American workers spend an average of 46 minutes to and from work in a typical da
ID: 3218347 • Letter: A
Question
American workers spend an average of 46 minutes to and from work in a typical day. In a sample of n = 104 employed adults in Indianapolis metropolitan area (IMA), the average commuting time to work was X^- = 40.5 minutes, with s = 21.6 minutes. Is there enough evidence, at alpha = 0.05, to conclude that the average commuting time in IMA is less than the national average? To perform a hypothesis test, the null and alternative hypotheses for this test are: A H_0: mu greaterthanorequalto 46 H_1: mu 46 D H_0: mu > 46 H_1: mu lessthanorequalto 46 For the previous question, the test statistic is |TS| (if negative, use the absolute value) A 2.01 DNR H_0 Conclude the average commuting time in IMA is not less than the national average. B 2.01 Reject H_0. Conclude the average commuting time in IMA is less than the national average. C 2.60 DNR H_0 Conclude the average commuting time in IMA is not less than the national average. D. 2.60 Reject H_0. Conclude the average commuting time in IMA is less than the national average. In the previous question the, p-value almost equal to A 0.050 B 0.042 C 0.024 D 0.005Explanation / Answer
Given that,
population mean(u)=46
sample mean, x =40.5
standard deviation, s =21.6
number (n)=104
null, Ho: =46
alternate, H1: <46
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.66
since our test is left-tailed
reject Ho, if to < -1.66
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =40.5-46/(21.6/sqrt(104))
to =-2.6
| to | =2.6
critical value
the value of |t | with n-1 = 103 d.f is 1.66
we got |to| =2.6 & | t | =1.66
make decision
hence value of | to | > | t | and here we reject Ho
p-value :left tail - Ha : ( p < -2.5967 ) = 0.00539
hence value of p0.05 > 0.00539,here we reject Ho
ANSWERS
---------------
null, Ho: =46
alternate, H1: <46
test statistic: -2.6
critical value: -1.66
decision: reject Ho
p-value: 0.00539 ~ 0.005
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