Answer:
The 99% confidence interval of the population standard deviation is 1.7047 < σ < 7.485
Step-by-step explanation:
Confidence interval of standard deviation is given as follows;
[tex]\sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{1-\alpha /2}^{}}}< \sigma < \sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{\alpha /2}^{}}}[/tex]
s = [tex]\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }[/tex]
Where:
[tex]\bar x[/tex] = Sample mean
s = Sample standard deviation
n = Sample size = 7
χ = Chi squared value at the given confidence level
[tex]\bar x[/tex] = ∑x/n = (62 + 58 + 58 + 56 + 60 +53 + 58)/7 = 57.857
The sample standard deviation s = [tex]\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }[/tex] = 2.854
The test statistic, derived through computation, = ±3.707
Which gives;
[tex]C. I. = 57.857 \pm 3.707 \times \dfrac{2.854}{\sqrt{7} }[/tex]
[tex]\sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{16.812}^{}}}< \sigma < \sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{0.872}}[/tex]
1.7047 < σ < 7.485
The 99% confidence interval of the population standard deviation = 1.7047 < σ < 7.485.
