3:51 PM webwork.gonzaga.edu o AT&T; 55% (1 pt) Randomly selected students were g

Question

3:51 PM webwork.gonzaga.edu o AT&T; 55% (1 pt) Randomly selected students were given five seconds to estimate the value of a product of numbers with the results shown below. Estimates from students given 1×2×3×4×5×6×7×8: 2040, 10000, 150, 45000, 5000, 200, 1000, 4000, 175, 560 Estimates from students given 8×7×6×5×4×3×2×1: 2000, 40000, 428, 3500, 1500, 25000, 50000, 23410, 400, 100000 Use a 0.05 significance level to test the claim that the two populations have the same mean. (a) The test statistic is (b) The conclusion is A. There is not sufficient evidence to warrant the rejection of the claim that the two populations have the same mean B. There is sufficient evidence to warrant the rejection of the claim that the two populations have the same mean Note: You can earn partial credit on this problem. Preview Answers Submit Answers

Explanation / Answer

Given that,
mean(x)=6848.5
standard deviation , s.d1=13760.44
number(n1)=10
y(mean)=28623.8
standard deviation, s.d2 =31373.47
number(n2)=10
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.26
since our test is two-tailed
reject Ho, if to < -2.26 OR if to > 2.26
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =6848.5-28623.8/sqrt((189349708.9936/10)+(984294619.8409/10))
to =-2.01
| to | =2.01
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.26
we got |to| = 2.01 & | t | = 2.26
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.01 ) = 0.075
hence value of p0.05 < 0.075,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -2.01
critical value: -2.26 , 2.26
decision: do not reject Ho
p-value: 0.075
we do not have enough evidence to support the claim that two populations have same mean

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