A national grocer’s magazine reports the typical shopper spends 9.5 minutes in l

Question

A national grocer’s magazine reports the typical shopper spends 9.5 minutes in line waiting to check out. A sample of 27 shoppers at the local Farmer Jack’s showed a mean of 9.1 minutes with a standard deviation of 2.2 minutes. Is the waiting time at the local Farmer Jack’s less than that reported in the national magazine? Use the 0.010 significance level.

What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

Reject H0: µ 9.5 when the test statistic is (Click to select)equal togreater thanless than .

The value of the test statistic is . (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

What is the decision rule? (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)

Explanation / Answer

Given that,
population mean(u)=9.5
sample mean, x =9.1
standard deviation, s =2.2
number (n)=27
null, Ho: =9.5
alternate, H1: <9.5
level of significance, = 0.01
from standard normal table,left tailed t /2 =2.479
since our test is left-tailed
reject Ho, if to < -2.479
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =9.1-9.5/(2.2/sqrt(27))
to =-0.945
| to | =0.945
critical value
the value of |t | with n-1 = 26 d.f is 2.479
we got |to| =0.945 & | t | =2.479
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :left tail - Ha : ( p < -0.9448 ) = 0.17674
hence value of p0.01 < 0.17674,here we do not reject Ho
ANSWERS
---------------
null, Ho: =9.5
alternate, H1: <9.5
test statistic: -0.945
critical value: -2.479
decision: do not reject Ho
p-value: 0.17674

we don't have evidence that Farmer Jack’s less than that reported in the national magazine

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