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KatRina [158]
3 years ago
11

A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl

e of 14 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29.
a) Find a 98% confidence interval for the true mean checking account balance for local customers.
b) Find a 95% confidence interval for the standard deviation.

Mathematics
1 answer:
Citrus2011 [14]3 years ago
7 0

Answer:

a) 98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

b)   95% confidence interval for the standard deviation.

(214.91 , 441.53)

Step-by-step explanation:

Given a size of sample 'n' =14

given mean of the sample x⁻ = $664.14

standard deviation of the sample 'S' = $297.29.

a)

<u>98% of confidence intervals</u>

<u></u>(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-}+ t_{\alpha }\frac{S}{\sqrt{n} } )<u></u>

The degrees of freedom γ=n-1 =14-1 =13

t₁₃ = 2.650 at 98% of confidence level of signification.

(664.14- 2.650\frac{297.29}{\sqrt{14} } , 664.14+ 2.650\frac{297.29}{\sqrt{14} } )

on calculation, we get

(664.14-210.553 , 664.14+210.553)

(453.586 , 874.693)

98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

<u>95% of confidence intervals</u>

({s\sqrt{\frac{n-1}{X^{2} _{(\frac{\alpha }{2} ,n-1) } } } ,s\sqrt{\frac{n-1}{X^2_{\frac{1-\alpha }{2},n-1 } } }  )

The degrees of freedom γ=n-1 =14-1 =13

X^2_{0.05,13} =22.36     (check table)

X^2_{0.95,13} = 5.892    (check table)

(297.29. (\sqrt{\frac{14-1}{X^2_{0.05,13}  } } ),297.29(\sqrt{\frac{14-1}{X^2_{0.95,13} } } )

(297.29. (\sqrt{\frac{14-1}{22.36  } } ),297.29(\sqrt{\frac{14-1}{5.892 } } )

(214.91 , 441.53)

95% confidence interval for the standard deviation.

(214.91 , 441.53)

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