Answer:
14.7 and 16.9
Step-by-step explanation:
We want to find the confidence interval for the mean when the population standard deviation [tex]\sigma[/tex], is known so we use the [tex]z[/tex] confidence interval for the mean.
The following assumptions are also met;
- The sample is a random sample
- [tex]n\ge 30[/tex]
The z confidence interval for the mean is given by:
[tex]\bar X-z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )\:<\:\mu\:<\bar X+z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )[/tex]
The appropriate z-value for 95% confidence interval is 1.96 (read from the standard normal z-distribution table)....See attachment.
From the question, we have [tex]n=49[/tex], [tex]\sigma=3.8[/tex] and [tex]\bar X=15.8[/tex]
We substitute all these values to get:
[tex]15.8-1.96(\frac{3.8}{\sqrt{49} } )\:<\:\mu\:<\bar 15.8+1.96(\frac{3.8}{\sqrt{49} } )[/tex]
[tex]15.8-1.96(\frac{3.8}{7 } )\:<\:\mu\:<15.8+1.96(\frac{3.8}{7} )[/tex]
[tex]14.7\:<\:\mu\:< 16.9[/tex] correct to one decimal place.
