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Answer:
Step-by-step explanation:
sample size = n = 60
Sample mean = x bar = 5.25
Sample sd = s = 0.55
Std error of sample = s/sqrt n = 0.071
(One tailed test)
Mean difference = 5.25-5.4 = -0.15
Test statistic t = Mean diff/Se = -2.11
df=59
(since population std dev is not known t test is used)
p value =0.01955
Since p <alpha =0.05 we reject null hypothesis.
The results suggest about the sample group that the sample is from a population with a mean less than 5.4