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

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) (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

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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

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