1. A large national survey shows that the waiting times for fast food restaurant
ID: 3179273 • Letter: 1
Question
1. A large national survey shows that the waiting times for fast food restaurants is approximately normally distributed with a mean of = 3.4 minutes and a standard deviation of = 0.6 minutes. A fast food company claims that their mean waiting time in line is less than 3.4 minutes. To test the claim, a random sample of 60 customers is selected, and is found to have a mean of 3.3 minutes. We will use this sample data to test the fast food outlet's claim at a significance level of = 0.05.
I entered in the answers that I have gotten and left what I am confused about blank:
a)Set up hypotheses Ho and Ha to test the company’s claim.
Claim: M < 3.4 Ho: less than or = to 3.4 Ha: <3.4
Go to STAT >> TESTS, and choose Z-test – select the STATS option and enter the given sample information.
b)State the test statistic and p-value given in the calculator.
Z= -1.219 P= 0.0984
c)Compare the p-value to = .05. State the appropriate decision
(Reject Ho or Fail to reject Ho), along with an explanation.
P=0.0984 < 0.05 Reject Ho
d)What can we conclude about the researcher’s claim?
Explanation / Answer
Given that,
population mean(u)=3.4
sample mean, x =3.3
standard deviation, s =0.6
number (n)=60
null, Ho: =3.4
alternate, H1: <3.4
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.671
since our test is left-tailed
reject Ho, if to < -1.671
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =3.3-3.4/(0.6/sqrt(60))
to =-1.291
| to | =1.291
critical value
the value of |t | with n-1 = 59 d.f is 1.671
we got |to| =1.291 & | t | =1.671
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -1.291 ) = 0.10087
hence value of p0.05 < 0.10087,here we do not reject Ho
ANSWERS
---------------
null, Ho: =3.4
alternate, H1: <3.4
test statistic: -1.291
critical value: -1.671
decision: do not reject Ho
p-value: 0.10087
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