Proctored Nonproctored A study was done on proctored and nonproctored tosts. The

Question

Proctored Nonproctored A study was done on proctored and nonproctored tosts. The results are shown in the table. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Use a 0.05 significance level for both parts n 34 77.47 86.95 0.84 a. Test the claim that students taking nonproctored tests get a higher mean score than those taking proctored tests What are the null and alternative hypotheses? The test statistic, t, is (Round to two decimal places as needed.)

Explanation / Answer

Given that,
mean(x)=77.47
standard deviation , s.d1=10.84
number(n1)=34
y(mean)=86.95
standard deviation, s.d2 =18.93
number(n2)=33
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, = 0.05
from standard normal table,left tailed t /2 =1.69
since our test is left-tailed
reject Ho, if to < -1.69
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =77.47-86.95/sqrt((117.5056/34)+(358.3449/33))
to =-2.51
| to | =2.51
critical value
the value of |t | with min (n1-1, n2-1) i.e 32 d.f is 1.69
we got |to| = 2.50561 & | t | = 1.69
make decision
hence value of | to | > | t | and here we reject Ho
p-value:left tail - Ha : ( p < -2.5056 ) = 0.00875
hence value of p0.05 > 0.00875,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.51
critical value: -1.69
decision: reject Ho
p-value: 0.00875

we have enough evidence to support the claim

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