The U.S. Department of Transportation provides the number of miles that resident

Question

The U.S. Department of Transportation provides the number of miles that residents of the 75 largest metropolitan areas travel per day in a car. Suppose a random sample of 55 Buffalo residents and 28 Boston residents yielded the following results. (See exercise 12 on page 449 of your textbook for a similar problem.) Buffalo Boston 28 18.9 7.9 bar = 22.7 8.5 bar = Based on the data, can you say the mean number of miles Boston residents travel per day is less than the mean number of miles Buffalo residents do at =0.025? State the hypothesis in terms of Boston-Buffalo. Question 2A Step 1: HO: Ha: Step 2: Step 3: Fill in row 88 if the hypothesis is one tailed. Reject HO if Fill in row 92 if the hypothesis is two tailed. The smaller number must be typed first. Reject HO if Step 4 Step S: Step 6: Step 7: HO Fill in row 102 if the Z table is appropriate. P-value Fill in row 106 if the t table is appropriate. The lower boundary must go on the left and the upper boundary on the right. P-value Question 28 Construct a 98% confidence interval for the difference in true means. State the C. in terms of Boston-Buffalo. Lower Endpoint Upper Endpoint

Explanation / Answer

PART A.
Given that,
mean(x)=22.7
standard deviation , s.d1=8.5
number(n1)=55
y(mean)=18.9
standard deviation, s.d2 =7.9
number(n2)=28
null, Ho: u1 < u2
alternate, H1: u1 > u2
level of significance, = 0.025
from standard normal table,right tailed t /2 =2.052
since our test is right-tailed
reject Ho, if to > 2.052
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =22.7-18.9/sqrt((72.25/55)+(62.41/28))
to =2.0189
| to | =2.0189
critical value
the value of |t | with min (n1-1, n2-1) i.e 27 d.f is 2.052
we got |to| = 2.01895 & | t | = 2.052
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value:right tail - Ha : ( p > 2.0189 ) = 0.02676
hence value of p0.025 < 0.02676,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 2.0189
critical value: 2.052
decision: do not reject Ho
p-value: 0.02676
we don't have evidence to support the claim

PART B.
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 = [( 22.7-18.9) ± t a/2 * sqrt((72.25/55)+(62.41/28)]
= [ (3.8) ± t a/2 * 1.882]
= [-0.855 , 8.455]

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