A recent study at a local college claimed that the proportion, p, of students who commute more than fifteen miles to school is no more than 15%. If a random sample of 260students at this college is selected, and it is found that 42 commute more than fifteen miles to school, can we reject the college's claim at the 0.1 level of significance?
Perform a onetailed test. Then fill in the table below.
Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)

Answer)
Null hypothesis Ho : P <= 0.15
Alternate hypothesis Ha : P > 0.15
N = 260
First we need to check the conditions of normality that is if n*p and n*(1p) both are greater than 5 or not
N*p = 260*0.15 = 39
N*(1p) = 221
Both the conditions are met so we can use standard normal z table to estimate the PValue
Z test
Test statistics z = (oberved p  claimed p)/standard error
Standard error = √{claimed p*(1claimed p)/√n
Observed p = 42/260
Claimed p = 0.15
N = 260
Z = 0.52
From z table, P(z>0.52) = 0.3015
As the obtained pvalue is > 0.1 (given significance)
We fail to reject the null hypothesis
So, no we cannot reject the claim
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