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pentagon [3]
3 years ago
9

Brainiest help asap 290-20

Mathematics
1 answer:
Lynna [10]3 years ago
5 0

Answer:

270

Step-by-step explanation:

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Find the 90% confidence interval for the variance and standard deviation of the ages of seniors at oak park college if a random
san4es73 [151]
Given that the standard deviation, s = 2.3 and that there are 24 students, thus the degree of freedom is 24 - 1 = 23.

For 90% confidence interval, \alpha =0.1 and \frac{ \alpha }{2} =0.05

From the chi-square table, the value of \chi^2_{1- \frac{ \alpha }{2} }=\chi^2_{0.95}=35.172 and the value of \chi^2_{ \frac{ \alpha }{2} }=\chi^2_{0.05}=13.091.

The 90% confidence interval for \sigma^2 is given by:

\frac{(n-1)s^2}{\chi^2_{1- \frac{ \alpha }{2} }} \ \textless \ \sigma^2\ \textless \ \frac{(n-1)s^2}{\chi^2_{ \frac{ \alpha }{2} }} \\  \\ = \frac{(24-1)(2.3)^2}{35.172} \ \textless \ \sigma^2\ \textless \ \frac{(24-1)(2.3)^2}{13.091} \\  \\ = \frac{23(5.29)}{35.172} \ \textless \ \sigma^2\ \textless \ \frac{23(5.29)}{13.091} = \frac{121.67}{35.172} \ \textless \ \sigma^2\ \textless \  \frac{121.67}{13.091}  \\  \\ =3.46\ \textless \ \sigma^2\ \textless \ 9.29

And the 90% confidence interval for \sigma is given by:

\sqrt{\frac{(n-1)s^2}{\chi^2_{1- \frac{ \alpha }{2} }}} \ \textless \ \sigma\ \textless \ \sqrt{\frac{(n-1)s^2}{\chi^2_{ \frac{ \alpha }{2} }}} \\  \\ = \sqrt{3.46} \ \textless \ \sigma\ \textless \  \sqrt{9.29} =1.86\ \textless \ \sigma\ \textless \ 3.05
3 0
3 years ago
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