A college claims that the proportion, p, of students who commute more than fifteen miles to school is less than 25%. A researcher wants to test this. A randomsample of 270 students at this college is selected, and it is found that 53 commute more than fifteen miles to school. Is there enough evidence to support thecollege's claim at the 0.10 level of significance?Perform a one-tailed test. Then complete the parts below.Carry your intermediate computations to three or more decimal places. (If necessary, consult a list of formulas.)(a) State the null hypothesis H and the alternative hypothesis H₁.H :DH₁ :0(b) Determine the type of test statistic to use.(Choose one) ▼(c) Find the value of the test statistic. (Round to three or more decimal places.)0(d) Find the critical value. (Round to three or more decimal places.)0(e) Is there enough evidence to support the claim that the proportion of studentswho commute more than fifteen miles to school is less than 25%?OYes NoHX0⁰0*0Oローロ OSOXSOOEspanol? 圖 □ □ □區Hامه

a) first of all we have to construct null and alternative hypotheses as follows we know that null hypothesis always contain sign ≤, ≥,= and alternative hypothesis always contain sign <,>,≠ the alternative hypothesis contains opposite sign to null hypothesis Null and alternative hypothesis can be stated as follows :- H0: p ≥ 0.25 H1:p<0.25 (claim) b) z test for signle population propotrtion using normal distribution . c) Next step is to find test statistic, formula for test statistics test statistic z=(p^−p)/√(pq) /n Where p^=x/n where x are favorable outcomes corresponding to sample data n is total sample size taken p^ = x n p^ = 53 270 p^ = 0.19629 Calculate our test statistic z: z = p^ - p √p(1 - p)/n z = 0.19629 - 0.25 √0.25(1 - 0.25)/270 z = -0.0537037 √0.25(0.75)/270 z = -0.05370 √0.000694 z = -0.05370 0.02635 z = -2.0379 d) Rule :- if alternative hypothesis contain sign <,> ,then test is left and right tailed respectively. Rejection region will be in left and right sides respectivelyif alternative hypothesis contain sign ≠ then test is called two tailed test. Rejection region will be in both tails. test is left-tailed . It means rejection region will be left side, we know that area of rejection region must be equal to level of significance. From normal table we have P( Z<-1.282) = 0.10 hence critical value is -1.282 e) RULE:- if absolute value of test statistic is greater than absolute value of critical value then we have to reject null hypothesis. so we reject null hypothesis.in Other words we can say that there is sufficient evidence to conclude that result is significant.or we can conclude that there is sufficient evidence to support alternative hypothesisor claim. yes.