The null and alternative hypotheses
Ho : d 0
H1 : d < 0
Test statistic
t = dbar /(Sd/√n)
t = 1.8/(3.4/√13)
t = 1.91
tcritical t0 for a = 0.01 , d.f = n -1= 12
to = t0.01,12 = 2.681
to = 2.68
Since the test statistic is lies in rejection region , fail to reject the null hypothesis . There is no statistically significant evidence to support the claim.
Test the claim below about the mean of the differences for a population of paired data...
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