Given Data
Sample mean is X = 56
Sample standard deviation is s = 7.5
Sample size = n = 28
significance level is = α = 0.002
Null and alternative hypotheses need to be tested
Ho: μ = 55.3
Ha: μ ≠ 55.3
This corresponds to a two-tailed test
At the significance level is α=0.002, and the critical value for a two-tailed test is
tc (lower)= - 3.421 using excel =T.INV(0.001,27)
tc (upper)= 3.421 using excel =T.INV(0.999,27)
Rejection region will be t<-3.421 and t>3.421
Test Statistics
(1) Test Statistics = t = 1.129
(2) p value is calculated as p = 0.2688, using excel =T.DIST.2T(1.129,27)
(3) p = 0.2688 > α=0.002,, This mean p value is greater than α.
(4) Since t test value is greater than tc lower and less than tc upper
-3.421<t = 1.129<3.421
So null hypothesis is not rejected
(5) As such Final conclusion is that There is sufficient evidence to warrant rejection of the claim that population mean is not equal to 55.3, So 1st option is correct
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