1) Calculate Pooled Variance, showing work and rounding your answer to the neare
ID: 3217547 • Letter: 1
Question
1) Calculate Pooled Variance, showing work and rounding your answer to the nearest two decimal places.
2) Based on the pooled variance, calculate the t statistic. Show work and round to the nearest two decimal places.
3) Based on the hypothesis test in the last question, you would decide to:
a) Reject null, because the t statistic is in the critical region.
b) Reject null, because the t statistic is NOT in the critical region.
c) Fail to reject null, because the t statistic is in the critical region.
d) Fail to reject null, because the t statistic is NOT in the critical region.
*Thank you for all of the help!*
Some safety researchers were interested in the effects of a loud beeping noise on seat belt wearing in automobiles. Two groups of participants were randomly selected to test drive a new car. For one of the groups the car made a loud beeping noise as soon as a key was put in the ignition to reinforce seat belt wearing. For the other group, no noise was made when the key was put in the ignition. The researchers measured time taken to put on a seat belt in seconds. Does the loud beeping noise influence the amount of time people spend before putting on a seat belt? NOTE: Treat this as a two-tailed test, with a p level of 0.O5. Calculate difference between the means as H1-H2.where H1 represents people who hear the beeping sound and U2 represents people who do not hear the beeping sound. Answer the follownig questions: Based on this data, calculate the pooled variance: Beeping Noise No Beeping Noise 15 16 12 36 10 22Explanation / Answer
Given that,
mean(x)=17.5
standard deviation , s.d1=10.1341
number(n1)=6
y(mean)=22.2
standard deviation, s.d2 =9.3915
number(n2)=5
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (5*102.7 + 4*88.2003) / (11- 2 )
s^2 = 96.2557
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=17.5-22.2/sqrt((96.2557( 1 /6+ 1/5 ))
to=-4.7/5.9409
to=-0.7911
| to | =0.7911
critical value
the value of |t | with (n1+n2-2) i.e 9 d.f is 2.262
we got |to| = 0.7911 & | t | = 2.262
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - ha : ( p != -0.7911 ) = 0.4472
hence value of p0.05 < 0.4472,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.7911
critical value: -2.262 , 2.262
decision: do not reject Ho
p-value: 0.4472
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