Bayus(1991) studied the mean numbers of auto dealers visited by ear y and late r

Question

Bayus(1991) studied the mean numbers of auto dealers visited by ear y and late replacement buyers. Leting be the mean number o dealers visited by all late replacement buyers, set up the null and alternative hypotheses needed if we wish to atte to pro de evidence that di ers rom 4 dealers. Ara dom sample of 100 late e la e er buyers ye sa mean and a standard deviation of the number of dealers visited of x-4.38 and s58. Using a oritical value and assuming approximate normality to test the hypotheses you set up by setting a equal to .10, 05, .01, and .001. Do we estimate that is less than 4 or greater than 47(Round your answers to 3 decimal places.) H0 : (Click to select, B 4 versus Ha : Click to select) B4. a/2-0.05 /9.005 la/2 -0,000 There is (click to select) evidence. (Click to select) 4.

Explanation / Answer

Given that,
population mean(u)=4
sample mean, x =4.38
standard deviation, s =0.58
number (n)=100
null, Ho: =4
alternate, H1: >4
level of significance, = 0.1
from standard normal table,right tailed t /2 =1.29
since our test is right-tailed
reject Ho, if to > 1.29
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.38-4/(0.58/sqrt(100))
to =6.5517
| to | =6.5517
critical value
the value of |t | with n-1 = 99 d.f is 1.29
we got |to| =6.5517 & | t | =1.29
make decision
hence value of | to | > | t | and here we reject Ho
p-value :right tail - Ha : ( p > 6.5517 ) = 0
hence value of p0.1 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: =4 , alternate, H1: >4
test statistic: 6.5517
critical value at .10 is : 1.29
critical value at .05 is : 1.66
critical value at .01 is : 2.365
critical value at .001 is : 3.175

decision: reject Ho
Uo is greater than 4

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