Question

A NY Times article claimed that working adults save an average of $300 per paycheck with a standard deviation of $25. A random sample of 25 students revealed an average savings of $290 per paycheck.

A hypothesis test is conducted using a 15% significance level to determine whether students on average save less than working adults. What is the p-value for this hypothesis test?

Express your answer rounded to four decimal places.

Answer 1

Given that,
population mean(u)=300
standard deviation, σ =25
sample mean, x =290
number (n)=25
null, Ho: μ=300
alternate, H1: μ<300
level of significance, α = 0.15
from standard normal table,left tailed z α/2 =1.036
since our test is left-tailed
reject Ho, if zo < -1.036
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 290-300/(25/sqrt(25)
zo = -2
| zo | = 2
critical value
the value of |z α| at los 15% is 1.036
we got |zo| =2 & | z α | = 1.036
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : left tail - ha : ( p < -2 ) = 0.023
hence value of p0.15 > 0.023, here we reject Ho
ANSWERS
---------------
null, Ho: μ=300
alternate, H1: μ<300
test statistic: -2
critical value: -1.036
decision: reject Ho
p-value: 0.023
we have enough evidence to support the claim that whether students on average save less than working adults.