Assuming that the population of tread wear indexes is normally distributed, construct a 90% confidence interval estimate of the population mean tread wear index for tires produced by this manufacturer under this brand name, round to two decimal places as needed.
Answer:
Given Data:
The Sample Mean $(\mu) = 190.7 $
The Sample Standard Deviation $(s)=29.3 $
The Sample Size $(n)=23 $
The Confidence Interval Level $(CI) = 90% $
Solution:
The Level Of Significance $(\alpha):$
$\alpha = 1 – 0.90$
$\therefore \alpha = 0.10$
The Degree Of Freedom $(df):$
$df = n – 1$
$df = 23 – 1$
$\therefore df = 22$
The Critical Value $(t_c):$
$t_c = t_{\frac{\alpha}{2}, df}$
$t_c = t_{\frac{0.10}{2}, 22}$
$t_c = t_{0.05, 22}$
$\therefore t_c \approx 1.717$
The Confidence Interval $(CI) :$
The 90% confidence interval is: $\bar{x} \pm t_c \times \frac{s}{\sqrt{n}}$
$= 190.7 \pm 1.717 \times \frac{29.3}{\sqrt{23}}$
$= 190.7 \pm 10.49$
Final Answer:
$\Rightarrow$ The 90% confidence interval $\approx (180.21, 201.19)$
