Please answer the questions the way I\'d have to fill them in. Thank you · Calbn
ID: 3340312 • Letter: P
Question
Please answer the questions the way I'd have to fill them in. Thank you
· Calbn112-A'a"8- 9-@wrap Tent General B 1 u.:L. O-A- E-EE Merge & Center. S'%, 12A A 406-Accent3 .8 Conditional Fable as 40% Accent 4 40% Accents mat Painter Formatting Table- ard Font Alignment Nunber Styles ne U.S. Department of Transportation provides the number of miles that residents of the 75 largest metropolitan areas travel per day in a car. uppose a random sample of 47 Buffalo residents and 29 Boston residents yielded the following results. (See exercise 12 on page 449 of your texth or a similar problem.) Buffalo Boston 47 22.2 8.6 29 18.7 7.8 xbar xbar Question 2A Can you conclude the mean number of miles Boston residents travel per day is less than the mean number of miles Buffalo residents do at a-0.025? State the hypothesis in terms of Boston-Buffalo. Step 1: HO: Ha: Step 2: 85 step 3: Fill in row 88 if the hypothesis is one tailed.
Explanation / Answer
Given that,
mean(x)=22.2
standard deviation , s.d1=8.6
number(n1)=47
y(mean)=18.7
standard deviation, s.d2 =7.8
number(n2)=29
null, Ho: ubuffalo - uboston < 0
alternate, H1: ubuffalo - uboston > 0
level of significance, = 0.025
from standard normal table,right tailed t /2 =2.048
since our test is right-tailed
reject Ho, if to > 2.048
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =22.2-18.7/sqrt((73.96/47)+(60.84/29))
to =1.827
| to | =1.827
critical value
the value of |t | with min (n1-1, n2-1) i.e 28 d.f is 2.048
we got |to| = 1.8266 & | t | = 2.048
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 1.8266 ) = 0.03922
hence value of p0.025 < 0.03922,here we do not reject Ho
ANSWERS
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null, Ho: ubuffalo - uboston < 0
alternate, H1: ubuffalo - uboston > 0
reject Ho, if to > 2.048
test statistic: 1.827
critical value: 2.048
decision: do not reject Ho
p-value: 0.03922
CI at 98%
given that,
mean(x)=22.2
standard deviation , s.d1=8.6
sample size, n1=47
y(mean)=18.7
standard deviation, s.d2 =7.8
sample size,n2 =29
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 22.2-18.7) ± t a/2 * sqrt((73.96/47)+(60.84/29)]
= [ (3.5) ± t a/2 * 1.916]
= [-1.227 , 8.227]