please help solve and give me a breakdown of how to solve A consultant for a lar

Question

please help solve and give me a breakdown of how to solve

A consultant for a large university studied the number of hours per week freshmen watch TV versus the number of hours seniors do. The resuits of this study follow. Is there enough evidence to show the mean number of hours per week freshmen watch TV is different from the mean number of hours seniors do at :0.001? Seniors 18.5 11.6 7 8740 3.9749 26 jFor the hypothesis stated above fin terms of Seniors- Freshmen) 27 23 29 Question 1What is/are the critical va luels)? Question 2 what is the decision? Question What is the p-value? Fil in only one of the totlowing statements aiue .if the Z table is appropriate, p 35 36 if the rrableis appropriste,

Explanation / Answer

Given that,
mean(x)=18.5
standard deviation , s.d1=7.874
number(n1)=8
y(mean)=11.6
standard deviation, s.d2 =3.9749
number(n2)=6
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.001
from standard normal table, two tailed t /2 =6.869
since our test is two-tailed
reject Ho, if to < -6.869 OR if to > 6.869
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =18.5-11.6/sqrt((61.99988/8)+(15.79983/6))
to =2.141
| to | =2.141
critical value
the value of |t | with min (n1-1, n2-1) i.e 5 d.f is 6.869
we got |to| = 2.14132 & | t | = 6.869
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.1413 ) = 0.085
hence value of p0.001 < 0.085,here we do not reject Ho
ANSWERS
---------------
crtical value = -6.869, +6.869
reject Ho, if to < -6.869 OR if to > 6.869
test statistic: 2.141
p-value: 0.085
decision: do not reject Ho

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