Important instructions: For all situations requiring a hypothesis test (one-samp
ID: 3310373 • Letter: I
Question
Important instructions: For all situations requiring a hypothesis test (one-sample t test, two-sample t test, one-sample variance test, two sample variance test, paired sample t- test, Welch's t test, non parametric tests) you must 1. Choose the appropriate test based on the information you are given 2. State null and alternative hypotheses 3. Choose a one or two tailed test and explain why you chose that test 4. Calculate the appropriate test statistic, showing all work neatly. This includes calculations of means, standard deviations, etc. You may use the “machine formulas", in fact, you are encouraged to 5. Draw the appropriate conclusions (i.e., do you reject or fail to reject Ho?)Explanation / Answer
5.
Given that,
mean(x)=6.375
standard deviation , s.d1=2.6125
number(n1)=8
y(mean)=7
standard deviation, s.d2 =2.2678
number(n2)=8
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.365
since our test is two-tailed
reject Ho, if to < -2.365 OR if to > 2.365
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =6.375-7/sqrt((6.82516/8)+(5.14292/8))
to =-0.511
| to | =0.511
critical value
the value of |t | with min (n1-1, n2-1) i.e 7 d.f is 2.365
we got |to| = 0.51099 & | t | = 2.365
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.511 ) = 0.625
hence value of p0.05 < 0.625,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.511
critical value: -2.365 , 2.365
decision: do not reject Ho
p-value: 0.625
we do not have enough evidence to support that bait differ in effectiveness