On average, university students eat out 4.0 times per week, with a standard devi
ID: 3170424 • Letter: O
Question
On average, university students eat out 4.0 times per week, with a standard deviation of 2.0. Random sampling of 25 Ohio University students had a mean of 5 times per week. Do Ohio University students eat out significantly more often than the average U.S. student? (use a-.01) State the null hypothesis in words. Set up the criteria for making a decision. That is, find the critical value(s). Compute the appropriate test statistic. Show your work. Based on your answer evaluate the null hypothesis. State your conclusion in words. Given your decision. what type of error could haw been committed?Explanation / Answer
Given that,
population mean(u)=4
standard deviation, =2
sample mean, x =5
number (n)=25
null, Ho: =4
alternate, H1: >4
level of significance, = 0.01
from standard normal table,right tailed z /2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 5-4/(2/sqrt(25)
zo = 2.5
| zo | = 2.5
critical value
the value of |z | at los 1% is 2.326
we got |zo| =2.5 & | z | = 2.326
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 2.5 ) = 0.00621
hence value of p0.01 > 0.00621, here we reject Ho
ANSWERS
---------------
null, Ho: =4
alternate, H1: >4
test statistic: 2.5
critical value: 2.326
decision: reject Ho
p-value: 0.00621
we have evidence to support the claim, i.e more often than average us student
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