normal approximation for the binomial distribution (without the continuity correction), was is thetest 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 =

It is given that 25% of the students in the element schools were classified as only children. Thus, the population proportion p is 0.25. The hypotheses are given below: Null hypothesis: H0:p=0.25. Alternative hypothesis: H1:p≠0.25. Since the alternative hypothesis has not equal sign, the test is two-tailed test. It is given that out of the 407 students sampled, 128 students are only children. Let p-cap be the sample proportion. The sample proportion, p-cap is, 0.314(=128/407). Here, the z-test for single proportion is used. The test statistic is given and calculated as follows.Computation of p-value: The p-value can be obtained by entering the formula “=2*(1-NORM.S.DIST(2.982,TRUE) ” in EXCEL. The p-value is 0.0029. Decision rule: If p-value ≤ α, then reject the null hypothesis. Otherwise, fail to reject the null hypothesis. Conclusion: The P-value is 0.0029 and the significance level is 0.01. Here, the P-value is less than the significance level. Hence, the null hypothesis is rejected. Thus, there is sufficient evidence to conclude that the proportion of only children in the special program is significantly different from the proportion for the district.p=0.25 z=2.982 p-value=0.0029