Given:
Sample mean = 65.4
Standard deviation = 1.2
Sample size = 45
Confidence level = 99%
To find:
The confidence interval.
Solution:
The formula for confidence interval is
[tex]CI=\overline{x}\pm z^*\dfrac{s}{\sqrt{n}}[/tex]
where, [tex]\overline{x}[/tex] is sample mean, z* is confidence value, s is standard deviation and n is sample size.
Confidence value or z-value at 99% = 2.58
Putting the given in the above formula, we get
[tex]CI=65.4\pm 2.58\times \dfrac{1.2}{\sqrt{45}}[/tex]
[tex]CI=65.4\pm 0.46[/tex]
[tex]CI=65.4-0.46\text{ and }CI=65.4+0.46[/tex]
[tex]CI=64.94\text{ and }65.86[/tex]
Therefore, the correct option is D.
