Sample size = n = 10
mean of the sample = μ_s = 678/10 = 67.8
standard deviation of the sample : s = 2.856
Mean of the population = μ = 66 kg
Significance level = 5% => Probability = 95%
Null hypothesis is that the population mean is μ = 66 kg.
student's t test t = (μ_s - μ) / (s/√n) => t = (67.8 - 66) / (2.856/√10)
t = 1.993
For a sample of 10 individuals, the number of degrees of freedom is 9. Consult the table of Student's t distribution. P(-2.262 < t < 2.262) = 95% . This is the two sided probability value, one has to read.
or, P( | t | > 2.262) = 5%
If the population mean and sample mean are with in : 2.262 * 2.856/√10 = 2.04 kg, then the sample belongs to the population. Or, the population mean given is in the acceptable range of the sample mean, with a confidence level of 95%.
So we will accept under 5% level of significance, a population mean in the range, 67.8 +- 2.04 kg, with a confidence of 95%. ie., 65.76 kg to 69.84 kg. Given value 66 kg is just inside this range.
Another way, as the computed t value is | 1.993 | < 2.262, the given population mean is possible. The claim is accepted.