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, 134 out of 385 students are hildren. 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 a=0.02 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 Hp0.31 Hop 0.31 HP 0.31 (A) (B) CHAD 0.31 HP 0.31 HP0. 31 (D) (E) Hp 0.348 HP 0.348 HP 0.348 P 0.348 HP0.348 HP 0.348 H (A) (B) (C) (D) (E) (F)

Answer 1

Step 1: To, test the population proportion of only children in the special program is significantly different from the proportion for the school district at the 2% level of significance. The null and alternative hypotheses are: H.: p=0.31 (The population proportion is equal to 0.31) versus 4, p > 0.31 The population proportion is greater than 0.31). Step2: The summary statistics: The sample proportion is po = 0.348052. The level of significance is a =0.02. --- ------ ------ The sample proportion is - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - _134 - - - - - - - - - - = 0.348052 px 0.348052 ----------------- Step3: The test statistic is given by 2=_p-pe Po(1-P) 0.348052-0.31 10.31(1-0.31) V 385 0.038051948 10.31(0.69) V 385 0.038051948 10.2139 V 385 0.038051948 70.000555584 0.038051948 0.023570838 = 1.614365499 zo - 1.614365499 --- ------ -------- -------- ------------ -------- Step 4: The P-value approach: P-value = 2P(2>z) = 2(1-P(z<z)) = 2(1-P(z<1.614365499)) = 2(1–0.946775915) = 2(0.053224085) = 0.106448171 P-value = 0.106448171 > a = 0.02). We do not reject the null hypothesis H, - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Step 5: Conclusion : Using, the z- test statistic is z = 1.6144 <20.02 = 2.326348 We do not reject the null hypothesis H. Using, the P-value approach, Because the P-value = 0.1064 > a = 0.02). We do not reject the null hypothesis H, There is not sufficient evidence to conclude that the population proportion of only children in the special program is significantly different from the proportion for the school district at the 2% level of significance.