A recent national survey found that high school students watched an average (mea

Question

A recent national survey found that high school students watched an average (mean) of 7.4 DVDs per month with a population standard deviation of .90. A random sample of 34 college students revealed that the mean number of DVDs watched last month was 6.90. At the .05 significance level, can we conclude that college students watch fewer DVDs a month than high school students?

b) State the decision rule.(choose one)

-Reject H0 if z < -1.65

-Reject H1 if z < -1.65

-Reject H1 if z > -1.65

-Reject H0 if z > -1.65

(d) What is your decision regarding H0 ?

-Cannot reject H0

-Reject H0

Explanation / Answer

Given that,
population mean(u)=7.4
standard deviation, =0.9
sample mean, x =6.9
number (n)=34
null, Ho: =7.4
alternate, H1: <7.4
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-u/(s.d/sqrt(n))
zo = 6.9-7.4/(0.9/sqrt(34)
zo = -3.24
| zo | = 3.24
critical value
the value of |z | at los 5% is 1.645
we got |zo| =3.24 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : left tail - ha : ( p < -3.24 ) = 0
hence value of p0.05 > 0, here we reject Ho
ANSWERS
---------------
reject Ho, if zo < -1.645

decision: reject Ho

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