Question

List below are the speeds (mi/h) measured from southbound traffic on I-280 near Cupertino, CA. Use the sample data to construct a 99% confidence interval estimate of the population standard deviation: speeds 62, 58, 58, 56, 60, 53, 58

Answer 1

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;

$\sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{1-\alpha /2}^{}}}< \sigma < \sqrt{\dfrac{\left (n-1 \right )s^{2}}{\chi _{\alpha /2}^{}}}$

s = $\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }$

Where:

$\bar x$ = Sample mean

s = Sample standard deviation

n = Sample size = 7

χ = Chi squared value at the given confidence level

$\bar x$ = ∑x/n = (62 + 58 + 58 + 56 + 60 +53 + 58)/7 = 57.857

The sample standard deviation s = $\sqrt{\dfrac{\Sigma (x - \bar x)^2}{n - 1} }$ = 2.854

The test statistic, derived through computation, = ±3.707

Which gives;

$C. I. = 57.857 \pm 3.707 \times \dfrac{2.854}{\sqrt{7} }$

$\sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{16.812}^{}}}< \sigma < \sqrt{\dfrac{\left (7-1 \right )2.854^{2}}{0.872}}$

1.7047 < σ < 7.485

The 99% confidence interval of the population standard deviation = 1.7047 < σ < 7.485.