Question

More time on the Internet: A researcher polled a sample of

1097

adults in the year

2010

, asking them how many hours per week they spent on the Internet. The sample mean was

9.42

with a standard deviation of

13.23

. A second sample of

1031

adults was taken in the year

2012

. For this sample, the mean was

10.63

with a standard deviation of

14.47

. Assume these are simple random samples from populations of adults. Can you conclude that the mean number of hours per week spent on the Internet increased between

2010

and

2012

? Let

μ1

denote the mean number of hours spent on the Internet in

2010

and

μ2

denote the mean number of hours spent on the Internet in

2012

. Use the  

=α0.05

level and the

P

-value method with the table.

Part 1 of 6

Your Answer is correct

State the appropriate null and alternate hypotheses.

H0:=μ1μ2
H1:<μ1μ2
This is a	▼left-tailed	test.

Part 2 of 6

Your Answer is correct

Compute the test statistic. Round the answer to three decimal places.

=t

  

−2.01

Part: 2 / 6

2 of 6 Parts Complete

Part 3 of 6

Your Answer is incorrect

How many degrees of freedom are there, using the simple method?

The degrees of freedom using the simple method is

Answer 1

Given that,
mean(x)=9.42
standard deviation , s.d1=13.23
number(n1)=1097
y(mean)=10.63
standard deviation, s.d2 =14.47
number(n2)=1031
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.646
since our test is left-tailed
reject Ho, if to < -1.646
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =9.42-10.63/sqrt((175.0329/1097)+(209.3809/1031))
to =-2.009
| to | =2.009
critical value
the value of |t α| with min (n1-1, n2-1) i.e 1030 d.f is 1.646
we got |to| = 2.00931 & | t α | = 1.646
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -2.0093 ) = 0.02238
hence value of p0.05 > 0.02238,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.009
critical value: -1.646
decision: reject Ho
p-value: 0.02238
we have enough evidence to support the claim that the mean number of hours spent on the Internet in 2010 is less than in 2012.