We have the sample size, sample mean and the sample standard deviation. Since the population standard deviation is not know, we will use t-distribution to find the confidence interval.
The critical t value for 95% confidence interval and 63 degrees of freedom is 1.998.
The 95% confidence for the population mean will be:
![(119-1.998 *(16)/( √(64) ) ,119+1.998 *(16)/( √(64) )) \\ \\ =(115.004, 122.996) \\ \\=(115,123)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/4qn3m8aabkgl99jrdbc6gpgv7z68fqvf7o.png)
Thus, the 95% confidence interval for the population mean will be (115,123)
So, option A is the correct answer