SOLUTION: Use the given information to find the number of degrees of​ freedom, the critical values χ2L and χ2R​, and the confidence interval estimate of σ. It is reasonable to assume

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Question 1169425: Use the given information to find the number of degrees of​ freedom, the critical values χ2L and χ2R​, and the confidence interval estimate of σ. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution.
Nicotine in menthol cigarettes 99​% ​confidence; n=20​, s=0.26 mg.
df=19
(Type a whole​ number.)
χ2L=
​(Round to three decimal places as​ needed.)
χ2R=
​(Round to three decimal places as​ needed.)
The confidence interval estimate of σ is mg<σ< mg.
​(Round to two decimal places as​ needed.)

Answer by CPhill(1987) About Me  (Show Source):
You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1. Degrees of Freedom (df)**
* The degrees of freedom (df) are calculated as $n - 1$.
* Given $n = 20$, then $df = 20 - 1 = 19$.
**2. Critical Values (χ²L and χ²R)**
* We need to find the critical chi-square values for a 99% confidence interval.
* The significance level ($\alpha$) is $1 - 0.99 = 0.01$.
* We divide the significance level by 2 to find the area in each tail: $\alpha/2 = 0.005$.
* For χ²L, we use the area to the right of the value: $1 - \alpha/2 = 1 - 0.005 = 0.995$.
* For χ²R, we use the area to the right of the value: $\alpha/2 = 0.005$.
* Using a chi-square distribution table or calculator with df = 19:
* χ²L (area to the right = 0.995) ≈ 6.844
* χ²R (area to the right = 0.005) ≈ 38.582
**3. Confidence Interval for σ**
* The formula for the confidence interval of σ is:
$$\sqrt{\frac{(n-1)s^2}{\chi^2_R}} < \sigma < \sqrt{\frac{(n-1)s^2}{\chi^2_L}}$$
* Given $n = 20$, $s = 0.26$, χ²L = 6.844, and χ²R = 38.582:
$$\sqrt{\frac{(20-1)(0.26)^2}{38.582}} < \sigma < \sqrt{\frac{(20-1)(0.26)^2}{6.844}}$$
* Let's calculate the values:
* $(n-1)s^2 = (19)(0.26)^2 = 19(0.0676) = 1.2844$
* Lower bound: $\sqrt{\frac{1.2844}{38.582}} \approx \sqrt{0.03328} \approx 0.1824$
* Upper bound: $\sqrt{\frac{1.2844}{6.844}} \approx \sqrt{0.18767} \approx 0.4332$
* Rounding to two decimal places:
* Lower bound: 0.18
* Upper bound: 0.43
**Final Answers**
* df = 19
* χ²L = 6.844
* χ²R = 38.582
* Confidence interval: 0.18 mg < σ < 0.43 mg