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 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 1.646. Correct Answer: The degrees of freedom using the simple method is 1030. Part: 3 / 63 of 6 Parts Complete Part 4 of 6 Estimate the P-value. Identify the form of the interval based on Critical Values for the Student's t Distribution Table. ≤p <≤p >p

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 =-2.0093
| to | =2.0093
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
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null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.0093 =-2.01
critical value: -1.646
decision: reject Ho
p-value: 0.02238
we have enough evidence to support the claim that mean of sample 1 is less than mean of sample 2.