Question
In: Statistics and Probability
a sample of 49 costumers for dinner at a new restaurant in town over a three-week...
a
sample of 49 costumers for dinner at a new restaurant in town over
a three-week period was randomly selected and the average amount
spent was 22.60. assume that the standard deviation is known to be
2.50.
using a 0.01 level of significance
Solutions
Expert Solution
Assumed data,
population mean(u)=21 because not given in the data,
Given that,
population mean(u)=21
sample mean, x =22.6
standard deviation, s =2.5
number (n)=49
null, Ho: μ=21
alternate, H1: μ!=21
level of significance, α = 0.01
from standard normal table, two tailed t α/2 =2.682
since our test is two-tailed
reject Ho, if to < -2.682 OR if to > 2.682
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =22.6-21/(2.5/sqrt(49))
to =4.48
| to | =4.48
critical value
the value of |t α| with n-1 = 48 d.f is 2.682
we got |to| =4.48 & | t α | =2.682
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 4.48 ) =
0
hence value of p0.01 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: μ=21
alternate, H1: μ!=21
test statistic: 4.48
critical value: -2.682 , 2.682
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that population mean
of dinner at a new restaurant in town over a three-week period was
22.
orchestra answered 2 years agoRelated Solutions
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