A cof fee manufacturer is interested in whether the mean daily consumption of re

Question

A cof fee manufacturer is interested in whether the mean daily consumption of regular-coffee drinkers is less than that of decaffeinated-coffee drinkers. Assume the population standard deviation for those drinkin regular coffee is 1.23 cups per day and 1.37 cups per day for those drinking decaffeinated coffee. A random sample of 46 regular-coffee drinkers showed a mean of 4.36 cups per day. A sample of 36 decaffeinated- coffee drinkers showed a mean of 5.64 cups per day Use the .05 significance level. (1) This is a [Click toselect"-tailed test. (2) The decision rule is to reject Ho: , d if Z

Explanation / Answer

Given that,
mean(x)=4.36
standard deviation , 1 =1.23
number(n1)=46
y(mean)=5.64
standard deviation, 2 =1.37
number(n2)=36
null, Ho: u1 > u2
alternate, H1: 1 < u2
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=4.36-5.64/sqrt((1.5129/46)+(1.8769/36))
zo =-4.39
| zo | =4.39
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =4.39 & | z | =1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -4.39 ) = 0.00001
hence value of p0.05 > 0.00001,here we reject Ho
ANSWERS
---------------
one tailed test
since our test is left-tailed, reject Ho, if zo < -1.645
test statistic: -4.39
decision: reject Ho
p-value: 0.00001

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