Auto Dail-Problem-181Jdsb-Excel Sign in nsert Page Layout FormulasDe Revie View
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Auto Dail-Problem-181Jdsb-Excel Sign in nsert Page Layout FormulasDe Revie View Developer Tell me what you went to do Normal 2 Normal 22 AutoSum , A Copy Format Painter B 1 u· -- Pasle Merge & Center . $. % , tas, conditional format as! Normel nsert Delete Fornat Sort & Find Filter Select · omatting Table Clipboard K35 12 13 A consultant for a large university studied the number of hours per week freshmen watch TV versus the 14 number of hours seniors do. The results of this study follow. Is there enough evidence to show the mean 15 number of hours per week freshmen watch TV is different from the mean number of hourss do at 16 -0.01? 21 11.6 18.7 7.8740 23 3.9749 26 For the hypothesis stated above (in terms of seniors-Freshmen): Question 1 What is/are the critical value(s)? t crit 169 Question 2 What is the decision? Fail to Reject 31 32 What is the p-value? Fill in only one of the following statements If the Z table is appropriate, If the ttable Is appropriate, Question 3 p-value 35 36 37 0.0200 o-value 0.0500 Daily Problem Ready 4:57 PM 4: 10/26/2017Explanation / Answer
Given that,
mean(x)=18.7
standard deviation , s.d1=7.874
number(n1)=9
y(mean)=11.6
standard deviation, s.d2 =3.9749
number(n2)=5
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.01
from standard normal table, two tailed t /2 =4.604
since our test is two-tailed
reject Ho, if to < -4.604 OR if to > 4.604
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =18.7-11.6/sqrt((61.99988/9)+(15.79983/5))
to =2.24
| to | =2.24
critical value
the value of |t | with min (n1-1, n2-1) i.e 4 d.f is 4.604
we got |to| = 2.23975 & | t | = 4.604
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.2398 ) = 0.089
hence value of p0.01 < 0.089,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.24
critical value: -4.604 , 4.604
decision: do not reject Ho
p-value: 0.089
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