1. The following paired observations show the number of traffic citations given

Question

1.

The following paired observations show the number of traffic citations given for speeding by Officer Dhondt and Officer Meredith of the Ontario Provincial Police (OPP) for the last five months:

                         

State the decision rule. (Negative answer should be indicated by a minus sign. Round the final answers to 3 decimal places.)

Compute the value of the test statistic. (Negative answer should be indicated by a minus sign.Round the final answer to 3 decimal places.)

Determine the p-value. (Round the final answer to 4 decimal places.)

2.

Clark Heter is an industrial engineer at Lyons Products. He would like to determine whether there are more units produced on the afternoon shift than on the day shift. A sample of 40 day-shift workers showed that the mean number of units produced was 341. A sample of 42 afternoon-shift workers showed that the mean number of units produced was 351, Assume that the population standard deviation for the number of units produced on the day shift is 16 and on the afternoon shift is 22.

  

The null and alternative hypotheses are:

Explanation / Answer

Q1.

Given that,
null, H0: Ud = 0
alternate, H1: Ud != 0
level of significance, = 0.01
from standard normal table, two tailed t /2 =4.604
since our test is two-tailed
reject Ho, if to < -4.604 OR if to > 4.604
we use Test Statistic  
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 2.8
We have d = 2.8
pooled variance = calculate value of Sd= S^2 = sqrt [ 502-(14^2/5 ] / 4 = 10.76
to = d/ (S/n) = 0.58
critical Value
the value of |t | with n-1 = 4 d.f is 4.604
we got |t o| = 0.58 & |t | =4.604
make Decision
hence Value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.5821 ) = 0.5917
hence value of p0.01 < 0.5917,here we do not reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud != 0
test statistic: 0.58
critical value: reject Ho, if to < -4.604 OR if to > 4.604
decision: Do not Reject Ho
p-value: 0.5917

X Y X-Y (X-Y)^2 17 28 -11 121 35 19 16 256 32 32 0 0 26 28 -2 4 26 15 11 121 14 502

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