Question 1086484
<font color="black" face="times" size="3">Given Info:
Confidence Level = C = 95% = 0.95
sample size = n = 11
sample standard deviation = s = 14.6

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Based on the sample size, the degrees of freedom (df) is
df = n-1 = 11-1 = 10


Let
L = left chi-square critical value
R = right chi-square critical value


We need to find the L & R values that make *[Tex \Large P(L \le \chi^2 \le R) = 0.95] true. The symbol *[Tex \Large \chi^2] is the Chi-Square symbol


Using a table or calculator those approximate values are
L = 3.247
R = 20.483
For this problem, I used a table. Specifically I used <a href="https://people.richland.edu/james/lecture/m170/tbl-chi.html">this table</a>. Look at the df = 10 row. Highlight the columns that have 0.975 and 0.025 up top. These values are areas to the right of the values marked in the table. So for example, *[Tex \Large P(\chi^2 \ge 3.247) = 0.975]. The middle area has 95% area. The two tails combine to 5% so each tail has area of 0.05/2 = 0.025

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Now we can compute the confidence interval


*[Tex \Large \sqrt{\frac{(n-1)*s^2}{R}} \le \sigma \le \sqrt{\frac{(n-1)*s^2}{L}}]


*[Tex \Large \sqrt{\frac{(11-1)*(14.6)^2}{20.483}} \le \sigma \le \sqrt{\frac{(11-1)*(14.6)^2}{3.247}}]


*[Tex \Large 10.2013130082227 \le \sigma \le 25.6219223966323]


*[Tex \Large 10.2 \le \sigma \le 25.6]
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The 95% confidence interval for the standard deviation is approximately <font color=red>(10.2, 25.6)</font>
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