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-> C | n https://tamu.bl 3832074-dt-content-rid-28776242 1courses/SCMT 303 501.173 0Ca.pdf Apes For quick acces place your boocmarks here on the bookmarks ar. Impo SCMT 303-IC BONUS We would lise to find if the average annual income far workers in Texas and Arizon are carrrerahk:the sienifiance level. I-order to make 1K: 'xmparison, we have ohlsetel sample stunistics fros ndom samples oe worker's nul incomes ( S10s in the 2 states we also km'uthal 'he Annual Inconc in buth sizies folkm bell-shaped distributio. Sample Size 30 25 The omgurison we would liea make is s se if the avng: ann al incme for warkers in Tesas is the sam as the avcrage annual income for workers in Arizona. Note : Since the question is to see i the average nual income for workersTexas is he same as the arerage annual income for workers ia Arizona, the test statements are: To hring the eqations to The ssndand form, Ist geinge vsrisble n the rigt This is the fen. iú which you are goin to coiiduct the testald alswer the questios bekow in the assessmemt for the IC Excrcis I. Coe the Psolad Varia-Sp 2. Compute the tes statistc. 3. What is the cricical valus tor his res? 4. What is the p value for this test? 5. What is yoar decision? Incerpeet your decisicn in tems of the peoblem stemea Noce 2: Do understand that instead of loving .nzms to the lett, ae could have moved The right, in ich case, the e wonld still be a 2-sided te, bu e est would be From a practice perspective, please recalculate and look at answering all the questioes (1- above and make sare thnt though some of the values (in particular, the signs) may change, the final conclusion is stll the same. ^ 4x 12/6/2017Explanation / Answer
Given that,
mean(x)=58.61
standard deviation , s.d1=16.57
number(n1)=30
y(mean)=50.28
standard deviation, s.d2 =9.02
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.1
from standard normal table, two tailed t /2 =1.67
since our test is two-tailed
reject Ho, if to < -1.67 OR if to > 1.67
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (29*274.5649 + 24*81.3604) / (55- 2 )
s^2 = 187.0761
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=58.61-50.28/sqrt((187.0761( 1 /30+ 1/25 ))
to=8.33/3.7039
to=2.249
| to | =2.249
critical value
the value of |t | with (n1+n2-2) i.e 53 d.f is 1.67
we got |to| = 2.249 & | t | = 1.67
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != 2.249 ) = 0.0286
hence value of p0.1 > 0.0286,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.249
critical value: -1.67 , 1.67
decision: reject Ho
p-value: 0.0286
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