Answer:
Step-by-step explanation:
This is a test of a single population mean since we are dealing with mean
For the null hypothesis,
µ = 60.8
For the alternative hypothesis,
µ ≠ 60.8
Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.
Since n = 8,
Degrees of freedom, df = n - 1 = 8 - 1 = 7
t = (x - µ)/(s/√n)
Where
x = sample mean = 66.9
µ = population mean = 60.8
s = samples standard deviation = 10.7
t = (66.9 - 60.8)/(10.7/√8) = 1.61
We would determine the p value using the t test calculator. It becomes
p = 0.076
Since alpha, 0.01 < than the p value, 0.076, then we would fail to reject the null hypothesis. Therefore, At a 1 % level of significance, the sample data showed that there is no significant evidence that μ ≠ 60.8
Therefore, this p-value leads to a decision to accept the null hypothesis
