Answer:
The given statement is false.
Step-by-step explanation:
We are given the following in the question:
95% confidence interval: (4.53,5.18)
Sample mean, [tex]\bar{x}[/tex] = 4.85 seconds
Sample size, n = 20
Alpha, α = 0.05
Sample standard deviation, σ = 1.47 seconds
Degree of freedom = n - 1 = 19
95% Confidence interval:
[tex]\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}[/tex]
Putting the values, we get,
[tex]t_{critical}\text{ at degree of freedom 4 and}~\alpha_{0.01} = \pm 2.09[/tex]
[tex]4.85 \pm 2.09(\dfrac{ 1.47}{\sqrt{20}} )\\\\ = 4.85 \pm 0.69 \\\\= (4.16 ,5.54)[/tex]
Thus, the calculated confidence interval is different from the given confidence interval.
Hence, the given statement is false.
